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++

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// 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);
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output: 

9

 

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

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