Square root of a number by Repeated Subtraction method
The Repeated Subtraction method for finding the square root of a number is an interesting approach. It involves repeatedly subtracting consecutive odd numbers from the number until you reach zero or a negative number. The number of times you can subtract is the approximate square root of the original number.
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++ 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 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 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# 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
<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)