Program for Square Root of Integer
Last Updated :
19 Jun, 2025
Improve
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
nis the answer.
#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;
}
#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;
}
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));
}
}
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))
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));
}
}
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
3
[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
nmust lie in a sorted (monotonic) range.
If a number's square is more thann, the square root must be smaller.
If it's less than or equal ton, the square root could be that number or greater.
Because of this pattern, we can apply binary search in the range1tonto efficiently find the square root.
#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;
}
#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;
}
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));
}
}
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))
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));
}
}
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
3
[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.
#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;
}
#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;
}
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));
}
}
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))
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));
}
}
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
3
[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.
#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;
}
#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;
}
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));
}
}
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))
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));
}
}
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
3
