Make n using 1s and 2s with minimum number of terms multiple of k

Given two positive integer n and k. n can be represented as the sum of 1s and 2s in many ways, using multiple numbers of terms. The task is to find the minimum number of terms of 1s and 2s use to make the sum n and also number of terms must be multiple of k. Print "-1", if no such number of terms exists. 
Examples : 
 

Input : n = 10, k = 2
Output : 6
10 can be represented as 2 + 2 + 2 + 2 + 1 + 1.
Number of terms used are 6 which is multiple of 2.

Input : n = 11, k = 4
Output : 8
10 can be represented as 2 + 2 + 2 + 1 + 1 + 1 + 1 + 1
Number of terms used are 8 which is multiple of 4.

Observe, the maximum number of terms used to represent n as the sum of 1s and 2s is n, when 1 are added n number of times. Also, the minimum number of terms will be n/2 times of 2s and n%2 times 1s are added. So, iterate from minimum number of terms to maximum number of terms and check if there is any multiple of k. 
 

// C++ program to find minimum multiple of k
// terms used to make sum n using 1s and 2s.
#include<bits/stdc++.h>
using namespace std;

// Return minimum multiple of k terms used to
// make sum n using 1s and 2s.
int minMultipleK(int n, int k)
{
    // Minimum number of terms required to make
    // sum n using 1s and 2s.
    int min = (n / 2) + (n % 2);

    // Iterate from Minimum to maximum to find
    // multiple of k. Maximum number of terns is
    // n (Sum of all 1s)
    for (int i = min; i <= n; i++)
        if (i % k == 0)
            return i;

    return -1;
}

// Driven Program
int main()
{
    int n = 10, k = 2;
    cout << minMultipleK(n, k) << endl;
    return 0;
}

// Java program to find minimum
// multiple of k terms used to 
// make sum n using 1s and 2s.
import java.io.*;

class GFG 
{
    
// Return minimum multiple of
// k terms used to make sum n
// using 1s and 2s.
static int minMultipleK(int n, 
                        int k)
{
    // Minimum number of terms 
    // required to make sum n 
    // using 1s and 2s.
    int min = (n / 2) + (n % 2);

    // Iterate from Minimum to 
    // maximum to findmultiple of k. 
    // Maximum number of terms is
    // n (Sum of all 1s)
    for (int i = min; i <= n; i++)
        if (i % k == 0)
            return i;

    return -1;
}

// Driver Code
public static void main (String[] args) 
{
    int n = 10, k = 2;
    System.out.println( minMultipleK(n, k));
}
}

// This code is contributed by anuj_67.

# Python3 program to find minimum multiple of k
# terms used to make sum n using 1s and 2s.

# Return minimum multiple of k terms 
# used to make sum n using 1s and 2s.
def minMultipleK( n, k):

    # Minimum number of terms required 
    # to make sum n using 1s and 2s.
    min = (n // 2) + (n % 2)

    # Iterate from Minimum to maximum to find
    #multiple of k. Maximum number of terns is
    # n (Sum of all 1s)
    for i in range(min, n + 1):
        if (i % k == 0):
            return i

    return -1

# Driver Code
if __name__=="__main__":

    n = 10
    k = 2
    print (minMultipleK(n, k))
    
# This code is contributed 
# by ChitraNayal

// C# program to find minimum
// multiple of k terms used to 
// make sum n using 1s and 2s.
using System;

class GFG 
{
    
// Return minimum multiple of
// k terms used to make sum n
// using 1s and 2s.
static int minMultipleK(int n, 
                        int k)
{
    // Minimum number of terms 
    // required to make sum n 
    // using 1s and 2s.
    int min = (n / 2) + (n % 2);

    // Iterate from Minimum to 
    // maximum to findmultiple of k. 
    // Maximum number of terms is
    // n (Sum of all 1s)
    for (int i = min; i <= n; i++)
        if (i % k == 0)
            return i;

    return -1;
}

// Driver Code
public static void Main () 
{
    int n = 10, k = 2;
    Console.WriteLine( minMultipleK(n, k));
}
}

// This code is contributed by anuj_67.

<?php
// PHP program to find minimum multiple of 
// k terms used to make sum n using 1s and 2s.

// Return minimum multiple of k terms used 
// to make sum n using 1s and 2s.
function minMultipleK($n, $k) 
                        
{
    // Minimum number of terms required 
    // to make sum n using 1s and 2s.
    $min = ($n / 2) + ($n % 2);

    // Iterate from Minimum to maximum to 
    // findmultiple of k. Maximum number 
    // of terms is n (Sum of all 1s)
    for ($i = $min; $i <= $n; $i++)
        if ($i % $k == 0)
            return $i;

    return -1;
}

// Driver Code
$n = 10; $k = 2;
echo(minMultipleK($n, $k));

// This code is contributed 
// by Mukul singh.

<script>

// JavaScript program for the above approach
   
// Return minimum multiple of 
// k terms used to make sum n 
// using 1s and 2s. 
function minMultipleK(n, k) 
{ 

    // Minimum number of terms  
    // required to make sum n  
    // using 1s and 2s. 
    let min = (n / 2) + (n % 2); 
  
    // Iterate from Minimum to  
    // maximum to findmultiple of k.  
    // Maximum number of terms is 
    // n (Sum of all 1s) 
    for (let i = min; i <= n; i++) 
        if (i % k == 0) 
            return i; 
  
    return -1; 
} 

// Driver Code
    
    let n = 10, k = 2; 
    document.write( minMultipleK(n, k)); 
       
       // This code is contributed by splevel62.
</script>

Output : 
 

6

Time complexity : O(n)

Space complexity : O(1)