0069 - Sqrt(x) (Easy)

Problem Link

https://leetcode.com/problems/sqrtx/

Problem Statement

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 <= 2^31 - 1

Approach 1: Binary Search

Prerequisite

• Binary Search

C++

Written by @wingkwong

class Solution {
public:
    int mySqrt(int x) {
        // init possible range
        long long l = 0, r = x;
        while (l < r) {
            long long m = l + (r - l + 1) / 2;
            // exclude m
            if (x < m * m) r = m - 1;
            // include m
            else l = m;
        }
        return l;
    }
};

Python

Written by @radojicic23

class Solution:
    def mySqrt(self, x: int) -> int:
        left = 0
        right = x
        while left <= right:
            mid = (left + right) // 2
            if x < mid * mid:
                right = mid - 1
            elif x > mid * mid:
                left = mid + 1
            else:
                return mid
        return right

JavaScript

Written by @radojicic23

/**
 * @param {number} x
 * @return {number}
 */
var mySqrt = function (x) {
  let left = 0;
  let right = x;
  while (left <= right) {
    let mid = Math.floor((left + right) / 2);
    if (x < mid * mid) {
      right = mid - 1;
    } else if (x > mid * mid) {
      left = mid + 1;
    } else {
      return mid;
    }
  }
  return right;
};