Square root of an integer

Difficulty Level : Easy
Last Updated : 05 Apr, 2021

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

Examples :

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;
 
    // Staring 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;
 
        // Staring 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
 
    # Staring 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;
 
        // Staring 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;
 
    // Staring 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;
 
    // Staring 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 less 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++

// 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;
 
    // Do Binary Search for floor(sqrt(x))
    int start = 1, end = x, ans;
    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;
        }
        else // If mid*mid is greater than x
            end = mid-1;        
    }
    return ans;
}
 
// Driver program
int main()
{    
    int x = 11;
    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))
        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;
            }
            else   // If mid*mid is greater than x
                end = mid-1;
        }
        return 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
?>

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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