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Maximum and minimum of an array using minimum number of comparisons

Program for Square Root of Integer

Last Updated : 19 Jun, 2025

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 = 4
Output: 2
Explanation: The square root of 4 is 2.
Input: n = 11
Output: 3
Explanation: 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

Start 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));

Output

[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));

Output

[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));

Output

[Alternate Approach] Using Formula Used by Pocket Calculators - O(1) Time and O(1) Space

The idea is to use mathematical formula √n = e^{1/2 * log(n)}to compute the square root of an integer n.

Let's say square root of n is x:
=> x = √n
Squaring both the sides:
=> x²=n
Taking 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 = e^{1/2 * log(n)}
x is the square root of n:
So, √n = e^{1/2 * 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));

Output

Maximum and minimum of an array using minimum number of comparisons

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