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 representationvoid 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 Codeint main(){ int N = 5; int K = 3; sumOddNumbers(N, K); return 0;} |
Java
// Java implementation to represent// N as sum of K even numbersimport java.util.*;class GFG{// Function to print the representationstatic 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 Codepublic 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 = 5K = 3; sumOddNumbers(N, K) # This code is contributed by PratikBasu |
C#
// C# implementation to represent// N as sum of K even numbersusing System;class GFG{// Function to print the representationstatic 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 Codepublic 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 representationfunction 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)
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