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Sum of Parts of An Array - JavaScript
How do JavaScript closures work?How do I remove a property from a JavaScript object?How do I check if an array includes an object in JavaScript?How to insert an item into an array at a specific index (JavaScript)?How do I include a JavaScript file in another JavaScript file?What does “use strict” do in JavaScript, and what is the reasoning behind it?How to check whether a string contains a substring in JavaScript?Loop through an array in JavaScriptHow do I remove a particular element from an array in JavaScript?For-each over an array in JavaScript?
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6
Trying to solve this challenge on codewars. According to the challenge, the parts of array:
ls = [0, 1, 3, 6, 10]
Are
ls = [0, 1, 3, 6, 10]
ls = [1, 3, 6, 10]
ls = [3, 6, 10]
ls = [6, 10]
ls = [10]
ls = []
And we need to return an array with the sums of those parts.
So my code is as follows:
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
console.log(partsSums([0, 1, 3, 6, 10]));The issue is that it wants us to add the last sum 0 when the array is empty. So we should be getting:
[ 20, 20, 19, 16, 10, 0 ]
Instead of
[ 20, 20, 19, 16, 10]
So I tried this:
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
arrayOfSums.push(0);
return arrayOfSums;
console.log(partsSums([0, 1, 3, 6, 10]));And this:
function partsSums(ls)
ls.push(0);
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
But these caused execution time-out errors on Codewars:
Execution Timed Out (12000 ms)
So I also tried:
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > -1)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
But now this causes a TypeError:
TypeError: Reduce of empty array with no initial value
I am not understanding the concept of how to get 0 into the array when all of the values have been shifted out. The challenge seems to want 0 as the final "sum" of the array, even when the array is empty. But you cannot reduce an empty array - what else can I do here?
EDIT: Tried adding initial value to the reduce method:
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b, 0);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
Unfortunately this still fails the basic test :
expected [] to deeply equal [ 0 ]
javascript arrays sum reduce
asked 8 hours ago
HappyHands31HappyHands31
1,3853 gold badges22 silver badges41 bronze badges
|
show 10 more comments
6
Trying to solve this challenge on codewars. According to the challenge, the parts of array:
ls = [0, 1, 3, 6, 10]
Are
ls = [0, 1, 3, 6, 10]
ls = [1, 3, 6, 10]
ls = [3, 6, 10]
ls = [6, 10]
ls = [10]
ls = []
And we need to return an array with the sums of those parts.
So my code is as follows:
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
console.log(partsSums([0, 1, 3, 6, 10]));The issue is that it wants us to add the last sum 0 when the array is empty. So we should be getting:
[ 20, 20, 19, 16, 10, 0 ]
Instead of
[ 20, 20, 19, 16, 10]
So I tried this:
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
arrayOfSums.push(0);
return arrayOfSums;
console.log(partsSums([0, 1, 3, 6, 10]));And this:
function partsSums(ls)
ls.push(0);
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
But these caused execution time-out errors on Codewars:
Execution Timed Out (12000 ms)
So I also tried:
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > -1)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
But now this causes a TypeError:
TypeError: Reduce of empty array with no initial value
I am not understanding the concept of how to get 0 into the array when all of the values have been shifted out. The challenge seems to want 0 as the final "sum" of the array, even when the array is empty. But you cannot reduce an empty array - what else can I do here?
EDIT: Tried adding initial value to the reduce method:
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b, 0);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
Unfortunately this still fails the basic test :
expected [] to deeply equal [ 0 ]
javascript arrays sum reduce
asked 8 hours ago
HappyHands31HappyHands31
1,3853 gold badges22 silver badges41 bronze badges
don't you think you should be solving it yourself?
– Arpit Pandey
8 hours ago
1
I have tried to solve it myself. I don't know what else to try. I am not understanding the concept of how to add the sum of an empty array to the final array. Because you can't use reduce on an empty array. Any other ideas?
– HappyHands31
8 hours ago
Cant you just push 0 after the loop?
– Kobe
8 hours ago
1
This: But you cannot reduce an empty array - that is not what that error message (TypeError: Reduce of empty array with no initial value) says. In fact, it is very specific: Please re-read the documentation for reduce: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… - and then reread the error message.
– Randy Casburn
8 hours ago
1
BINGO! But consider the iterator point too.
– Randy Casburn
7 hours ago
|
show 10 more comments
6
6
6
1
Trying to solve this challenge on codewars. According to the challenge, the parts of array:
ls = [0, 1, 3, 6, 10]
Are
ls = [0, 1, 3, 6, 10]
ls = [1, 3, 6, 10]
ls = [3, 6, 10]
ls = [6, 10]
ls = [10]
ls = []
And we need to return an array with the sums of those parts.
So my code is as follows:
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
console.log(partsSums([0, 1, 3, 6, 10]));The issue is that it wants us to add the last sum 0 when the array is empty. So we should be getting:
[ 20, 20, 19, 16, 10, 0 ]
Instead of
[ 20, 20, 19, 16, 10]
So I tried this:
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
arrayOfSums.push(0);
return arrayOfSums;
console.log(partsSums([0, 1, 3, 6, 10]));And this:
function partsSums(ls)
ls.push(0);
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
But these caused execution time-out errors on Codewars:
Execution Timed Out (12000 ms)
So I also tried:
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > -1)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
But now this causes a TypeError:
TypeError: Reduce of empty array with no initial value
I am not understanding the concept of how to get 0 into the array when all of the values have been shifted out. The challenge seems to want 0 as the final "sum" of the array, even when the array is empty. But you cannot reduce an empty array - what else can I do here?
EDIT: Tried adding initial value to the reduce method:
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b, 0);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
Unfortunately this still fails the basic test :
expected [] to deeply equal [ 0 ]
javascript arrays sum reduce
asked 8 hours ago
HappyHands31HappyHands31
1,3853 gold badges22 silver badges41 bronze badges
Trying to solve this challenge on codewars. According to the challenge, the parts of array:
ls = [0, 1, 3, 6, 10]
Are
ls = [0, 1, 3, 6, 10]
ls = [1, 3, 6, 10]
ls = [3, 6, 10]
ls = [6, 10]
ls = [10]
ls = []
And we need to return an array with the sums of those parts.
So my code is as follows:
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
console.log(partsSums([0, 1, 3, 6, 10]));The issue is that it wants us to add the last sum 0 when the array is empty. So we should be getting:
[ 20, 20, 19, 16, 10, 0 ]
Instead of
[ 20, 20, 19, 16, 10]
So I tried this:
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
arrayOfSums.push(0);
return arrayOfSums;
console.log(partsSums([0, 1, 3, 6, 10]));And this:
function partsSums(ls)
ls.push(0);
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
But these caused execution time-out errors on Codewars:
Execution Timed Out (12000 ms)
So I also tried:
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > -1)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
But now this causes a TypeError:
TypeError: Reduce of empty array with no initial value
I am not understanding the concept of how to get 0 into the array when all of the values have been shifted out. The challenge seems to want 0 as the final "sum" of the array, even when the array is empty. But you cannot reduce an empty array - what else can I do here?
EDIT: Tried adding initial value to the reduce method:
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b, 0);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
Unfortunately this still fails the basic test :
expected [] to deeply equal [ 0 ]
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
console.log(partsSums([0, 1, 3, 6, 10]));function partsSums(ls)
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
console.log(partsSums([0, 1, 3, 6, 10]));function partsSums(ls)
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
arrayOfSums.push(0);
return arrayOfSums;
console.log(partsSums([0, 1, 3, 6, 10]));function partsSums(ls)
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
arrayOfSums.push(0);
return arrayOfSums;
console.log(partsSums([0, 1, 3, 6, 10]));function partsSums(ls)
ls.push(0);
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
function partsSums(ls)
ls.push(0);
let arrayOfSums = [];
while(ls.length > 0)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > -1)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
function partsSums(ls)
let arrayOfSums = [];
while(ls.length > -1)
let sum = ls.reduce((a, b) => a + b);
arrayOfSums.push(sum);
ls.shift();
return arrayOfSums;
javascript arrays sum reduce
javascript arrays sum reduce
asked 8 hours ago
HappyHands31HappyHands31
1,3853 gold badges22 silver badges41 bronze badges
asked 8 hours ago
HappyHands31HappyHands31
1,3853 gold badges22 silver badges41 bronze badges
edited 7 hours ago
HappyHands31
asked 8 hours ago
HappyHands31HappyHands31
1,3853 gold badges22 silver badges41 bronze badges
asked 8 hours ago
HappyHands31HappyHands31
1,3853 gold badges22 silver badges41 bronze badges
asked 8 hours ago
HappyHands31HappyHands31
1,3853 gold badges22 silver badges41 bronze badges
1,3853 gold badges22 silver badges41 bronze badges
don't you think you should be solving it yourself?
– Arpit Pandey
8 hours ago
1
I have tried to solve it myself. I don't know what else to try. I am not understanding the concept of how to add the sum of an empty array to the final array. Because you can't use reduce on an empty array. Any other ideas?
– HappyHands31
8 hours ago
Cant you just push 0 after the loop?
– Kobe
8 hours ago
1
This: But you cannot reduce an empty array - that is not what that error message (TypeError: Reduce of empty array with no initial value) says. In fact, it is very specific: Please re-read the documentation for reduce: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… - and then reread the error message.
– Randy Casburn
8 hours ago
1
BINGO! But consider the iterator point too.
– Randy Casburn
7 hours ago
|
show 10 more comments
don't you think you should be solving it yourself?
– Arpit Pandey
8 hours ago
1
I have tried to solve it myself. I don't know what else to try. I am not understanding the concept of how to add the sum of an empty array to the final array. Because you can't use reduce on an empty array. Any other ideas?
– HappyHands31
8 hours ago
Cant you just push 0 after the loop?
– Kobe
8 hours ago
1
This: But you cannot reduce an empty array - that is not what that error message (TypeError: Reduce of empty array with no initial value) says. In fact, it is very specific: Please re-read the documentation for reduce: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… - and then reread the error message.
– Randy Casburn
8 hours ago
1
BINGO! But consider the iterator point too.
– Randy Casburn
7 hours ago
don't you think you should be solving it yourself?
– Arpit Pandey
8 hours ago
don't you think you should be solving it yourself?
– Arpit Pandey
8 hours ago
1
1
I have tried to solve it myself. I don't know what else to try. I am not understanding the concept of how to add the sum of an empty array to the final array. Because you can't use reduce on an empty array. Any other ideas?
– HappyHands31
8 hours ago
I have tried to solve it myself. I don't know what else to try. I am not understanding the concept of how to add the sum of an empty array to the final array. Because you can't use reduce on an empty array. Any other ideas?
– HappyHands31
8 hours ago
Cant you just push 0 after the loop?
– Kobe
8 hours ago
Cant you just push 0 after the loop?
– Kobe
8 hours ago
1
1
This: But you cannot reduce an empty array - that is not what that error message (TypeError: Reduce of empty array with no initial value) says. In fact, it is very specific: Please re-read the documentation for reduce: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… - and then reread the error message.
– Randy Casburn
8 hours ago
This: But you cannot reduce an empty array - that is not what that error message (TypeError: Reduce of empty array with no initial value) says. In fact, it is very specific: Please re-read the documentation for reduce: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… - and then reread the error message.
– Randy Casburn
8 hours ago
1
1
BINGO! But consider the iterator point too.
– Randy Casburn
7 hours ago
BINGO! But consider the iterator point too.
– Randy Casburn
7 hours ago
|
show 10 more comments
6 Answers
6
active
oldest
votes
7
There is no reason to compute the sum over and over. On a long array this will be very inefficient ( O(n²) ) and might explain your timeout errors. Compute the sum at the beginning and then subtract each element from it in a loop.
ls = [0, 1, 3, 6, 10]
function partsSums(ls)
let sum = ls.reduce((sum, n) => sum + n, 0)
res = [sum]
for (let i = 1; i <= ls.length; i++)
sum -= ls[i-1]
res.push(sum )
return res
console.log(partsSums(ls))answered 7 hours ago
Mark MeyerMark Meyer
47.6k3 gold badges41 silver badges74 bronze badges
It works. There are some things about it that I don't understand. What is( O(n^2) )? XOR, right?
– HappyHands31
7 hours ago
O(n²)means that the number of calculations required grows exponentially as the length of your array grows. Thereduceneeds to look at every element in the array and you do that for every element in the array.
– Mark Meyer
7 hours ago
still some overhead ...
– Nina Scholz
7 hours ago
1
@NinaScholz Are you talking about the initialreduce()? At least on chrome the overhead of the initial sum is much smaller than the penalty forunshift()in my tests.
– Mark Meyer
7 hours ago
1
So, to break this down a bit more, we have the first sum as20, then within the for-loop, we create the "result" array, initializing it to20. Then we subtractls[i - 1]from sum. So the first sum to get pushed into the result array from the for-loop will also be20, becausels[i - 1]in this case is0. Then 20 - 1, then 19 - 3, etc. Makes sense - thank you.
– HappyHands31
7 hours ago
|
show 1 more comment
3
Another solution that passed all of the tests:
function partsSums(ls)
let result = [0],
l = ls.length - 1;
for (let i = l; i >= 0; i--)
result.push(ls[i] + result[ l - i]);
return result.reverse();
console.log(partsSums([]));
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));answered 7 hours ago
FractionFraction
1,8702 silver badges16 bronze badges
1
That is very clever yet simple. Thx.
– HappyHands31
6 hours ago
You could also doresult.unshift(ls[i] + result[0])and avoidreverse.
– georg
6 hours ago
1
@georgunshiftdoesn't pass the test because it is slower thanpush+reverse
– Fraction
6 hours ago
add a comment |
1
You could use for loop with slice and when i == 0 you can slice len + 1 which is going to return you empty array and sum will be 0.
function partsSums(arr)
const res = [], len = arr.length
for (let i = len; i > -1; i--)
res.push(arr.slice(-i
return res;
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));You can also use two double reduce and if there is no next element push zero.
function partsSums(arr)
const sum = arr => arr.reduce((r, e) => r + e, 0);
return arr.reduce((r, e, i, a) =>
const res = sum(a.slice(i, a.length));
return r.concat(!a[i + 1] ? [res, 0] : res)
, [])
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));answered 7 hours ago
Nenad VracarNenad Vracar
76.6k12 gold badges65 silver badges88 bronze badges
add a comment |
1
try this with recursion :
function partsSums(ls)
let sum = ls.reduce((a, b) => a + b, 0);
return ls.length > 0 ? [sum].concat(partsSums(ls.slice(1))) : [0];
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));answered 7 hours ago
Elma CherbElma Cherb
1961 silver badge10 bronze badges
add a comment |
1
Here's one thing you could do
function partsSums(ls)
if(!ls.length) return [0];
let prevTotal = ls.reduce((a,b) => a + b);
return [prevTotal, ...ls.map(val => prevTotal -= val)]
console.log(partsSums([0, 1, 3, 6, 10]));answered 7 hours ago
Alexandre FradetteAlexandre Fradette
799 bronze badges
I like this also - can you please help me understand how theif(!ls.length) return [0]gets added to the end of the returned array? There aren't any loops that are reducing the length ofls?
– HappyHands31
7 hours ago
add a comment |
1
You could iterate from the end and take this value plus the last inserted value of the result set.
This approach works with a single loop and without calculating the maximum sum in advance.
function partsSums(ls)
var result = [0],
i = ls.length;
while (i--)
result.unshift(ls[i] + result[0]);
return result;
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([]));.as-console-wrapper max-height: 100% !important; top: 0; With push and reverse.
function partsSums(ls)
var result = [0],
l = 0,
i = ls.length;
while (i--) result.push(l += ls[i]);
return result.reverse();
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([]));.as-console-wrapper max-height: 100% !important; top: 0; answered 7 hours ago
Nina ScholzNina Scholz
212k16 gold badges127 silver badges192 bronze badges
add a comment |
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6 Answers
6
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6 Answers
6
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7
There is no reason to compute the sum over and over. On a long array this will be very inefficient ( O(n²) ) and might explain your timeout errors. Compute the sum at the beginning and then subtract each element from it in a loop.
ls = [0, 1, 3, 6, 10]
function partsSums(ls)
let sum = ls.reduce((sum, n) => sum + n, 0)
res = [sum]
for (let i = 1; i <= ls.length; i++)
sum -= ls[i-1]
res.push(sum )
return res
console.log(partsSums(ls))answered 7 hours ago
Mark MeyerMark Meyer
47.6k3 gold badges41 silver badges74 bronze badges
It works. There are some things about it that I don't understand. What is( O(n^2) )? XOR, right?
– HappyHands31
7 hours ago
O(n²)means that the number of calculations required grows exponentially as the length of your array grows. Thereduceneeds to look at every element in the array and you do that for every element in the array.
– Mark Meyer
7 hours ago
still some overhead ...
– Nina Scholz
7 hours ago
1
@NinaScholz Are you talking about the initialreduce()? At least on chrome the overhead of the initial sum is much smaller than the penalty forunshift()in my tests.
– Mark Meyer
7 hours ago
1
So, to break this down a bit more, we have the first sum as20, then within the for-loop, we create the "result" array, initializing it to20. Then we subtractls[i - 1]from sum. So the first sum to get pushed into the result array from the for-loop will also be20, becausels[i - 1]in this case is0. Then 20 - 1, then 19 - 3, etc. Makes sense - thank you.
– HappyHands31
7 hours ago
|
show 1 more comment
7
There is no reason to compute the sum over and over. On a long array this will be very inefficient ( O(n²) ) and might explain your timeout errors. Compute the sum at the beginning and then subtract each element from it in a loop.
ls = [0, 1, 3, 6, 10]
function partsSums(ls)
let sum = ls.reduce((sum, n) => sum + n, 0)
res = [sum]
for (let i = 1; i <= ls.length; i++)
sum -= ls[i-1]
res.push(sum )
return res
console.log(partsSums(ls))answered 7 hours ago
Mark MeyerMark Meyer
47.6k3 gold badges41 silver badges74 bronze badges
It works. There are some things about it that I don't understand. What is( O(n^2) )? XOR, right?
– HappyHands31
7 hours ago
O(n²)means that the number of calculations required grows exponentially as the length of your array grows. Thereduceneeds to look at every element in the array and you do that for every element in the array.
– Mark Meyer
7 hours ago
still some overhead ...
– Nina Scholz
7 hours ago
1
@NinaScholz Are you talking about the initialreduce()? At least on chrome the overhead of the initial sum is much smaller than the penalty forunshift()in my tests.
– Mark Meyer
7 hours ago
1
So, to break this down a bit more, we have the first sum as20, then within the for-loop, we create the "result" array, initializing it to20. Then we subtractls[i - 1]from sum. So the first sum to get pushed into the result array from the for-loop will also be20, becausels[i - 1]in this case is0. Then 20 - 1, then 19 - 3, etc. Makes sense - thank you.
– HappyHands31
7 hours ago
|
show 1 more comment
7
7
7
There is no reason to compute the sum over and over. On a long array this will be very inefficient ( O(n²) ) and might explain your timeout errors. Compute the sum at the beginning and then subtract each element from it in a loop.
ls = [0, 1, 3, 6, 10]
function partsSums(ls)
let sum = ls.reduce((sum, n) => sum + n, 0)
res = [sum]
for (let i = 1; i <= ls.length; i++)
sum -= ls[i-1]
res.push(sum )
return res
console.log(partsSums(ls))answered 7 hours ago
Mark MeyerMark Meyer
47.6k3 gold badges41 silver badges74 bronze badges
There is no reason to compute the sum over and over. On a long array this will be very inefficient ( O(n²) ) and might explain your timeout errors. Compute the sum at the beginning and then subtract each element from it in a loop.
ls = [0, 1, 3, 6, 10]
function partsSums(ls)
let sum = ls.reduce((sum, n) => sum + n, 0)
res = [sum]
for (let i = 1; i <= ls.length; i++)
sum -= ls[i-1]
res.push(sum )
return res
console.log(partsSums(ls))ls = [0, 1, 3, 6, 10]
function partsSums(ls)
let sum = ls.reduce((sum, n) => sum + n, 0)
res = [sum]
for (let i = 1; i <= ls.length; i++)
sum -= ls[i-1]
res.push(sum )
return res
console.log(partsSums(ls))ls = [0, 1, 3, 6, 10]
function partsSums(ls)
let sum = ls.reduce((sum, n) => sum + n, 0)
res = [sum]
for (let i = 1; i <= ls.length; i++)
sum -= ls[i-1]
res.push(sum )
return res
console.log(partsSums(ls))answered 7 hours ago
Mark MeyerMark Meyer
47.6k3 gold badges41 silver badges74 bronze badges
edited 7 hours ago
answered 7 hours ago
Mark MeyerMark Meyer
47.6k3 gold badges41 silver badges74 bronze badges
answered 7 hours ago
Mark MeyerMark Meyer
47.6k3 gold badges41 silver badges74 bronze badges
answered 7 hours ago
Mark MeyerMark Meyer
47.6k3 gold badges41 silver badges74 bronze badges
47.6k3 gold badges41 silver badges74 bronze badges
It works. There are some things about it that I don't understand. What is( O(n^2) )? XOR, right?
– HappyHands31
7 hours ago
O(n²)means that the number of calculations required grows exponentially as the length of your array grows. Thereduceneeds to look at every element in the array and you do that for every element in the array.
– Mark Meyer
7 hours ago
still some overhead ...
– Nina Scholz
7 hours ago
1
@NinaScholz Are you talking about the initialreduce()? At least on chrome the overhead of the initial sum is much smaller than the penalty forunshift()in my tests.
– Mark Meyer
7 hours ago
1
So, to break this down a bit more, we have the first sum as20, then within the for-loop, we create the "result" array, initializing it to20. Then we subtractls[i - 1]from sum. So the first sum to get pushed into the result array from the for-loop will also be20, becausels[i - 1]in this case is0. Then 20 - 1, then 19 - 3, etc. Makes sense - thank you.
– HappyHands31
7 hours ago
|
show 1 more comment
It works. There are some things about it that I don't understand. What is( O(n^2) )? XOR, right?
– HappyHands31
7 hours ago
O(n²)means that the number of calculations required grows exponentially as the length of your array grows. Thereduceneeds to look at every element in the array and you do that for every element in the array.
– Mark Meyer
7 hours ago
still some overhead ...
– Nina Scholz
7 hours ago
1
@NinaScholz Are you talking about the initialreduce()? At least on chrome the overhead of the initial sum is much smaller than the penalty forunshift()in my tests.
– Mark Meyer
7 hours ago
1
So, to break this down a bit more, we have the first sum as20, then within the for-loop, we create the "result" array, initializing it to20. Then we subtractls[i - 1]from sum. So the first sum to get pushed into the result array from the for-loop will also be20, becausels[i - 1]in this case is0. Then 20 - 1, then 19 - 3, etc. Makes sense - thank you.
– HappyHands31
7 hours ago
It works. There are some things about it that I don't understand. What is
( O(n^2) )? XOR, right?– HappyHands31
7 hours ago
It works. There are some things about it that I don't understand. What is
( O(n^2) )? XOR, right?– HappyHands31
7 hours ago
O(n²) means that the number of calculations required grows exponentially as the length of your array grows. The reduce needs to look at every element in the array and you do that for every element in the array.– Mark Meyer
7 hours ago
O(n²) means that the number of calculations required grows exponentially as the length of your array grows. The reduce needs to look at every element in the array and you do that for every element in the array.– Mark Meyer
7 hours ago
still some overhead ...
– Nina Scholz
7 hours ago
still some overhead ...
– Nina Scholz
7 hours ago
1
1
@NinaScholz Are you talking about the initial
reduce()? At least on chrome the overhead of the initial sum is much smaller than the penalty for unshift() in my tests.– Mark Meyer
7 hours ago
@NinaScholz Are you talking about the initial
reduce()? At least on chrome the overhead of the initial sum is much smaller than the penalty for unshift() in my tests.– Mark Meyer
7 hours ago
1
1
So, to break this down a bit more, we have the first sum as
20, then within the for-loop, we create the "result" array, initializing it to 20. Then we subtract ls[i - 1] from sum. So the first sum to get pushed into the result array from the for-loop will also be 20, because ls[i - 1] in this case is 0. Then 20 - 1, then 19 - 3, etc. Makes sense - thank you.– HappyHands31
7 hours ago
So, to break this down a bit more, we have the first sum as
20, then within the for-loop, we create the "result" array, initializing it to 20. Then we subtract ls[i - 1] from sum. So the first sum to get pushed into the result array from the for-loop will also be 20, because ls[i - 1] in this case is 0. Then 20 - 1, then 19 - 3, etc. Makes sense - thank you.– HappyHands31
7 hours ago
|
show 1 more comment
3
Another solution that passed all of the tests:
function partsSums(ls)
let result = [0],
l = ls.length - 1;
for (let i = l; i >= 0; i--)
result.push(ls[i] + result[ l - i]);
return result.reverse();
console.log(partsSums([]));
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));answered 7 hours ago
FractionFraction
1,8702 silver badges16 bronze badges
1
That is very clever yet simple. Thx.
– HappyHands31
6 hours ago
You could also doresult.unshift(ls[i] + result[0])and avoidreverse.
– georg
6 hours ago
1
@georgunshiftdoesn't pass the test because it is slower thanpush+reverse
– Fraction
6 hours ago
add a comment |
3
Another solution that passed all of the tests:
function partsSums(ls)
let result = [0],
l = ls.length - 1;
for (let i = l; i >= 0; i--)
result.push(ls[i] + result[ l - i]);
return result.reverse();
console.log(partsSums([]));
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));answered 7 hours ago
FractionFraction
1,8702 silver badges16 bronze badges
1
That is very clever yet simple. Thx.
– HappyHands31
6 hours ago
You could also doresult.unshift(ls[i] + result[0])and avoidreverse.
– georg
6 hours ago
1
@georgunshiftdoesn't pass the test because it is slower thanpush+reverse
– Fraction
6 hours ago
add a comment |
3
3
3
Another solution that passed all of the tests:
function partsSums(ls)
let result = [0],
l = ls.length - 1;
for (let i = l; i >= 0; i--)
result.push(ls[i] + result[ l - i]);
return result.reverse();
console.log(partsSums([]));
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));answered 7 hours ago
FractionFraction
1,8702 silver badges16 bronze badges
Another solution that passed all of the tests:
function partsSums(ls)
let result = [0],
l = ls.length - 1;
for (let i = l; i >= 0; i--)
result.push(ls[i] + result[ l - i]);
return result.reverse();
console.log(partsSums([]));
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));function partsSums(ls)
let result = [0],
l = ls.length - 1;
for (let i = l; i >= 0; i--)
result.push(ls[i] + result[ l - i]);
return result.reverse();
console.log(partsSums([]));
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));function partsSums(ls)
let result = [0],
l = ls.length - 1;
for (let i = l; i >= 0; i--)
result.push(ls[i] + result[ l - i]);
return result.reverse();
console.log(partsSums([]));
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));answered 7 hours ago
FractionFraction
1,8702 silver badges16 bronze badges
edited 7 hours ago
answered 7 hours ago
FractionFraction
1,8702 silver badges16 bronze badges
answered 7 hours ago
FractionFraction
1,8702 silver badges16 bronze badges
answered 7 hours ago
FractionFraction
1,8702 silver badges16 bronze badges
1,8702 silver badges16 bronze badges
1
That is very clever yet simple. Thx.
– HappyHands31
6 hours ago
You could also doresult.unshift(ls[i] + result[0])and avoidreverse.
– georg
6 hours ago
1
@georgunshiftdoesn't pass the test because it is slower thanpush+reverse
– Fraction
6 hours ago
add a comment |
1
That is very clever yet simple. Thx.
– HappyHands31
6 hours ago
You could also doresult.unshift(ls[i] + result[0])and avoidreverse.
– georg
6 hours ago
1
@georgunshiftdoesn't pass the test because it is slower thanpush+reverse
– Fraction
6 hours ago
1
1
That is very clever yet simple. Thx.
– HappyHands31
6 hours ago
That is very clever yet simple. Thx.
– HappyHands31
6 hours ago
You could also do
result.unshift(ls[i] + result[0]) and avoid reverse.– georg
6 hours ago
You could also do
result.unshift(ls[i] + result[0]) and avoid reverse.– georg
6 hours ago
1
1
@georg
unshift doesn't pass the test because it is slower than push+reverse– Fraction
6 hours ago
@georg
unshift doesn't pass the test because it is slower than push+reverse– Fraction
6 hours ago
add a comment |
1
You could use for loop with slice and when i == 0 you can slice len + 1 which is going to return you empty array and sum will be 0.
function partsSums(arr)
const res = [], len = arr.length
for (let i = len; i > -1; i--)
res.push(arr.slice(-i
return res;
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));You can also use two double reduce and if there is no next element push zero.
function partsSums(arr)
const sum = arr => arr.reduce((r, e) => r + e, 0);
return arr.reduce((r, e, i, a) =>
const res = sum(a.slice(i, a.length));
return r.concat(!a[i + 1] ? [res, 0] : res)
, [])
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));answered 7 hours ago
Nenad VracarNenad Vracar
76.6k12 gold badges65 silver badges88 bronze badges
add a comment |
1
You could use for loop with slice and when i == 0 you can slice len + 1 which is going to return you empty array and sum will be 0.
function partsSums(arr)
const res = [], len = arr.length
for (let i = len; i > -1; i--)
res.push(arr.slice(-i
return res;
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));You can also use two double reduce and if there is no next element push zero.
function partsSums(arr)
const sum = arr => arr.reduce((r, e) => r + e, 0);
return arr.reduce((r, e, i, a) =>
const res = sum(a.slice(i, a.length));
return r.concat(!a[i + 1] ? [res, 0] : res)
, [])
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));answered 7 hours ago
Nenad VracarNenad Vracar
76.6k12 gold badges65 silver badges88 bronze badges
add a comment |
1
1
1
You could use for loop with slice and when i == 0 you can slice len + 1 which is going to return you empty array and sum will be 0.
function partsSums(arr)
const res = [], len = arr.length
for (let i = len; i > -1; i--)
res.push(arr.slice(-i
return res;
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));You can also use two double reduce and if there is no next element push zero.
function partsSums(arr)
const sum = arr => arr.reduce((r, e) => r + e, 0);
return arr.reduce((r, e, i, a) =>
const res = sum(a.slice(i, a.length));
return r.concat(!a[i + 1] ? [res, 0] : res)
, [])
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));answered 7 hours ago
Nenad VracarNenad Vracar
76.6k12 gold badges65 silver badges88 bronze badges
You could use for loop with slice and when i == 0 you can slice len + 1 which is going to return you empty array and sum will be 0.
function partsSums(arr)
const res = [], len = arr.length
for (let i = len; i > -1; i--)
res.push(arr.slice(-i
return res;
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));You can also use two double reduce and if there is no next element push zero.
function partsSums(arr)
const sum = arr => arr.reduce((r, e) => r + e, 0);
return arr.reduce((r, e, i, a) =>
const res = sum(a.slice(i, a.length));
return r.concat(!a[i + 1] ? [res, 0] : res)
, [])
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));function partsSums(arr)
const res = [], len = arr.length
for (let i = len; i > -1; i--)
res.push(arr.slice(-i
return res;
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));function partsSums(arr)
const res = [], len = arr.length
for (let i = len; i > -1; i--)
res.push(arr.slice(-i
return res;
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));function partsSums(arr)
const sum = arr => arr.reduce((r, e) => r + e, 0);
return arr.reduce((r, e, i, a) =>
const res = sum(a.slice(i, a.length));
return r.concat(!a[i + 1] ? [res, 0] : res)
, [])
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));function partsSums(arr)
const sum = arr => arr.reduce((r, e) => r + e, 0);
return arr.reduce((r, e, i, a) =>
const res = sum(a.slice(i, a.length));
return r.concat(!a[i + 1] ? [res, 0] : res)
, [])
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));answered 7 hours ago
Nenad VracarNenad Vracar
76.6k12 gold badges65 silver badges88 bronze badges
edited 7 hours ago
answered 7 hours ago
Nenad VracarNenad Vracar
76.6k12 gold badges65 silver badges88 bronze badges
answered 7 hours ago
Nenad VracarNenad Vracar
76.6k12 gold badges65 silver badges88 bronze badges
answered 7 hours ago
Nenad VracarNenad Vracar
76.6k12 gold badges65 silver badges88 bronze badges
76.6k12 gold badges65 silver badges88 bronze badges
add a comment |
add a comment |
1
try this with recursion :
function partsSums(ls)
let sum = ls.reduce((a, b) => a + b, 0);
return ls.length > 0 ? [sum].concat(partsSums(ls.slice(1))) : [0];
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));answered 7 hours ago
Elma CherbElma Cherb
1961 silver badge10 bronze badges
add a comment |
1
try this with recursion :
function partsSums(ls)
let sum = ls.reduce((a, b) => a + b, 0);
return ls.length > 0 ? [sum].concat(partsSums(ls.slice(1))) : [0];
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));answered 7 hours ago
Elma CherbElma Cherb
1961 silver badge10 bronze badges
add a comment |
1
1
1
try this with recursion :
function partsSums(ls)
let sum = ls.reduce((a, b) => a + b, 0);
return ls.length > 0 ? [sum].concat(partsSums(ls.slice(1))) : [0];
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));answered 7 hours ago
Elma CherbElma Cherb
1961 silver badge10 bronze badges
try this with recursion :
function partsSums(ls)
let sum = ls.reduce((a, b) => a + b, 0);
return ls.length > 0 ? [sum].concat(partsSums(ls.slice(1))) : [0];
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));function partsSums(ls)
let sum = ls.reduce((a, b) => a + b, 0);
return ls.length > 0 ? [sum].concat(partsSums(ls.slice(1))) : [0];
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));function partsSums(ls)
let sum = ls.reduce((a, b) => a + b, 0);
return ls.length > 0 ? [sum].concat(partsSums(ls.slice(1))) : [0];
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([1, 2, 3, 4, 5, 6]));
console.log(partsSums([744125, 935, 407, 454, 430, 90, 144, 6710213, 889, 810, 2579358]));answered 7 hours ago
Elma CherbElma Cherb
1961 silver badge10 bronze badges
answered 7 hours ago
Elma CherbElma Cherb
1961 silver badge10 bronze badges
answered 7 hours ago
Elma CherbElma Cherb
1961 silver badge10 bronze badges
answered 7 hours ago
Elma CherbElma Cherb
1961 silver badge10 bronze badges
1961 silver badge10 bronze badges
add a comment |
add a comment |
1
Here's one thing you could do
function partsSums(ls)
if(!ls.length) return [0];
let prevTotal = ls.reduce((a,b) => a + b);
return [prevTotal, ...ls.map(val => prevTotal -= val)]
console.log(partsSums([0, 1, 3, 6, 10]));answered 7 hours ago
Alexandre FradetteAlexandre Fradette
799 bronze badges
I like this also - can you please help me understand how theif(!ls.length) return [0]gets added to the end of the returned array? There aren't any loops that are reducing the length ofls?
– HappyHands31
7 hours ago
add a comment |
1
Here's one thing you could do
function partsSums(ls)
if(!ls.length) return [0];
let prevTotal = ls.reduce((a,b) => a + b);
return [prevTotal, ...ls.map(val => prevTotal -= val)]
console.log(partsSums([0, 1, 3, 6, 10]));answered 7 hours ago
Alexandre FradetteAlexandre Fradette
799 bronze badges
I like this also - can you please help me understand how theif(!ls.length) return [0]gets added to the end of the returned array? There aren't any loops that are reducing the length ofls?
– HappyHands31
7 hours ago
add a comment |
1
1
1
Here's one thing you could do
function partsSums(ls)
if(!ls.length) return [0];
let prevTotal = ls.reduce((a,b) => a + b);
return [prevTotal, ...ls.map(val => prevTotal -= val)]
console.log(partsSums([0, 1, 3, 6, 10]));answered 7 hours ago
Alexandre FradetteAlexandre Fradette
799 bronze badges
Here's one thing you could do
function partsSums(ls)
if(!ls.length) return [0];
let prevTotal = ls.reduce((a,b) => a + b);
return [prevTotal, ...ls.map(val => prevTotal -= val)]
console.log(partsSums([0, 1, 3, 6, 10]));function partsSums(ls)
if(!ls.length) return [0];
let prevTotal = ls.reduce((a,b) => a + b);
return [prevTotal, ...ls.map(val => prevTotal -= val)]
console.log(partsSums([0, 1, 3, 6, 10]));function partsSums(ls)
if(!ls.length) return [0];
let prevTotal = ls.reduce((a,b) => a + b);
return [prevTotal, ...ls.map(val => prevTotal -= val)]
console.log(partsSums([0, 1, 3, 6, 10]));answered 7 hours ago
Alexandre FradetteAlexandre Fradette
799 bronze badges
answered 7 hours ago
Alexandre FradetteAlexandre Fradette
799 bronze badges
answered 7 hours ago
Alexandre FradetteAlexandre Fradette
799 bronze badges
answered 7 hours ago
Alexandre FradetteAlexandre Fradette
799 bronze badges
799 bronze badges
I like this also - can you please help me understand how theif(!ls.length) return [0]gets added to the end of the returned array? There aren't any loops that are reducing the length ofls?
– HappyHands31
7 hours ago
add a comment |
I like this also - can you please help me understand how theif(!ls.length) return [0]gets added to the end of the returned array? There aren't any loops that are reducing the length ofls?
– HappyHands31
7 hours ago
I like this also - can you please help me understand how the
if(!ls.length) return [0] gets added to the end of the returned array? There aren't any loops that are reducing the length of ls?– HappyHands31
7 hours ago
I like this also - can you please help me understand how the
if(!ls.length) return [0] gets added to the end of the returned array? There aren't any loops that are reducing the length of ls?– HappyHands31
7 hours ago
add a comment |
1
You could iterate from the end and take this value plus the last inserted value of the result set.
This approach works with a single loop and without calculating the maximum sum in advance.
function partsSums(ls)
var result = [0],
i = ls.length;
while (i--)
result.unshift(ls[i] + result[0]);
return result;
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([]));.as-console-wrapper max-height: 100% !important; top: 0; With push and reverse.
function partsSums(ls)
var result = [0],
l = 0,
i = ls.length;
while (i--) result.push(l += ls[i]);
return result.reverse();
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([]));.as-console-wrapper max-height: 100% !important; top: 0; answered 7 hours ago
Nina ScholzNina Scholz
212k16 gold badges127 silver badges192 bronze badges
add a comment |
1
You could iterate from the end and take this value plus the last inserted value of the result set.
This approach works with a single loop and without calculating the maximum sum in advance.
function partsSums(ls)
var result = [0],
i = ls.length;
while (i--)
result.unshift(ls[i] + result[0]);
return result;
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([]));.as-console-wrapper max-height: 100% !important; top: 0; With push and reverse.
function partsSums(ls)
var result = [0],
l = 0,
i = ls.length;
while (i--) result.push(l += ls[i]);
return result.reverse();
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([]));.as-console-wrapper max-height: 100% !important; top: 0; answered 7 hours ago
Nina ScholzNina Scholz
212k16 gold badges127 silver badges192 bronze badges
add a comment |
1
1
1
You could iterate from the end and take this value plus the last inserted value of the result set.
This approach works with a single loop and without calculating the maximum sum in advance.
function partsSums(ls)
var result = [0],
i = ls.length;
while (i--)
result.unshift(ls[i] + result[0]);
return result;
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([]));.as-console-wrapper max-height: 100% !important; top: 0; With push and reverse.
function partsSums(ls)
var result = [0],
l = 0,
i = ls.length;
while (i--) result.push(l += ls[i]);
return result.reverse();
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([]));.as-console-wrapper max-height: 100% !important; top: 0; answered 7 hours ago
Nina ScholzNina Scholz
212k16 gold badges127 silver badges192 bronze badges
You could iterate from the end and take this value plus the last inserted value of the result set.
This approach works with a single loop and without calculating the maximum sum in advance.
function partsSums(ls)
var result = [0],
i = ls.length;
while (i--)
result.unshift(ls[i] + result[0]);
return result;
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([]));.as-console-wrapper max-height: 100% !important; top: 0; With push and reverse.
function partsSums(ls)
var result = [0],
l = 0,
i = ls.length;
while (i--) result.push(l += ls[i]);
return result.reverse();
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([]));.as-console-wrapper max-height: 100% !important; top: 0; function partsSums(ls)
var result = [0],
i = ls.length;
while (i--)
result.unshift(ls[i] + result[0]);
return result;
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([]));.as-console-wrapper max-height: 100% !important; top: 0; function partsSums(ls)
var result = [0],
i = ls.length;
while (i--)
result.unshift(ls[i] + result[0]);
return result;
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([]));.as-console-wrapper max-height: 100% !important; top: 0; function partsSums(ls)
var result = [0],
l = 0,
i = ls.length;
while (i--) result.push(l += ls[i]);
return result.reverse();
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([]));.as-console-wrapper max-height: 100% !important; top: 0; function partsSums(ls)
var result = [0],
l = 0,
i = ls.length;
while (i--) result.push(l += ls[i]);
return result.reverse();
console.log(partsSums([0, 1, 3, 6, 10]));
console.log(partsSums([]));.as-console-wrapper max-height: 100% !important; top: 0; answered 7 hours ago
Nina ScholzNina Scholz
212k16 gold badges127 silver badges192 bronze badges
edited 7 hours ago
answered 7 hours ago
Nina ScholzNina Scholz
212k16 gold badges127 silver badges192 bronze badges
answered 7 hours ago
Nina ScholzNina Scholz
212k16 gold badges127 silver badges192 bronze badges
answered 7 hours ago
Nina ScholzNina Scholz
212k16 gold badges127 silver badges192 bronze badges
212k16 gold badges127 silver badges192 bronze badges
add a comment |
add a comment |
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don't you think you should be solving it yourself?
– Arpit Pandey
8 hours ago
1
I have tried to solve it myself. I don't know what else to try. I am not understanding the concept of how to add the sum of an empty array to the final array. Because you can't use reduce on an empty array. Any other ideas?
– HappyHands31
8 hours ago
Cant you just push 0 after the loop?
– Kobe
8 hours ago
1
This: But you cannot reduce an empty array - that is not what that error message (TypeError: Reduce of empty array with no initial value) says. In fact, it is very specific: Please re-read the documentation for reduce: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… - and then reread the error message.
– Randy Casburn
8 hours ago
1
BINGO! But consider the iterator point too.
– Randy Casburn
7 hours ago