Make n using 1s and 2s with minimum number of terms multiple of k
Last Updated :
22 Feb, 2023
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++
#include<bits/stdc++.h>
using namespace std;
int minMultipleK( int n, int k)
{
int min = (n / 2) + (n % 2);
for ( int i = min; i <= n; i++)
if (i % k == 0)
return i;
return -1;
}
int main()
{
int n = 10, k = 2;
cout << minMultipleK(n, k) << endl;
return 0;
}
|
Java
import java.io.*;
class GFG
{
static int minMultipleK( int n,
int k)
{
int min = (n / 2 ) + (n % 2 );
for ( int i = min; i <= n; i++)
if (i % k == 0 )
return i;
return - 1 ;
}
public static void main (String[] args)
{
int n = 10 , k = 2 ;
System.out.println( minMultipleK(n, k));
}
}
|
Python3
def minMultipleK( n, k):
min = (n / / 2 ) + (n % 2 )
for i in range ( min , n + 1 ):
if (i % k = = 0 ):
return i
return - 1
if __name__ = = "__main__" :
n = 10
k = 2
print (minMultipleK(n, k))
|
C#
using System;
class GFG
{
static int minMultipleK( int n,
int k)
{
int min = (n / 2) + (n % 2);
for ( int i = min; i <= n; i++)
if (i % k == 0)
return i;
return -1;
}
public static void Main ()
{
int n = 10, k = 2;
Console.WriteLine( minMultipleK(n, k));
}
}
|
PHP
<?php
function minMultipleK( $n , $k )
{
$min = ( $n / 2) + ( $n % 2);
for ( $i = $min ; $i <= $n ; $i ++)
if ( $i % $k == 0)
return $i ;
return -1;
}
$n = 10; $k = 2;
echo (minMultipleK( $n , $k ));
|
Javascript
<script>
function minMultipleK(n, k)
{
let min = (n / 2) + (n % 2);
for (let i = min; i <= n; i++)
if (i % k == 0)
return i;
return -1;
}
let n = 10, k = 2;
document.write( minMultipleK(n, k));
</script>
|
Output :
6
Time complexity : O(n)
Space complexity : O(1)
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