# Count the minimum steps to reach 0 from the given integer N

Given **two integers N and K** where K represents the number of jumps that we are allowed to make directly from N reducing N to N – K, our task is to count minimum steps to reach 0 following the given operations:

- We can jump by a amount of K from N that is N = N – K
- Decrement N by 1 that is N = N -1.

**Examples:**

Input:N = 11, K = 4Output:5Explanation:

For the given value N we can perform the operation in the given sequence: 11 -> 7 -> 3 -> 2 -> 1 -> 0

Input:N = 6, K = 3Output:2Explanation:

For the given value N we can perform the operation in the given sequence: 6 -> 3 -> 0.

**Approach:**

To solve the problem mentioned above we know that it will take **N / K** steps to directly jump from value N to least divisible value with K and **N % K** steps to decrement it by 1 such as to reduce the count to 0. So the total number of steps required to **reach 0 from N will be **

(N / K) + (N % K)

Below is the implementation of the above approach:

## C++

`// C++ program to Count the minimum steps` `// to reach 0 from the given integer N` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function returns min step` `// to reach 0 from N` `int` `getMinSteps(` `int` `n, ` `int` `jump)` `{` ` ` `// Direct possible` ` ` `// reduction of value N` ` ` `int` `quotient = n / jump;` ` ` `// Remaining steps needs` ` ` `// to be reduced by 1` ` ` `int` `remainder = n % jump;` ` ` `// Summation of both the values` ` ` `int` `steps = quotient + remainder;` ` ` `// Return the final answer` ` ` `return` `steps;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `N = 6, K = 3;` ` ` `cout << getMinSteps(N, K);` ` ` `return` `0;` `}` |

## Java

`// Java program to count the minimum steps` `// to reach 0 from the given integer N` `class` `GFG{` `// Function returns min step` `// to reach 0 from N` `static` `int` `getMinSteps(` `int` `n, ` `int` `jump)` `{` ` ` ` ` `// Direct possible` ` ` `// reduction of value N` ` ` `int` `quotient = n / jump;` ` ` `// Remaining steps needs` ` ` `// to be reduced by 1` ` ` `int` `remainder = n % jump;` ` ` `// Summation of both the values` ` ` `int` `steps = quotient + remainder;` ` ` `// Return the final answer` ` ` `return` `steps;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `6` `, K = ` `3` `;` ` ` `System.out.print(getMinSteps(N, K));` `}` `}` `// This code is contributed by Rohit_ranjan` |

## Python3

`# Python3 program to Count the minimum steps` `# to reach 0 from the given integer N` `# Function returns min step` `# to reach 0 from N` `def` `getMinSteps(n, jump):` ` ` `# Direct possible` ` ` `# reduction of value N` ` ` `quotient ` `=` `int` `(n ` `/` `jump)` ` ` `# Remaining steps needs` ` ` `# to be reduced by 1` ` ` `remainder ` `=` `n ` `%` `jump` ` ` `# Summation of both the values` ` ` `steps ` `=` `quotient ` `+` `remainder` ` ` `# Return the final answer` ` ` `return` `steps` `# Driver code` `N ` `=` `6` `K ` `=` `3` `print` `(getMinSteps(N, K))` `# This code is contributed by PratikBasu` |

## C#

`// C# program to count the minimum steps` `// to reach 0 from the given integer N` `using` `System;` `class` `GFG{` `// Function returns min step` `// to reach 0 from N` `static` `int` `getMinSteps(` `int` `n, ` `int` `jump)` `{` ` ` ` ` `// Direct possible` ` ` `// reduction of value N` ` ` `int` `quotient = n / jump;` ` ` `// Remaining steps needs` ` ` `// to be reduced by 1` ` ` `int` `remainder = n % jump;` ` ` `// Summation of both the values` ` ` `int` `steps = quotient + remainder;` ` ` `// Return the final answer` ` ` `return` `steps;` `}` `// Driver code` `public` `static` `void` `Main(` `string` `[] args)` `{` ` ` `int` `N = 6, K = 3;` ` ` `Console.Write(getMinSteps(N, K));` `}` `}` `// This code is contributed by rutvik_56` |

## Javascript

`<script>` `// JavaScript program to Count the minimum steps` `// to reach 0 from the given integer N` `// Function returns min step` `// to reach 0 from N` `function` `getMinSteps(n, jump)` `{` ` ` `// Direct possible` ` ` `// reduction of value N` ` ` `let quotient = Math.floor(n / jump);` ` ` `// Remaining steps needs` ` ` `// to be reduced by 1` ` ` `let remainder = n % jump;` ` ` `// Summation of both the values` ` ` `let steps = quotient + remainder;` ` ` `// Return the final answer` ` ` `return` `steps;` `}` `// Driver code` ` ` `let N = 6, K = 3;` ` ` `document.write(getMinSteps(N, K));` `// This code is contributed by Surbhi Tyagi.` `</script>` |

**Output:**

2

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