**LeetCode Problem | LeetCode Problems For Beginners | LeetCode Problems & Solutions | Improve Problem Solving Skills | LeetCode Problems Java | LeetCode Solutions in C++**

Hello **Programmers/Coders,** Today we are going to share ** solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python**. At Each Problem with Successful submission with

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In this post, you will find the solution for the **Sqrt(x) in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Sqrt(x)– LeetCode Problem

Sqrt(x)– LeetCode Problem

**Problem:**

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 = 4Output:2

**Example 2:**

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

**Constraints:**

`0 <= x <= 2`

^{31}- 1

Sqrt(x)– LeetCode Solutions

Sqrt(x) in C++:

class Solution { public: int mySqrt(int x) { unsigned l = 1; unsigned r = x + 1u; while (l < r) { const unsigned m = l + (r - l) / 2; if (m > x / m) r = m; else l = m + 1; } // l: smallest number s.t. l * l > x return l - 1; } };

Sqrt(x) in Java:

class Solution { public int mySqrt(long x) { long l = 1; long r = x + 1; while (l < r) { final long m = l + (r - l) / 2; if (m > x / m) r = m; else l = m + 1; } // l: smallest number s.t. l * l > x return (int) l - 1; } }

Sqrt(x) in Python:

class Solution: def mySqrt(self, x: int) -> int: l = 1 r = x + 1 while l < r: m = (l + r) // 2 if m * m > x: r = m else: l = m + 1 # l: smallest number s.t. l * l > x return l - 1