diff --git a/LeetCode/Sqrt(x)/C++/Screenshot.png b/LeetCode/Sqrt(x)/C++/Screenshot.png
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diff --git a/LeetCode/Sqrt(x)/C++/main.cpp b/LeetCode/Sqrt(x)/C++/main.cpp
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+#include <iostream>
+using namespace std;
+
+class Solution {
+public:
+    
+    long long int binarySearch(int n){
+        int s=0;
+        int e=n;
+        long long int mid = s+(e-s)/2;
+        long long int ans = -1;
+        
+        while(s<=e){
+            long long int square = mid*mid;
+            if(square == n){
+                return mid;
+            }
+            if(square < n){
+                ans = mid;
+                s=mid+1;
+            }
+            else{
+                e=mid-1;
+            }
+            mid = s+(e-s)/2;
+        }
+        return ans;
+    }
+    int mySqrt(int x) {
+        return(binarySearch(x));
+    }
+};
+
+int main(){
+
+    Solution solution;
+
+    int num = 8;
+
+    long long result = solution.mySqrt(num);
+
+    if (result)
+    {
+        cout<<result;
+    }
+    else
+    {
+        cout << "No solution found." << endl;
+    }
+
+    return 0;
+}
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diff --git a/LeetCode/Sqrt(x)/PROBLEM.md b/LeetCode/Sqrt(x)/PROBLEM.md
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+# [69. Sqrt(x)](https://leetcode.com/problems/sqrtx/description/)
+
+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++ or x ** 0.5 in python.
+
+
+ 
+```example 1
+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.
+```
+
+```constrants
+Constraints:
+
+0 <= x <= 231 - 1
+```
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