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Return only the number of paired values in array javascript

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7

My purpose is to display the number of paired values in the array.

for instance I have this array :[10,10,10,10,20,20,20,30,50]

I would like to display 3 because we have 3 pairs of number.

Any help will be greatly appreciated.

function pairNumber(arr) 
 var sorted_arr = arr.sort();
 var i;
 var results = [];
 for (i = 0; i < sorted_arr.length; i++) 
 if (sorted_arr[i + 1] == sorted_arr[i]) 
 results.push(sorted_arr[i]);
 

 
 return results.length;

console.log(pairNumber([10, 10, 10, 10, 20, 20, 20, 30, 50]))

javascript

share|improve this question

edited 8 hours ago

CertainPerformance

130k1717 gold badges8686 silver badges116116 bronze badges

asked 8 hours ago

DiaslineDiasline

1741212 bronze badges

add a comment | 

7

My purpose is to display the number of paired values in the array.

for instance I have this array :[10,10,10,10,20,20,20,30,50]

I would like to display 3 because we have 3 pairs of number.

Any help will be greatly appreciated.

function pairNumber(arr) 
 var sorted_arr = arr.sort();
 var i;
 var results = [];
 for (i = 0; i < sorted_arr.length; i++) 
 if (sorted_arr[i + 1] == sorted_arr[i]) 
 results.push(sorted_arr[i]);
 

 
 return results.length;

console.log(pairNumber([10, 10, 10, 10, 20, 20, 20, 30, 50]))

javascript

share|improve this question

edited 8 hours ago

CertainPerformance

130k1717 gold badges8686 silver badges116116 bronze badges

asked 8 hours ago

DiaslineDiasline

1741212 bronze badges

add a comment | 

My purpose is to display the number of paired values in the array.

for instance I have this array :[10,10,10,10,20,20,20,30,50]

I would like to display 3 because we have 3 pairs of number.

Any help will be greatly appreciated.

function pairNumber(arr) 
 var sorted_arr = arr.sort();
 var i;
 var results = [];
 for (i = 0; i < sorted_arr.length; i++) 
 if (sorted_arr[i + 1] == sorted_arr[i]) 
 results.push(sorted_arr[i]);
 

 
 return results.length;

console.log(pairNumber([10, 10, 10, 10, 20, 20, 20, 30, 50]))

javascript

share|improve this question

edited 8 hours ago

CertainPerformance

130k1717 gold badges8686 silver badges116116 bronze badges

asked 8 hours ago

DiaslineDiasline

1741212 bronze badges

function pairNumber(arr) 
 var sorted_arr = arr.sort();
 var i;
 var results = [];
 for (i = 0; i < sorted_arr.length; i++) 
 if (sorted_arr[i + 1] == sorted_arr[i]) 
 results.push(sorted_arr[i]);
 

 
 return results.length;

console.log(pairNumber([10, 10, 10, 10, 20, 20, 20, 30, 50]))

function pairNumber(arr) 
 var sorted_arr = arr.sort();
 var i;
 var results = [];
 for (i = 0; i < sorted_arr.length; i++) 
 if (sorted_arr[i + 1] == sorted_arr[i]) 
 results.push(sorted_arr[i]);
 

 
 return results.length;

console.log(pairNumber([10, 10, 10, 10, 20, 20, 20, 30, 50]))

function pairNumber(arr) 
 var sorted_arr = arr.sort();
 var i;
 var results = [];
 for (i = 0; i < sorted_arr.length; i++) 
 if (sorted_arr[i + 1] == sorted_arr[i]) 
 results.push(sorted_arr[i]);
 

 
 return results.length;

console.log(pairNumber([10, 10, 10, 10, 20, 20, 20, 30, 50]))

share|improve this question

edited 8 hours ago

CertainPerformance

130k1717 gold badges8686 silver badges116116 bronze badges

asked 8 hours ago

DiaslineDiasline

1741212 bronze badges

share|improve this question

edited 8 hours ago

CertainPerformance

130k1717 gold badges8686 silver badges116116 bronze badges

asked 8 hours ago

DiaslineDiasline

1741212 bronze badges

edited 8 hours ago

CertainPerformance

130k1717 gold badges8686 silver badges116116 bronze badges

edited 8 hours ago

CertainPerformance

130k1717 gold badges8686 silver badges116116 bronze badges

asked 8 hours ago

DiaslineDiasline

1741212 bronze badges

asked 8 hours ago

DiaslineDiasline

1741212 bronze badges

3 Answers
3

active

oldest

votes

4

Here is another approach using a Set:

function pairNumbers(arr) 
 let count = 0;
 const set = new Set();

 for (let i = 0; i < arr.length; i++) 
 if (set.has(arr[i])) 
 count++;
 set.delete(arr[i])
 else 
 set.add(arr[i])
 
 

 return count;

console.log(pairNumbers([10, 10, 10, 10, 20, 20, 20, 30, 50])) // 3

share|improve this answer

edited 8 hours ago

georg

166k3737 gold badges221221 silver badges322322 bronze badges

answered 8 hours ago

Matt AftMatt Aft

5,47588 silver badges2424 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  this is a smart solution!
  
  – georg
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @georg thanks for the edit, missed that :)
  
  – Matt Aft
  8 hours ago
  

  
  
  
  

add a comment | 

8

I'd reduce into an object, counting up the number of occurrences of each number. Then reduce again on the Object.values of the object to count up the number of pairs, adding Math.floor(count / 2) to the accumulator on each iteration:

function pairNumber(arr) 
 const itemCounts = arr.reduce((a, item) => 
 a[item] = (a[item] , );
 return Object.values(itemCounts)
 .reduce((pairsSoFar, count) => pairsSoFar + Math.floor(count / 2), 0);

console.log(pairNumber([10, 10, 10, 10, 20, 20, 20, 30, 50]))

Probably better to avoid .sort if possible - that increases the computational complexity from O(n) (minimum) to O(n log n).

share|improve this answer

answered 8 hours ago

CertainPerformanceCertainPerformance

130k1717 gold badges8686 silver badges116116 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I wonder if building itemCounts doesn't add the same O(nlogn) back? How do objects work under the hood? Hashtables? Balanced trees?
  
  – Vilx-
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  For any remotely sane implementation, it'll be O(1), luckily: stackoverflow.com/questions/7700987/… . If using an object causes an issue, could use a Map instead
  
  – CertainPerformance
  8 hours ago
  
  

  
  
  
  

add a comment | 

1

If I understood the question well then this can be simplified even further by relying on sort initially...

Increment i to the next position after finding the pair and let the for loop increment it once again.

 function pairNumber(arr) 
 var sorted_arr = [...arr].sort(); // disallowing array mutation
 let cnt = 0;
 for (let i = 0; i < sorted_arr.length; i++) 
 if (sorted_arr[i + 1] === sorted_arr[i]) 
 cnt++;
 i = i + 1;
 

 
 return cnt;
 
 console.log(pairNumber([10, 10, 10, 10, 10, 20, 20, 20, 20, 30, 30, 50]))
 // 5 --> 2 pairs of 10, 2 pairs of 20, 1 pair of 30
 console.log(pairNumbers([10, 10, 10, 10, 20, 20, 20, 30, 50])) 
 // 3 --> 2 pairs of 10 one pair of 20

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Eugene SunicEugene Sunic

3,82844 gold badges3232 silver badges5050 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I believe sort will mutate the original array, so a side-effect would be that the array being passed in will be sorted after calling this function.
  
  – Matt Aft
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MattAft tnx, edited my answer
  
  – Eugene Sunic
  7 hours ago
  

  
  
  
  

add a comment | 

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4

Here is another approach using a Set:

function pairNumbers(arr) 
 let count = 0;
 const set = new Set();

 for (let i = 0; i < arr.length; i++) 
 if (set.has(arr[i])) 
 count++;
 set.delete(arr[i])
 else 
 set.add(arr[i])
 
 

 return count;

console.log(pairNumbers([10, 10, 10, 10, 20, 20, 20, 30, 50])) // 3

share|improve this answer

edited 8 hours ago

georg

166k3737 gold badges221221 silver badges322322 bronze badges

answered 8 hours ago

Matt AftMatt Aft

5,47588 silver badges2424 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  this is a smart solution!
  
  – georg
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @georg thanks for the edit, missed that :)
  
  – Matt Aft
  8 hours ago
  

  
  
  
  

add a comment | 

4

Here is another approach using a Set:

function pairNumbers(arr) 
 let count = 0;
 const set = new Set();

 for (let i = 0; i < arr.length; i++) 
 if (set.has(arr[i])) 
 count++;
 set.delete(arr[i])
 else 
 set.add(arr[i])
 
 

 return count;

console.log(pairNumbers([10, 10, 10, 10, 20, 20, 20, 30, 50])) // 3

share|improve this answer

edited 8 hours ago

georg

166k3737 gold badges221221 silver badges322322 bronze badges

answered 8 hours ago

Matt AftMatt Aft

5,47588 silver badges2424 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  this is a smart solution!
  
  – georg
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @georg thanks for the edit, missed that :)
  
  – Matt Aft
  8 hours ago
  

  
  
  
  

add a comment | 

Here is another approach using a Set:

function pairNumbers(arr) 
 let count = 0;
 const set = new Set();

 for (let i = 0; i < arr.length; i++) 
 if (set.has(arr[i])) 
 count++;
 set.delete(arr[i])
 else 
 set.add(arr[i])
 
 

 return count;

console.log(pairNumbers([10, 10, 10, 10, 20, 20, 20, 30, 50])) // 3

share|improve this answer

edited 8 hours ago

georg

166k3737 gold badges221221 silver badges322322 bronze badges

answered 8 hours ago

Matt AftMatt Aft

5,47588 silver badges2424 bronze badges

function pairNumbers(arr) 
 let count = 0;
 const set = new Set();

 for (let i = 0; i < arr.length; i++) 
 if (set.has(arr[i])) 
 count++;
 set.delete(arr[i])
 else 
 set.add(arr[i])
 
 

 return count;

console.log(pairNumbers([10, 10, 10, 10, 20, 20, 20, 30, 50])) // 3

share|improve this answer

edited 8 hours ago

georg

166k3737 gold badges221221 silver badges322322 bronze badges

answered 8 hours ago

Matt AftMatt Aft

5,47588 silver badges2424 bronze badges

edited 8 hours ago

georg

166k3737 gold badges221221 silver badges322322 bronze badges

edited 8 hours ago

georg

166k3737 gold badges221221 silver badges322322 bronze badges

answered 8 hours ago

Matt AftMatt Aft

5,47588 silver badges2424 bronze badges

answered 8 hours ago

Matt AftMatt Aft

5,47588 silver badges2424 bronze badges

8

I'd reduce into an object, counting up the number of occurrences of each number. Then reduce again on the Object.values of the object to count up the number of pairs, adding Math.floor(count / 2) to the accumulator on each iteration:

function pairNumber(arr) 
 const itemCounts = arr.reduce((a, item) => 
 a[item] = (a[item] , );
 return Object.values(itemCounts)
 .reduce((pairsSoFar, count) => pairsSoFar + Math.floor(count / 2), 0);

console.log(pairNumber([10, 10, 10, 10, 20, 20, 20, 30, 50]))

Probably better to avoid .sort if possible - that increases the computational complexity from O(n) (minimum) to O(n log n).

share|improve this answer

answered 8 hours ago

CertainPerformanceCertainPerformance

130k1717 gold badges8686 silver badges116116 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I wonder if building itemCounts doesn't add the same O(nlogn) back? How do objects work under the hood? Hashtables? Balanced trees?
  
  – Vilx-
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  For any remotely sane implementation, it'll be O(1), luckily: stackoverflow.com/questions/7700987/… . If using an object causes an issue, could use a Map instead
  
  – CertainPerformance
  8 hours ago
  
  

  
  
  
  

add a comment | 

8

I'd reduce into an object, counting up the number of occurrences of each number. Then reduce again on the Object.values of the object to count up the number of pairs, adding Math.floor(count / 2) to the accumulator on each iteration:

function pairNumber(arr) 
 const itemCounts = arr.reduce((a, item) => 
 a[item] = (a[item] , );
 return Object.values(itemCounts)
 .reduce((pairsSoFar, count) => pairsSoFar + Math.floor(count / 2), 0);

console.log(pairNumber([10, 10, 10, 10, 20, 20, 20, 30, 50]))

Probably better to avoid .sort if possible - that increases the computational complexity from O(n) (minimum) to O(n log n).

share|improve this answer

answered 8 hours ago

CertainPerformanceCertainPerformance

130k1717 gold badges8686 silver badges116116 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I wonder if building itemCounts doesn't add the same O(nlogn) back? How do objects work under the hood? Hashtables? Balanced trees?
  
  – Vilx-
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  For any remotely sane implementation, it'll be O(1), luckily: stackoverflow.com/questions/7700987/… . If using an object causes an issue, could use a Map instead
  
  – CertainPerformance
  8 hours ago
  
  

  
  
  
  

add a comment | 

I'd reduce into an object, counting up the number of occurrences of each number. Then reduce again on the Object.values of the object to count up the number of pairs, adding Math.floor(count / 2) to the accumulator on each iteration:

function pairNumber(arr) 
 const itemCounts = arr.reduce((a, item) => 
 a[item] = (a[item] , );
 return Object.values(itemCounts)
 .reduce((pairsSoFar, count) => pairsSoFar + Math.floor(count / 2), 0);

console.log(pairNumber([10, 10, 10, 10, 20, 20, 20, 30, 50]))

Probably better to avoid .sort if possible - that increases the computational complexity from O(n) (minimum) to O(n log n).

share|improve this answer

answered 8 hours ago

CertainPerformanceCertainPerformance

130k1717 gold badges8686 silver badges116116 bronze badges

function pairNumber(arr) 
 const itemCounts = arr.reduce((a, item) => 
 a[item] = (a[item] , );
 return Object.values(itemCounts)
 .reduce((pairsSoFar, count) => pairsSoFar + Math.floor(count / 2), 0);

console.log(pairNumber([10, 10, 10, 10, 20, 20, 20, 30, 50]))

function pairNumber(arr) 
 const itemCounts = arr.reduce((a, item) => 
 a[item] = (a[item] , );
 return Object.values(itemCounts)
 .reduce((pairsSoFar, count) => pairsSoFar + Math.floor(count / 2), 0);

console.log(pairNumber([10, 10, 10, 10, 20, 20, 20, 30, 50]))

function pairNumber(arr) 
 const itemCounts = arr.reduce((a, item) => 
 a[item] = (a[item] , );
 return Object.values(itemCounts)
 .reduce((pairsSoFar, count) => pairsSoFar + Math.floor(count / 2), 0);

console.log(pairNumber([10, 10, 10, 10, 20, 20, 20, 30, 50]))

share|improve this answer

answered 8 hours ago

CertainPerformanceCertainPerformance

130k1717 gold badges8686 silver badges116116 bronze badges

answered 8 hours ago

CertainPerformanceCertainPerformance

130k1717 gold badges8686 silver badges116116 bronze badges

answered 8 hours ago

CertainPerformanceCertainPerformance

130k1717 gold badges8686 silver badges116116 bronze badges

1

If I understood the question well then this can be simplified even further by relying on sort initially...

Increment i to the next position after finding the pair and let the for loop increment it once again.

 function pairNumber(arr) 
 var sorted_arr = [...arr].sort(); // disallowing array mutation
 let cnt = 0;
 for (let i = 0; i < sorted_arr.length; i++) 
 if (sorted_arr[i + 1] === sorted_arr[i]) 
 cnt++;
 i = i + 1;
 

 
 return cnt;
 
 console.log(pairNumber([10, 10, 10, 10, 10, 20, 20, 20, 20, 30, 30, 50]))
 // 5 --> 2 pairs of 10, 2 pairs of 20, 1 pair of 30
 console.log(pairNumbers([10, 10, 10, 10, 20, 20, 20, 30, 50])) 
 // 3 --> 2 pairs of 10 one pair of 20

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Eugene SunicEugene Sunic

3,82844 gold badges3232 silver badges5050 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I believe sort will mutate the original array, so a side-effect would be that the array being passed in will be sorted after calling this function.
  
  – Matt Aft
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MattAft tnx, edited my answer
  
  – Eugene Sunic
  7 hours ago
  

  
  
  
  

add a comment | 

1

If I understood the question well then this can be simplified even further by relying on sort initially...

Increment i to the next position after finding the pair and let the for loop increment it once again.

 function pairNumber(arr) 
 var sorted_arr = [...arr].sort(); // disallowing array mutation
 let cnt = 0;
 for (let i = 0; i < sorted_arr.length; i++) 
 if (sorted_arr[i + 1] === sorted_arr[i]) 
 cnt++;
 i = i + 1;
 

 
 return cnt;
 
 console.log(pairNumber([10, 10, 10, 10, 10, 20, 20, 20, 20, 30, 30, 50]))
 // 5 --> 2 pairs of 10, 2 pairs of 20, 1 pair of 30
 console.log(pairNumbers([10, 10, 10, 10, 20, 20, 20, 30, 50])) 
 // 3 --> 2 pairs of 10 one pair of 20

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Eugene SunicEugene Sunic

3,82844 gold badges3232 silver badges5050 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I believe sort will mutate the original array, so a side-effect would be that the array being passed in will be sorted after calling this function.
  
  – Matt Aft
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MattAft tnx, edited my answer
  
  – Eugene Sunic
  7 hours ago
  

  
  
  
  

add a comment | 

If I understood the question well then this can be simplified even further by relying on sort initially...

Increment i to the next position after finding the pair and let the for loop increment it once again.

 function pairNumber(arr) 
 var sorted_arr = [...arr].sort(); // disallowing array mutation
 let cnt = 0;
 for (let i = 0; i < sorted_arr.length; i++) 
 if (sorted_arr[i + 1] === sorted_arr[i]) 
 cnt++;
 i = i + 1;
 

 
 return cnt;
 
 console.log(pairNumber([10, 10, 10, 10, 10, 20, 20, 20, 20, 30, 30, 50]))
 // 5 --> 2 pairs of 10, 2 pairs of 20, 1 pair of 30
 console.log(pairNumbers([10, 10, 10, 10, 20, 20, 20, 30, 50])) 
 // 3 --> 2 pairs of 10 one pair of 20

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Eugene SunicEugene Sunic

3,82844 gold badges3232 silver badges5050 bronze badges

 function pairNumber(arr) 
 var sorted_arr = [...arr].sort(); // disallowing array mutation
 let cnt = 0;
 for (let i = 0; i < sorted_arr.length; i++) 
 if (sorted_arr[i + 1] === sorted_arr[i]) 
 cnt++;
 i = i + 1;
 

 
 return cnt;
 
 console.log(pairNumber([10, 10, 10, 10, 10, 20, 20, 20, 20, 30, 30, 50]))
 // 5 --> 2 pairs of 10, 2 pairs of 20, 1 pair of 30
 console.log(pairNumbers([10, 10, 10, 10, 20, 20, 20, 30, 50])) 
 // 3 --> 2 pairs of 10 one pair of 20

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Eugene SunicEugene Sunic

3,82844 gold badges3232 silver badges5050 bronze badges

answered 7 hours ago

Eugene SunicEugene Sunic

3,82844 gold badges3232 silver badges5050 bronze badges

answered 7 hours ago

Eugene SunicEugene Sunic

3,82844 gold badges3232 silver badges5050 bronze badges