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 numberint 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 Codeint 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 numberdef 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 CodeN = 81print(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)
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