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The requirement is to find the square root of a positive integer using binary search and the math property that square root of a number n is between 0 and n/2, and the required answer is "floored", meaning mySqrt(8) is to return 2.

Please comment on the efficiency, and if possible, the loop invariants in terms of correctness:

class Solution(object):
 def mySqrt(self, x):
 """
 :type x: int
 :rtype: int

 Loop invariant: 

 The answer is always in the range [low, high] inclusive,
 except possibly:

 1) low == high == mid, and mid * mid == x and 
 any of low, high, or mid can be returned as the answer.

 2) if there is no exact answer and the floor is to be 
 returned, then low > high by 1. Since sq != x,
 so either low or high is set inside the loop.

 If low gets set and gets pushed up, it is pushed up too much. 
 So when low > high by 1, low - 1 is the answer and it is the same 
 as high, because low > high by 1.

 If high gets set and gets pushed down, high can be
 the correct answer. When low > high, it is by 1,
 and high is the correct floor value to be returned. 
 (since there is no perfect square root and the floor is required)

 0 <= low <= answer <= high <= n//2 + 1

 where answer is floor(sqrt(x)) to be found,
 except if low > high and the loop will exit.

 Each loop iteration always makes the range smaller.

 If the range is empty at the end, low is > high by 1, and high is 
 the correct floored value, and low is the ceiling value, so high is returned.
 """

 low = 0;
 high = x//2 + 1;

 while (low <= high):
 mid = low + (high - low) // 2;
 sq = mid * mid;
 if (sq == x):
 return mid;
 elif (sq > x):
 high = mid - 1; # sq exceeds target, so mid cannot be the answer floored, but when high is set to mid - 1, then it can be the answer
 else:
 low = mid + 1; # (here sq < x, and mid might be the answer floored, so when low is set to mid + 1, then low might be too big, while high is correct)

 return high;

• Your comments on the elif / else part are too long to be just after the statements




• Don't use semicolons (;) in Python.




• This is a refactored version of the code

import math

class Solution(object):
 def mySqrt(self, x):
 """
 :type x: int
 :rtype: int

 Returns floor(sqrt(x))
 """
 low = 0
 high = x//2 + 1

 """
 It is proved that 0 <= sqrt(x) <= x/2, so
 we run a dichotomic in [0, x/2] to find floor(sqrt(x))

 Loop analysis:
 * Initialization: low = 0 and high = x/2 + 1 
 * Termination: |high-low| is reduced each iteration,
 as shown in lines high = mid - 1 and low = mid + 1.
 * Invariant: low <= floor(sqrt(x)) <= high.
 Let mid be (low + high)/2.
 - If mid^2 <= x < (mid+1)^2,
 then mid is floor(sqrt(x)) and just return it.
 - If mid^2 > x, search for values smaller than mid.
 - Otherwise, if mid^2 < x, search within higher values.
 """
 while (low <= high):
 mid = (low + high) // 2
 sq = mid * mid
 sq_next = (mid+1)*(mid+1)
 if (sq <= x < sq_next):
 return mid
 elif (sq > x):
 high = mid - 1
 else:
 low = mid + 1

for i in range(1, 26):
 assert(math.floor(math.sqrt(i)) == Solution().mySqrt(i))