Square root of an integer

Difficulty Level : Easy
Last Updated : 30 Mar, 2020

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.

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 
?> 

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/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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

Related Articles

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C/C++

Java

Python3

C#

PHP