Square root of a number by Repeated Subtraction method

Namachila November 19, 2025

0 0 3 minutes read

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:

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

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