Square root of a number by Repeated Subtraction method

Given an integer N, the task is to find its perfect square root by repeated subtraction only.
Examples:

Input: N = 25 
Output: 5
Input: N = 841 
Output: 29 

Babylonian Method and Binary Search Approach: Refer to Square root of an integer for the approaches based on Babylonian Method and Binary Search.
Repeated Subtraction Approach:
Follow the steps below to solve the problem: 

• Sum of the first N odd natural numbers is equal to N². 
  
• Based on the fact mentioned above, repetitive subtraction of odd numbers starting from 1, until N becomes 0 needs to be performed. 
  
• The count of odd numbers, used in this process, will give the square root of the number N. 
  

Illustration:
N = 81
Step 1: 81-1=80 
Step 2: 80-3=77 
Step 3: 77-5=72 
Step 4: 72-7=65 
Step 5: 65-9=56 
Step 6: 56-11=45 
Step 7: 45-13=32 
Step 8: 32-15=17 
Step 9: 17-17=0
Since, 9 odd numbers were used, hence the square root of 81 is 9. 

Below is the implementation of the above approach.

C++

// C++ implementation of
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the square
// root of the given number
int SquareRoot(int num)
{
    int count = 0;

    for (int n = 1; n <= num; n += 2) {

        // Subtract n-th odd number
        num = num - n;
        count += 1;
        if (num == 0)
            break;
    }

    // Return the result
    return count;
}

// Driver Code
int main()
{
    int N = 81;
    cout << SquareRoot(N);
}

Java

// Java implementation of
// the above approach
class GFG{

// Function to return the square
// root of the given number
public static int SquareRoot(int num)
{
    int count = 0;

    for(int n = 1; n <= num; n += 2)
    {

       // Subtract n-th odd number
       num = num - n;
       count += 1;
       if (num == 0)
           break;
    }

    // Return the result
    return count;
}

// Driver code
public static void main(String[] args)
{
    int N = 81;
    System.out.println(SquareRoot(N));
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 implementation of the
# above approach

# Function to return the square
# root of the given number
def SquareRoot(num):

    count = 0
    for n in range(1, num + 1, 2):

        # Subtract n-th odd number
        num = num - n
        count = count + 1
        if (num == 0):
            break

    # Return the result
    return count

# Driver Code
N = 81
print(SquareRoot(N))

# This code is contributed by Sanjit_Prasad

C#

// C# implementation of
// the above approach
using System;

class GFG{

// Function to return the square
// root of the given number
public static int SquareRoot(int num)
{
    int count = 0;

    for(int n = 1; n <= num; n += 2)
    {

        // Subtract n-th odd number
        num = num - n;
        count += 1;
        if (num == 0)
            break;
    }

    // Return the result
    return count;
}

// Driver code
public static void Main()
{
    int N = 81;

    Console.Write(SquareRoot(N));
}
}

// This code is contributed by chitranayal

Javascript

<script>

// Javascript implementation of
// the above approach

// Function to return the square
// root of the given number
function SquareRoot(num)
{
    let count = 0;

    for (let n = 1; n <= num; n += 2) {

        // Subtract n-th odd number
        num = num - n;
        count += 1;
        if (num == 0)
            break;
    }

    // Return the result
    return count;
}

// Driver Code

    let N = 81;
    document.write(SquareRoot(N));

// This code is contributed by Mayank Tyagi

</script>

Output:

9

Time Complexity: O(N) 
Auxiliary Space: O(1)