Last updated on October 25th, 2024 at 10:26 pm
Here, We see Sqrt(x) LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.
List of all LeetCode Solution
Topics
Binary Search, Math
Companies
Apple, Bloomberg, Facebook
Level of Question
Easy
Sqrt(x) LeetCode Solution
Table of Contents
Problem Statement
Given a non-negative integer x
, return the square root of x
rounded down to the nearest integer. The returned integer should be non-negative as well.
You must not use any built-in exponent function or operator.
- For example, do not use
pow(x, 0.5)
in c++ orx ** 0.5
in python.
Example 1: Input: x = 4 Output: 2 Explanation: The square root of 4 is 2, so we return 2. Example 2: Input: x = 8 Output: 2 Explanation: The square root of 8 is 2.82842..., and since we round it down to the nearest integer, 2 is returned.
1. Sqrt(x) Leetcode Solution C++
class Solution { public: int mySqrt(int x) { long long s=0,e=INT_MAX,ans=0; while(s<=e){ long long m=s+(e-s)/2; if(m*m<=x){ ans=m; s=m+1; } else e=m-1; } return ans; } };
2. Sqrt(x) Leetcode Solution Java
class Solution { public int mySqrt(int x) { long r = x; while (r*r > x) r = (r + x/r) / 2; return (int) r; } }
3. Sqrt(x) Leetcode Solution JavaScript
var mySqrt = function(x) { r = x; while (r*r > x) r = ((r + x/r) / 2) | 0; return r; };
4. Sqrt(x) Solution Python
class Solution(object): def mySqrt(self, x): r = x while r*r > x: r = (r + x/r) / 2 return r