Square root of an integer

Difficulty Level : Easy
Last Updated : 19 Oct, 2021

Given an integer x, find it’s square root. If x is not a perfect square, then return floor(√x).

Examples :

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Input: x = 4
Output: 2
Explanation:  The square root of 4 is 2.

Input: x = 11
Output: 3
Explanation:  The square root of 11 lies in between
3 and 4 so floor of the square root is 3.

There can be many ways to solve this problem. For example Babylonian Method is one way.
Simple Approach: To find the floor of the square root, try with all-natural numbers starting from 1. Continue incrementing the number until the square of that number is greater than the given number.

Algorithm:
1. Create a variable (counter) i and take care of some base cases, i.e when the given number is 0 or 1.
2. Run a loop until i*i <= n , where n is the given number. Increment i by 1.
3. The floor of the square root of the number is i – 1
Implementation:

C++

// A C++ program to find floor(sqrt(x)
#include<bits/stdc++.h>
using namespace std;
 
// Returns floor of square root of x
int floorSqrt(int x)
{
    // Base cases
    if (x == 0 || x == 1)
    return x;
 
    // Starting from 1, try all numbers until
    // i*i is greater than or equal to x.
    int i = 1, result = 1;
    while (result <= x)
    {
      i++;
      result = i * i;
    }
    return i - 1;
}
 
// Driver program
int main()
{
    int x = 11;
    cout << floorSqrt(x) << endl;
    return 0;
}

Java

// A Java program to find floor(sqrt(x))
 
class GFG {
     
    // Returns floor of square root of x
    static int floorSqrt(int x)
    {
        // Base cases
        if (x == 0 || x == 1)
            return x;
 
        // Starting from 1, try all numbers until
        // i*i is greater than or equal to x.
        int i = 1, result = 1;
         
        while (result <= x) {
            i++;
            result = i * i;
        }
        return i - 1;
    }
 
    // Driver program
    public static void main(String[] args)
    {
        int x = 11;
        System.out.print(floorSqrt(x));
    }
}
 
// This code is contributed by Smitha Dinesh Semwal.

Python3

# Python3 program to find floor(sqrt(x)
 
# Returns floor of square root of x
def floorSqrt(x):
 
    # Base cases
    if (x == 0 or x == 1):
        return x
 
    # Starting from 1, try all numbers until
    # i*i is greater than or equal to x.
    i = 1; result = 1
    while (result <= x):
     
        i += 1
        result = i * i
     
    return i - 1
 
# Driver Code
x = 11
print(floorSqrt(x))
 
# This code is contributed by Smitha Dinesh Semwal.

C#

// A C# program to
// find floor(sqrt(x))
using System;
 
class GFG
{
    // Returns floor of
    // square root of x
    static int floorSqrt(int x)
    {
        // Base cases
        if (x == 0 || x == 1)
            return x;
 
        // Starting from 1, try all
        // numbers until i*i is
        // greater than or equal to x.
        int i = 1, result = 1;
         
        while (result <= x)
        {
            i++;
            result = i * i;
        }
        return i - 1;
    }
 
    // Driver Code
    static public void Main ()
    {
        int x = 11;
        Console.WriteLine(floorSqrt(x));
    }
}
 
// This code is contributed by ajit

PHP

<?php
// A PHP program to find floor(sqrt(x)
 
// Returns floor of square root of x
function floorSqrt($x)
{
    // Base cases
    if ($x == 0 || $x == 1)
    return $x;
 
    // Starting from 1, try all
    // numbers until i*i is
    // greater than or equal to x.
    $i = 1;
    $result = 1;
    while ($result <= $x)
    {
        $i++;
        $result = $i * $i;
    }
    return $i - 1;
}
 
// Driver Code
$x = 11;
echo floorSqrt($x), "\n";
 
// This code is contributed by ajit
?>

Javascript

<script>
 
// A Javascript program to find floor(sqrt(x)
 
// Returns floor of square root of x
function floorSqrt(x)
{
     
    // Base cases
    if (x == 0 || x == 1)
        return x;
 
    // Starting from 1, try all
    // numbers until i*i is
    // greater than or equal to x.
    let i = 1;
    let result = 1;
     
    while (result <= x)
    {
        i++;
        result = i * i;
    }
    return i - 1;
}
 
// Driver Code
let x = 11;
document.write(floorSqrt(x));
 
// This code is contributed by mohan
 
</script>

Output :

Complexity Analysis:
- Time Complexity: O(√ n).
  Only one traversal of the solution is needed, so the time complexity is O(√ n).
- Space Complexity: O(1).
  Constant extra space is needed.

Thanks Fattepur Mahesh for suggesting this solution.
Better Approach: The idea is to find the largest integer i whose square is less than or equal to the given number. The idea is to use Binary Search to solve the problem. The values of i * i is monotonically increasing, so the problem can be solved using binary search.

Algorithm:
1. Take care of some base cases, i.e when the given number is 0 or 1.
2. Create some variables, lowerbound l = 0, upperbound r = n, where n is the given number, mid and ans to store the answer.
3. Run a loop until l <= r , the search space vanishes
4. Check if the square of mid (mid = (l + r)/2 ) is less than or equal to n, If yes then search for a larger value in second half of search space, i.e l = mid + 1, update ans = mid
5. Else if the square of mid is more than n then search for a smaller value in first half of search space, i.e r = mid – 1
6. Print the value of answer ( ans)
Implementation:

C++

#include <iostream>
 
using namespace std;
int floorSqrt(int x)
{
    // Base cases
    if (x == 0 || x == 1)
        return x;
 
    // Do Binary Search for floor(sqrt(x))
    int start = 1, end = x/2, ans;
    while (start <= end) {
        int mid = (start + end) / 2;
 
        // If x is a perfect square
        int sqr = mid * mid;
        if (sqr == x)
            return mid;
 
        // Since we need floor, we update answer when
        // mid*mid is smaller than x, and move closer to
        // sqrt(x)
 
        /*
           if(mid*mid<=x)
                   {
                           start = mid+1;
                           ans = mid;
                   }
            Here basically if we multiply mid with itself so
           there will be integer overflow which will throw
           tle for larger input so to overcome this
           situation we can use long or we can just divide
            the number by mid which is same as checking
           mid*mid < x
 
           */
        if (sqr <= x) {
            start = mid + 1;
            ans = mid;
        }
        else // If mid*mid is greater than x
            end = mid - 1;
    }
    return ans;
}
 
// Driver program
int main()
{
    int x = 20221;
    cout << floorSqrt(x) << endl;
    return 0;
}

Java

// A Java program to find floor(sqrt(x)
public class Test
{
    public static int floorSqrt(int x)
    {
        // Base Cases
        if (x == 0 || x == 1)
            return x;
 
 
        // Do Binary Search for floor(sqrt(x))
        long start = 1, end = x, ans=0;
        while (start <= end)
        {
            int mid = (start + end) / 2;
 
            // If x is a perfect square
            if (mid*mid == x)
                return (int)mid;
 
            // Since we need floor, we update answer when mid*mid is
            // smaller than x, and move closer to sqrt(x)
            if (mid*mid < x)
            {
                start = mid + 1;
                ans = mid;
            }
            else   // If mid*mid is greater than x
                end = mid-1;
        }
        return (int)ans;
    }
 
    // Driver Method
    public static void main(String args[])
    {
        int x = 11;
        System.out.println(floorSqrt(x));
    }
}
// Contributed by InnerPeace

Python3

# Python 3 program to find floor(sqrt(x)
 
# Returns floor of square root of x        
def floorSqrt(x) :
 
    # Base cases
    if (x == 0 or x == 1) :
        return x
  
    # Do Binary Search for floor(sqrt(x))
    start = 1
    end = x  
    while (start <= end) :
        mid = (start + end) // 2
         
        # If x is a perfect square
        if (mid*mid == x) :
            return mid
             
        # Since we need floor, we update
        # answer when mid*mid is smaller
        # than x, and move closer to sqrt(x)
        if (mid * mid < x) :
            start = mid + 1
            ans = mid
             
        else :
             
            # If mid*mid is greater than x
            end = mid-1
             
    return ans
 
# driver code   
x = 11
print(floorSqrt(x))
     
# This code is contributed by Nikita Tiwari.

C#

// A C# program to
// find floor(sqrt(x)
using System;
 
class GFG
{
    public static int floorSqrt(int x)
    {
        // Base Cases
        if (x == 0 || x == 1)
            return x;
 
        // Do Binary Search
        // for floor(sqrt(x))
        int start = 1, end = x, ans = 0;
        while (start <= end)
        {
            int mid = (start + end) / 2;
 
            // If x is a
            // perfect square
            if (mid * mid == x)
                return mid;
 
            // Since we need floor, we
            // update answer when mid *
            // mid is smaller than x,
            // and move closer to sqrt(x)
            if (mid * mid < x)
            {
                start = mid + 1;
                ans = mid;
            }
             
            // If mid*mid is
            // greater than x
            else
                end = mid-1;
        }
        return ans;
    }
 
    // Driver Code
    static public void Main ()
    {
        int x = 11;
        Console.WriteLine(floorSqrt(x));
    }
}
 
// This code is Contributed by m_kit

PHP

<?php
// A PHP program to find floor(sqrt(x)
 
// Returns floor of
// square root of x        
function floorSqrt($x)
{
    // Base cases
    if ($x == 0 || $x == 1)
    return $x;
 
    // Do Binary Search
    // for floor(sqrt(x))
    $start = 1; $end = $x; $ans;
    while ($start <= $end)
    {
        $mid = ($start + $end) / 2;
 
        // If x is a perfect square
        if ($mid * $mid == $x)
            return $mid;
 
        // Since we need floor, we update
        // answer when mid*mid is  smaller
        // than x, and move closer to sqrt(x)
        if ($mid * $mid < $x)
        {
            $start = $mid + 1;
            $ans = $mid;
        }
         
        // If mid*mid is
        // greater than x
        else
            $end = $mid-1;    
    }
    return $ans;
}
 
// Driver Code
$x = 11;
echo floorSqrt($x), "\n";
 
// This code is contributed by ajit
?>

Javascript

<script>
    // A Javascript program to find floor(sqrt(x)
 
// Returns floor of
// square root of x        
function floorSqrt(x)
{
    // Base cases
    if (x == 0 || x == 1)
    return x;
 
    // Do Binary Search
    // for floor(sqrt(x))
    let start = 1;
    let end = x;
    let ans;
    while (start <= end)
    {
        let mid = (start + end) / 2;
 
        // If x is a perfect square
        if (mid * mid == x)
            return mid;
 
        // Since we need floor, we update
        // answer when mid*mid is  smaller
        // than x, and move closer to sqrt(x)
        if (mid * mid < x)
        {
            start = mid + 1;
            ans = mid;
        }
         
        // If mid*mid is
        // greater than x
        else
            end = mid-1;    
    }
    return ans;
}
 
// Driver Code
let x = 11;
document.write(floorSqrt(x) +  "<br>");
 
// This code is contributed by _saurabh_jaiswal
</script>

Output :

Complexity Analysis:
- Time complexity: O(log n).
  The time complexity of binary search is O(log n).
- Space Complexity: O(1).
  Constant extra space is needed.

Thanks to Gaurav Ahirwar for suggesting above method.
Note: The Binary Search can be further optimized to start with ‘start’ = 0 and ‘end’ = x/2. Floor of square root of x cannot be more than x/2 when x > 1.
Thanks to vinit for suggesting above optimization.
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Square root of an integer

C++

Java

Python3

C#

PHP

Javascript

C++

Java

Python3

C#

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