Square root of an integer

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

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My Personal Notes

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

C++

Java

Python3

C#

PHP

C/C++

Java

Python3

C#

PHP

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