Maximum and minimum of an array using minimum number of comparisons

Program for Square Root of Integer

Last Updated : 14 Feb, 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
[Expected Approach] Using Binary Search - O(logn) Time and O(1) Space
[Alternate Approach] Using Built In functions - O(logn) 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

As, we know that square root of a positive integer is always greater than or equal to one. So we start iterating from 1 and calculate the square of each number. We continue the iteration until we reach to an integer whose square is greater than the given integer, then the integer just before it will be our answer.

C++

// C++ program to find the square root of 
// given integer using a loop

#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

// C program to find the square root of 
// given integer using a loop

#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

// Java program to find the square root of 
// given integer using a loop

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

# Python program to find the square root of 
# given integer using a loop

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#

// C# program to find the square root of 
// given integer using a loop

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

// JavaScript program to find the square root of 
// given integer using a loop

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

Time Complexity - O(sqrt(n))
Auxiliary Space - O(1)

[Expected Approach] Using Binary Search

The square root of an integer follows a monotonic pattern, because as we increase any number, it's square also increases. If the square of a number is greater than given integer, then square root will definitely exist before this number. Conversely, if the square of a number is less than or equal to n, then either this number is the square root or it lies after this number.
Therefore, we can use binary search to find the square root of n. Initial search space will be 1 to the given integer itself, because square root of any positive integer always exists within this range.

C++

// C++ program to find the square root of given integer
// using binary search

#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

// C program to find the square root of given integer
// using binary search

#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

// Java program to find the square root of given integer
// using binary search

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

# Python program to find the square root of given integer
# using binary search

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#

// C# program to find the square root of given integer
// using binary search

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

// JavaScript program to find the square root of given integer
// using binary search

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;
}

let n = 11;
console.log(floorSqrt(n));

Output

Time Complexity - O(logn)
Auxiliary Space - O(1)

[Alternate Approach] Using Built In functions

We can directly use built in functions to find square root of an integer.

C++

// C++ program to find the square root of given
// integer using built in square root function

#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

// C program to find the square root of given
// integer using built in square root function

#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

// Java program to find the square root of given
// integer using built in square root function

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

# Python program to find the square root of given
# integer using built in square root function

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#

// C# program to find the square root of given
// integer using built in square root function

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

// JavaScript program to find the square root of given
// integer using built in square root function

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;
}

let n = 11;
console.log(floorSqrt(n));

Output

Time Complexity - O(logn)
Auxiliary Space - O(1)

[Alternate Approach] Using Formula Used by Pocket Calculators

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

Let's say square root of n is x:
x = √n
Squaring both the sides:
x^{2 =}n
Taking log on both the sides:
=> ln(x²) = ln(n)
=> 2*ln(x) = ln(n)
=> ln(x) = 1/2 * ln(n)
To isolate x, exponentiate both sides with base e:
=> x = e^{1/2 * ln(n)}
x is the square root of n:
√n = e^{1/2 * ln(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++

// C++ program to find the square root of given integer
// using mathematical formula 

#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

// C program to find the square root of given integer
// using mathematical formula 

#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

// Java program to find the square root of given integer
// using mathematical formula 

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

# Python program to find the square root of given integer
# using mathematical formula 

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#

// C# program to find the square root of given integer
// using mathematical formula 

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

// JavaScript program to find the square root of given integer
// using mathematical formula 

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;
}

let n = 11;
console.log(floorSqrt(n));

Output

Time Complexity - O(1) Time
Auxiliary Space - O(1)

Maximum and minimum of an array using minimum number of comparisons

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