Program for Square Root of Integer (original) (raw)

Last Updated : 12 Aug, 2025

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.

Try It Yourself

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) 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 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));

[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 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));

[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 #include 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);
  // use sqrtl for big numbers
  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));

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

The 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 = √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 = 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 #include 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));