Number of times the largest perfect square number can be subtracted from N
Given a number N. At every step, subtract the largest perfect square( ≤ N) from N. Repeat this step while N > 0. The task is to count the number of steps that can be performed.
Examples:
Input: N = 85
Output: 2
First step, 85 – (9 * 9) = 4
Second step 4 – (2 * 2) = 0Input: N = 114
Output: 4
First step, 114 – (10 * 10) = 14
Second step 14 – (3 * 3) = 5
Third step 5 – (2 * 2) = 1
Fourth step 1 – (1 * 1) = 0
Approach: Iteratively subtract the largest perfect square (≤ N) from N while N > 0 and count the number of steps.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the count of steps int countSteps(int n) { // Variable to store the count of steps int steps = 0; // Iterate while N > 0 while (n) { // Get the largest perfect square // and subtract it from N int largest = sqrt(n); n -= (largest * largest); // Increment steps steps++; } // Return the required count return steps; } // Driver code int main() { int n = 85; cout << countSteps(n); return 0; } |
Java
// Java implementation of the approach import java.lang.Math; public class GfG{ // Function to return the count of steps static int countSteps(int n) { // Variable to store the count of steps int steps = 0; // Iterate while N > 0 while (n > 0) { // Get the largest perfect square // and subtract it from N int largest = (int)Math.sqrt(n); n -= (largest * largest); // Increment steps steps++; } // Return the required count return steps; } public static void main(String []args){ int n = 85; System.out.println(countSteps(n)); } } // This code is contributed by Rituraj Jain |
Python3
# Python3 implementation of the approach from math import sqrt # Function to return the count of steps def countSteps(n) : # Variable to store the count of steps steps = 0; # Iterate while N > 0 while (n) : # Get the largest perfect square # and subtract it from N largest = int(sqrt(n)); n -= (largest * largest); # Increment steps steps += 1; # Return the required count return steps; # Driver code if __name__ == "__main__" : n = 85; print(countSteps(n)); # This code is contributed by Ryuga |
C#
// C# implementation of the approach using System; class GfG { // Function to return the count of steps static int countSteps(int n) { // Variable to store the count of steps int steps = 0; // Iterate while N > 0 while (n > 0) { // Get the largest perfect square // and subtract it from N int largest = (int)Math.Sqrt(n); n -= (largest * largest); // Increment steps steps++; } // Return the required count return steps; } // Driver code public static void Main() { int n = 85; Console.WriteLine(countSteps(n)); } } // This code is contributed by Code_Mech. |
PHP
<?php // PHP implementation of the approach // Function to return the count of steps function countSteps($n) { // Variable to store the count of steps $steps = 0; // Iterate while N > 0 while ($n) { // Get the largest perfect square // and subtract it from N $largest = (int)sqrt($n); $n -= ($largest * $largest); // Increment steps $steps++; } // Return the required count return $steps; } // Driver code $n = 85; echo countSteps($n); // This code is contributed // by Akanksha Rai ?> |
Output:
2
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