# Form N by adding 1 or 2 in minimum number of operations X where X is divisible by M

• Difficulty Level : Easy
• Last Updated : 01 Mar, 2022

Given a number N, the task is to form N (from 0) by adding 1 or 2 in the minimum number of operations X such that X is divisible by M.
Examples:

Input: N = 10, M = 2
Output: X = 6
Explanation:
Taken operations are 2 2 2 2 1 1
X = 6 which is divisible by 2
Input: N = 17, M = 4
Output: 12

Approach:

• As we can take either 1 or 2 steps at a time, so we can say that minimum no. of steps taken is n/2, and the maximum no. of steps is n, irrespective of that the no. of steps are divisible by m.
• So we have to count n/2 steps to get a minimum number of steps. Now if n is even, then a minimum number of steps will be n/2, but if it is odd, then it will be n/2+1, irrespective of that the no. of steps are divisible by m. To make minimum steps of a multiple of m we can do floor((minimum_steps + m – 1)/m) * m
• Also if n is less than m, then it is not possible to find the minimum number of steps, and in that case, we will return -1.

Below is the implementation of above approach:

## C++

 `// C++ program to find minimum``// number of steps to cover distance x` `#include ``using` `namespace` `std;` `// Function to calculate the minimum number of steps required``// total steps taken is divisible``// by m and only 1 or 2 steps can be taken at // a time``int` `minsteps(``int` `n, ``int` `m)``{` `    ``// If m > n ans is -1``    ``if` `(m > n) {``        ``return` `-1;``    ``}``    ``// else discussed above approach``    ``else` `{``        ``return` `((n + 1) / 2 + m - 1) / m * m;``    ``}``}` `// Driver code``int` `main()``{``    ``int` `n = 17, m = 4;``    ``int` `ans = minsteps(n, m);``    ``cout << ans << ``'\n'``;` `    ``return` `0;``}`

## Java

 `// Java program to find minimum``// number of steps to cover distance x``class` `GFG``{` `    ``// Function to calculate the``    ``// minimum number of steps required``    ``// total steps taken is divisible``    ``// by m and only 1 or 2 steps can be``    ``// taken at // a time``    ``static` `int` `minsteps(``int` `n, ``int` `m)``    ``{``    ` `        ``// If m > n ans is -1``        ``if` `(m > n)``        ``{``            ``return` `-``1``;``        ``}``        ` `        ``// else discussed above approach``        ``else``        ``{``            ``return` `((n + ``1``) / ``2` `+ m - ``1``) / m * m;``        ``}``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `n = ``17``, m = ``4``;``        ``int` `ans = minsteps(n, m);``        ``System.out.println(ans);``    ``}``}` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 program to find minimum``# number of steps to cover distance x` `# Function to calculate the minimum number of``# steps required total steps taken is divisible``# by m and only 1 or 2 steps can be taken at a time``def` `minsteps(n, m):` `    ``# If m > n ans is -1``    ``if` `(m > n):``        ``return` `-``1``        ` `    ``# else discussed above approach``    ``else` `:``        ``return` `((n ``+` `1``) ``/``/` `2` `+` `m ``-` `1``) ``/``/` `m ``*` `m;` `# Driver code``n ``=` `17``m ``=` `4``ans ``=` `minsteps(n, m)``print``(ans)` `# This code is contributed by Mohit Kumar`

## C#

 `// C# program to find minimum``// number of steps to cover distance x``using` `System;``    ` `class` `GFG``{` `    ``// Function to calculate the``    ``// minimum number of steps required``    ``// total steps taken is divisible``    ``// by m and only 1 or 2 steps can be``    ``// taken at // a time``    ``static` `int` `minsteps(``int` `n, ``int` `m)``    ``{``    ` `        ``// If m > n ans is -1``        ``if` `(m > n)``        ``{``            ``return` `-1;``        ``}``        ` `        ``// else discussed above approach``        ``else``        ``{``            ``return` `((n + 1) / 2 + m - 1) / m * m;``        ``}``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main (String[] args)``    ``{``        ``int` `n = 17, m = 4;``        ``int` `ans = minsteps(n, m);``        ``Console.WriteLine(ans);``    ``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

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Output:

`12`

Time Complexity: O(1)

Auxiliary Space: O(1)

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