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

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 <bits/stdc++.h>
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

<script>

// javascript 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
function 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

var n = 17, m = 4;
var ans = minsteps(n, m);
document.write(ans);

// This code contributed by shikhasingrajput

</script>

Output:

12

Time Complexity: O(1)

Auxiliary Space: O(1)