Represent N as sum of K odd numbers with repetitions allowed

Given two integers N and K, the task is to represent N as sum of K odd numbers. If it is not possible to create the sum then output -1.
Note: The representation may contain duplicate odd numbers.
Examples:

Input: N = 5, K = 3 
Output: 1, 1, 3 
Explanation:
The given number N can be represented as 1 + 1 + 3 = 5
Input: N = 7, K = 5 
Output: 1, 1, 1, 1, 3 
Explanation:
The given number N can be represented as 1 + 1 + 1 + 1 + 3 = 7 

Approach:
To solve the problem mentioned above a simple solution is to maximise the occurrence of 1 which is the possible smallest odd number. Necessary conditions for representing the number N as K odd numbers are: 

• (K – 1) must be less than N.
• N – (K – 1) must be a Odd number.

Below is the implementation of the above approach:

C++

// C++ implementation to represent
// N as sum of K even numbers

#include <bits/stdc++.h>

using namespace std;

// Function to print the representation
void sumOddNumbers(int N, int K)
{
    int check = N - (K - 1);

    // N must be greater than equal to 2*K
    // and must be odd
    if (check > 0 && check % 2 == 1) {
        for (int i = 0; i < K - 1; i++) {
            cout << "1 ";
        }
        cout << check;
    }
    else
        cout << "-1";
}

// Driver Code
int main()
{

    int N = 5;
    int K = 3;

    sumOddNumbers(N, K);
    return 0;
}

Java

// Java implementation to represent
// N as sum of K even numbers
import java.util.*;

class GFG{

// Function to print the representation
static void sumOddNumbers(int N, int K)
{
    int check = N - (K - 1);

    // N must be greater than equal
    // to 2*K and must be odd
    if (check > 0 && check % 2 == 1)
    {
        for(int i = 0; i < K - 1; i++)
        {
           System.out.print("1 ");
        }
        System.out.print(+check);
    }
    else
        System.out.println("-1 ");
}

// Driver Code
public static void main(String args[])
{
    int N = 5;
    int K = 3;

    sumOddNumbers(N, K);
}
}

// This code is contributed by AbhiThakur

Python3

# Python3 implementation to represent
# N as sum of K even numbers

# Function to print the representation
def sumOddNumbers(N, K):

    check = N - (K - 1)

    # N must be greater than equal
    # to 2*K and must be odd
    if (check > 0 and check % 2 == 1):
        for i in range(0, K - 1):
            print("1", end = " ")

        print(check, end = " ")

    else:
        print("-1")

# Driver Code
N = 5
K = 3;

sumOddNumbers(N, K)

# This code is contributed by PratikBasu

C#

// C# implementation to represent
// N as sum of K even numbers
using System;

class GFG{

// Function to print the representation
static void sumOddNumbers(int N, int K)
{
    int check = N - (K - 1);

    // N must be greater than equal
    // to 2*K and must be odd
    if (check > 0 && check % 2 == 1)
    {
        for(int i = 0; i < K - 1; i++)
        {
           Console.Write("1 ");
        }
        Console.Write(+check);
    }
    else
        Console.WriteLine("-1 ");
}

// Driver Code
public static void Main()
{
    int N = 5;
    int K = 3;

    sumOddNumbers(N, K);
}
}

// This code is contributed by Code_Mech

JavaScript

<script>

// Javascript implementation to represent
// N as sum of K even numbers

// Function to print the representation
function sumOddNumbers(N, K)
{
    var check = N - (K - 1);
    var i;

    // N must be greater than equal to 2*K
    // and must be odd
    if (check > 0 && check % 2 == 1) {
        for (i = 0; i < K - 1; i++) {
             document.write("1,"+" ");
        }
        document.write(check + " ");
    }
    else
        document.write("-1");
}

// Driver Code
    var N = 5;
    var K = 3;

    sumOddNumbers(N, K);

</script>

Output:

1 1 3

Time Complexity: O(k)

Auxiliary Space: O(1)