Find any K distinct odd integers such that their sum is equal to N

Given two integers N and K, the task is to find any K distinct odd integers such that their sum is equal to N. If no such integers exists, print -1.

Examples:

Input: N = 10, K = 2
Output: 1, 9
Explanation:
There are two possible distinct odd integers, such that their sum is equal to N.
Possible K integers can be – {(1, 9), (3, 7)}

Input: N = 5, K = 4
Output: -1
Explanation:
There are no such 4 distinct odd integers such that their sum is 5.

Approach:



  • The key observation in this problem is if N and K have different parity then it is not possible to find K such distinct integers such that their sum is equal to N,
  • Otherwise such K – 1 integers will consist of first K-1 odd positive integers
  • The Kth odd number will be equal to (N – the sum of first (K-1) odd integers)
    Kth Odd number  = N - sum of first K-1 integer
    

Below is the implementation of the above approach:

C++

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// C++ implementation to find k
// odd integers such that their sum is N
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find K odd integers
// such that their sum is N
void oddIntegers(int n, int k)
{
    // Condition to check if there
    // exist such K integers
    if (n % 2 != k % 2) {
        cout << "-1"
             << "\n";
        return;
    }
  
    int sum = 0;
    int i = 1;
    int j = 1;
  
    // Loop to find first K-1
    // distinct odd integers
    while (j < k) {
        sum = sum + i;
        cout << i << " ";
        i = i + 2;
        j++;
    }
  
    // Final Kth odd number
    int finalOdd = n - sum;
  
    cout << finalOdd << "\n";
}
  
// Driver code
int main()
{
    int n = 10;
    int k = 2;
  
    oddIntegers(n, k);
  
    return 0;
}

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Java

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// Java implementation to find k
// odd integers such that their sum is N
class GFG
{
  
// Function to find K odd integers
// such that their sum is N
static void oddIntegers(int n, int k)
{
    // Condition to check if there
    // exist such K integers
    if (n % 2 != k % 2) {
        System.out.println("-1");
        return;
    }
  
    int sum = 0;
    int i = 1;
    int j = 1;
  
    // Loop to find first K-1
    // distinct odd integers
    while (j < k) {
        sum = sum + i;
        System.out.print(i+" ");
        i = i + 2;
        j++;
    }
  
    // Final Kth odd number
    int finalOdd = n - sum;
      
    System.out.println(finalOdd);
}
  
// Driver code
public static void main (String[] args)
{
    int n = 10;
    int k = 2;
  
    oddIntegers(n, k);
}
}
  
// This code is contributed by shubhamsingh

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Python3

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# Python3 implementation to find k 
# odd integers such that their sum is N 
  
# Function to find K odd integers 
# such that their sum is N 
def oddIntegers(n, k) : 
  
    # Condition to check if there 
    # exist such K integers 
    if (n % 2 != k % 2) :
        print("-1"); 
          
        return
  
    sum = 0
    i = 1
    j = 1
  
    # Loop to find first K-1 
    # distinct odd integers 
    while (j < k) :
        sum += i; 
        print(i,end= " "); 
        i += 2
        j += 1
  
    # Final Kth odd number 
    finalOdd = n - sum
  
    print(finalOdd); 
  
# Driver code 
if __name__ == "__main__"
  
    n = 10
    k = 2
  
    oddIntegers(n, k); 
      
# This code is contributed by AnkitRai01

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C#

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// C# implementation to find k
// odd integers such that their sum is N
using System;
  
class GFG
{
   
// Function to find K odd integers
// such that their sum is N
static void oddints(int n, int k)
{
    // Condition to check if there
    // exist such K integers
    if (n % 2 != k % 2) {
        Console.WriteLine("-1");
        return;
    }
   
    int sum = 0;
    int i = 1;
    int j = 1;
   
    // Loop to find first K-1
    // distinct odd integers
    while (j < k) {
        sum = sum + i;
        Console.Write(i+" ");
        i = i + 2;
        j++;
    }
   
    // Final Kth odd number
    int finalOdd = n - sum;
       
    Console.WriteLine(finalOdd);
}
   
// Driver code
public static void Main(String[] args)
{
    int n = 10;
    int k = 2;
   
    oddints(n, k);
}
}
  
// This code is contributed by PrinciRaj1992

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Output:

1 9

Performance Analysis:

  • Time Complexity: As in the above approach, There is a loop to find such K odd integers which takes O(K) time in worst case. Hence the Time Complexity will be O(K).
  • Auxiliary Space Complexity: As in the above approach, There is no extra space used. Hence the auxiliary space complexity will be O(1).

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