Sqrt(x) LeetCode Solution

Problem – Sqrt(x)

Given a non-negative integer x, compute and return the square root of x.

Since the return type is an integer, the decimal digits are truncated, and only the integer part of the result is returned.

Note: You are not allowed to use any built-in exponent function or operator, such as pow(x, 0.5) or x ** 0.5.

Example 1:

Input: x = 4
Output: 2

Example 2:

Input: x = 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since the decimal part is truncated, 2 is returned.

Constraints:

  • 0 <= x <= 231 - 1

Sqrt(x) LeetCode Solution in Python

    r = x
    while r*r > x:
        r = (r + x/r) / 2
    return r

Sqrt(x) LeetCode Solution in Java

public int mySqrt(int x) {
	if (x == 0) return 0;
	int start = 1, end = x;
	while (start < end) { 
		int mid = start + (end - start) / 2;
		if (mid <= x / mid && (mid + 1) > x / (mid + 1))// Found the result
			return mid; 
		else if (mid > x / mid)// Keep checking the left part
			end = mid;
		else
			start = mid + 1;// Keep checking the right part
	}
	return start;
}

Sqrt(x) LeetCode Solution in C++

long long s=0, e=x, ans, mid;   //long long due to some of test cases overflows integer limit.
        while(s<=e){             
            mid=(s+e)/2;
            if(mid*mid==x) return mid;     //if the 'mid' value ever gives the result, we simply return it.
            else if(mid*mid<x){             
                s=mid+1;        //if 'mid' value encounterted gives lower result, we simply discard all the values lower than mid.
                ans=mid;        //an extra pointer 'ans' is maintained to keep track of only lowest 'mid' value. 
            }
            else e=mid-1;       //if 'mid' value encountered gives greater result, we simply discard all the values greater than mid. 
        }
        return ans;   
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