Número de veces que el número cuadrado perfecto más grande se puede restar de N

Dado un número N. En cada paso, reste el cuadrado perfecto más grande (≤ N) de N. Repita este paso mientras N > 0 . La tarea es contar el número de pasos que se pueden realizar. 

Ejemplos: 

Entrada: N = 85 
Salida: 2 
Primer paso, 85 – (9 * 9) = 4 
Segundo paso 4 – (2 * 2) = 0

Entrada: N = 114 
Salida: 4 
Primer paso, 114 – (10 * 10) = 14 
Segundo paso 14 – (3 * 3) = 5 
Tercer paso 5 – (2 * 2) = 1 
Cuarto paso 1 – (1 * 1 ) = 0 

Enfoque : reste iterativamente el cuadrado perfecto más grande (≤ N) de N mientras N > 0 y cuente el número de pasos. 
A continuación se muestra la implementación del enfoque anterior:  

// 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 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 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# 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 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
?>

<script>
 
// JavaScript implementation of the approach
 
// Function to return the count of steps
function countSteps(n)
{
     
    // Variable to store the count of steps
    let steps = 0;
 
    // Iterate while N > 0
    while (n)
    {
         
        // Get the largest perfect square
        // and subtract it from N
        let largest = Math.floor(Math.sqrt(n));
        n -= (largest * largest);
 
        // Increment steps
        steps++;
    }
 
    // Return the required count
    return steps;
}
 
// Driver code
 
let n = 85;
 
document.write(countSteps(n));
 
// This code is contributed by Manoj.
 
</script>

2

Complejidad de tiempo: O(1)

Espacio Auxiliar: O(1), ya que no se ha ocupado ningún espacio extra.