Program for Square Root of Integer Last Updated : 19 Jun, 2025 Comments Improve Suggest changes Like Article Like Report Try it on GfG Practice Given a positive integer n, find its square root. If n is not a perfect square, then return floor of √n.Examples : Input: n = 4Output: 2Explanation: The square root of 4 is 2.Input: n = 11Output: 3Explanation: The square root of 11 lies in between 3 and 4 so floor of the square root is 3.Table of Content[Naive Approach] Using a loop - O(sqrt(n)) Time and O(1) Space[Expected Approach] Using Binary Search - O(log(n)) Time and O(1) Space [Alternate Approach] Using Built In functions - O(log(n)) Time and O(1) Space[Alternate Approach] Using Formula Used by Pocket Calculators - O(1) Time and O(1) Space[Naive Approach] Using a loop - O(sqrt(n)) Time and O(1) SpaceStart from 1 and square each number until the square exceeds the given number. The last number whose square is less than or equal to n is the answer. C++ #include <iostream> using namespace std; int floorSqrt(int n) { // Start iteration from 1 until the // square of a number exceeds n int res = 1; while(res*res <= n){ res++; } // return the largest integer whose // square is less than or equal to n return res - 1; } int main() { int n = 11; cout << floorSqrt(n); return 0; } C #include <stdio.h> int floorSqrt(int n) { // Start iteration from 1 until the // square of a number exceeds n int res = 1; while (res * res <= n) { res++; } // return the largest integer whose // square is less than or equal to n return res - 1; } int main() { int n = 11; printf("%d", floorSqrt(n)); return 0; } Java class GfG { static int floorSqrt(int n) { // Start iteration from 1 until the // square of a number exceeds n int res = 1; while (res * res <= n) { res++; } // return the largest integer whose // square is less than or equal to n return res - 1; } public static void main(String[] args) { int n = 11; System.out.println(floorSqrt(n)); } } Python def floorSqrt(n): # Start iteration from 1 until the # square of a number exceeds n res = 1 while res * res <= n: res += 1 # return the largest integer whose # square is less than or equal to n return res - 1 if __name__ == "__main__": n = 11 print(floorSqrt(n)) C# using System; class GfG { static int floorSqrt(int n) { // Start iteration from 1 until the // square of a number exceeds n int res = 1; while (res * res <= n) { res++; } // return the largest integer whose // square is less than or equal to n return res - 1; } static void Main() { int n = 11; Console.WriteLine(floorSqrt(n)); } } JavaScript function floorSqrt(n) { // Start iteration from 1 until the // square of a number exceeds n let res = 1; while (res * res <= n) { res++; } // return the largest integer whose // square is less than or equal to n return res - 1; } // Driver Code let n = 11; console.log(floorSqrt(n)); Output3[Expected Approach] Using Binary Search - O(log(n)) Time and O(1) Space The square of a number increases as the number increases, so the square root of n must lie in a sorted (monotonic) range.If a number's square is more than n, the square root must be smaller.If it's less than or equal to n, the square root could be that number or greater.Because of this pattern, we can apply binary search in the range 1 to n to efficiently find the square root. C++ #include <iostream> using namespace std; int floorSqrt(int n) { // Initial search space int lo = 1, hi = n; int res = 1; while(lo <= hi) { int mid = lo + (hi - lo)/2; // If square of mid is less than or equal to n // update the result and search in upper half if(mid*mid <= n) { res = mid; lo = mid + 1; } // If square of mid exceeds n, // search in the lower half else { hi = mid - 1; } } return res; } int main() { int n = 11; cout << floorSqrt(n); return 0; } C #include <stdio.h> int floorSqrt(int n) { // Initial search space int lo = 1, hi = n; int res = 1; while (lo <= hi) { int mid = lo + (hi - lo) / 2; // If square of mid is less than or equal to n // update the result and search in upper half if (mid * mid <= n) { res = mid; lo = mid + 1; } // If square of mid exceeds n, // search in the lower half else { hi = mid - 1; } } return res; } int main() { int n = 11; printf("%d", floorSqrt(n)); return 0; } Java class GfG { static int floorSqrt(int n) { // Initial search space int lo = 1, hi = n; int res = 1; while (lo <= hi) { int mid = lo + (hi - lo) / 2; // If square of mid is less than or equal to n // update the result and search in upper half if (mid * mid <= n){ res = mid; lo = mid + 1; } // If square of mid exceeds n, // search in the lower half else { hi = mid - 1; } } return res; } public static void main(String[] args) { int n = 11; System.out.println(floorSqrt(n)); } } Python def floorSqrt(n): # Initial search space lo = 1 hi = n res = 1 while lo <= hi: mid = lo + (hi - lo) // 2 # If square of mid is less than or equal to n # update the result and search in upper half if mid * mid <= n: res = mid lo = mid + 1 # If square of mid exceeds n, # search in the lower half else: hi = mid - 1 return res if __name__ == "__main__": n = 11 print(floorSqrt(n)) C# using System; class GfG { static int floorSqrt(int n) { // Initial search space int lo = 1, hi = n; int res = 1; while (lo <= hi) { int mid = lo + (hi - lo) / 2; // If square of mid is less than or equal to n // update the result and search in upper half if (mid * mid <= n) { res = mid; lo = mid + 1; } // If square of mid exceeds n, // search in the lower half else { hi = mid - 1; } } return res; } static void Main() { int n = 11; Console.WriteLine(floorSqrt(n)); } } JavaScript function floorSqrt(n) { // Initial search space let lo = 1, hi = n; let res = 1; while (lo <= hi) { let mid = lo + Math.floor((hi - lo) / 2); // If square of mid is less than or equal to n // update the result and search in upper half if (mid * mid <= n) { res = mid; lo = mid + 1; } // If square of mid exceeds n, // search in the lower half else { hi = mid - 1; } } return res; } // Driver Code let n = 11; console.log(floorSqrt(n)); Output3[Alternate Approach] Using Built In functions - O(log(n)) Time and O(1) Space We can directly use built in functions to find square root of an integer. C++ #include <iostream> #include <cmath> using namespace std; int floorSqrt(int n) { // Square root using sqrt function, it returns // the double value, which is casted to integer int res = sqrt(n); return res; } int main() { int n = 11; cout << floorSqrt(n); return 0; } C #include <stdio.h> #include <math.h> int floorSqrt(int n) { // Square root using sqrt function, it returns // the double value, which is casted to integer int res = sqrt(n); return res; } int main() { int n = 11; printf("%d", floorSqrt(n)); return 0; } Java class GfG { static int floorSqrt(int n) { // Square root using sqrt function, it returns // the double value, which is casted to integer int res = (int)Math.sqrt(n); return res; } public static void main(String[] args) { int n = 11; System.out.println(floorSqrt(n)); } } Python import math def floorSqrt(n): # Square root using sqrt function, it returns # the double value, which is casted to integer res = int(math.sqrt(n)) return res if __name__ == "__main__": n = 11 print(floorSqrt(n)) C# using System; class GfG { static int floorSqrt(int n) { // Square root using sqrt function, it returns // the double value, which is casted to integer int res = (int)Math.Sqrt(n); return res; } static void Main() { int n = 11; Console.WriteLine(floorSqrt(n)); } } JavaScript function floorSqrt(n) { // Square root using sqrt function, it returns // the double value, which is casted to integer let res = Math.floor(Math.sqrt(n)); return res; } // Driver Code let n = 11; console.log(floorSqrt(n)); Output3[Alternate Approach] Using Formula Used by Pocket Calculators - O(1) Time and O(1) SpaceThe idea is to use mathematical formula √n = e1/2 * log(n) to compute the square root of an integer n. Let's say square root of n is x: => x = √nSquaring both the sides: => x2 =nTaking log on both the sides:=> log(x2) = log(n) => 2*log(x) = log(n)=> log(x) = 1/2 * log(n)To isolate x, exponentiate both sides with base e:=> x = e1/2 * log(n)x is the square root of n: So, √n = e1/2 * log(n)Because of the way computations are done in computers in case of decimals, the result from the expression may be slightly less than the actual square root. Therefore, we will also consider the next integer after the calculated result as a potential answer. C++ #include <iostream> #include <cmath> using namespace std; int floorSqrt(int n) { // Calculating square root using mathematical formula int res = exp(0.5 * log(n)); // If square of res + 1 is less than or equal to n // then, it will be our answer if ((res + 1) * (res + 1) <= n) { res++; } return res; } int main() { int n = 11; cout << floorSqrt(n); return 0; } C #include <stdio.h> #include <math.h> int floorSqrt(int n) { // Calculating square root using mathematical formula int res = exp(0.5 * log(n)); // If square of res + 1 is less than or equal to n // then, it will be our answer if ((res + 1) * (res + 1) <= n) { res++; } return res; } int main() { int n = 11; printf("%d", floorSqrt(n)); return 0; } Java class GfG { static int floorSqrt(int n) { // Calculating square root using mathematical formula int res = (int)Math.exp(0.5 * Math.log(n)); // If square of res + 1 is less than or equal to n // then, it will be our answer if ((res + 1) * (res + 1) <= n) { res++; } return res; } public static void main(String[] args) { int n = 11; System.out.println(floorSqrt(n)); } } Python import math def floorSqrt(n): # Calculating square root using mathematical formula res = int(math.exp(0.5 * math.log(n))) # If square of res + 1 is less than or equal to n # then, it will be our answer if (res + 1) ** 2 <= n: res += 1 return res if __name__ == "__main__": n = 11 print(floorSqrt(n)) C# using System; class GfG { static int floorSqrt(int n) { // Calculating square root using mathematical formula int res = (int)Math.Exp(0.5 * Math.Log(n)); // If square of res + 1 is less than or equal to n // then, it will be our answer if ((long)(res + 1) * (res + 1) <= n) { res++; } return res; } static void Main() { int n = 11; Console.WriteLine(floorSqrt(n)); } } JavaScript function floorSqrt(n) { // Calculating square root using mathematical formula let res = Math.floor(Math.exp(0.5 * Math.log(n))); // If square of res + 1 is less than or equal to n // then, it will be our answer if ((res + 1) * (res + 1) <= n) { res++; } return res; } // Driver Code let n = 11; console.log(floorSqrt(n)); Output3 Comment More infoAdvertise with us Next Article Maximum and minimum of an array using minimum number of comparisons kartik Follow Improve Article Tags : Divide and Conquer Mathematical DSA Microsoft Amazon Snapdeal Accolite Ola Cabs Binary Search +5 More Practice Tags : AccoliteAmazonMicrosoftOla CabsSnapdealBinary SearchDivide and ConquerMathematical +4 More Similar Reads Divide and Conquer Algorithm Divide and Conquer algorithm is a problem-solving strategy that involves. 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