# 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) = 0

Input: 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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

 ` 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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