# Find K distinct positive odd integers with sum N

Given two integers N and K, the task is to find K distinct positive odd integers such that their sum is equal to the given number N.

Examples:

Input: N = 10, K = 2
Output: 1 9
Explanation:
Two odd positive integers such that their sum is 10 can be (1, 9) or (3, 7).

Input: N = 10, K = 4
Output: NO
Explanation:
There does not exists four odd positive integers with sum 10.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

The number N can be represented as the sum of K positive odd integers only is the following two conditions satisfies:

1. If the square of K is less than or equal to N and,
2. If the sum of N and K is an even number.

If these conditions are satisfied then there exist K positive odd integers whose sum is N.

To generate K such odd numbers:

• Print first K-1 odd numbers starting from 1, i.e. 1, 3, 5, 7, 9…….
• The last odd number will be : N – (Sum of first K-1 odd positive integers)

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find K  ` `// odd positive integers such that ` `// their sum is equal to given number ` ` `  `#include ` `using` `namespace` `std; ` ` `  `#define ll long long int ` ` `  `// Function to find K odd positive ` `// integers such that their sum is N ` `void` `findDistinctOddSum(ll n, ll k) ` `{ ` `    ``// Condition to check if there ` `    ``// are enough values to check ` `    ``if` `((k * k) <= n &&  ` `        ``(n + k) % 2 == 0){ ` `        ``int` `val = 1; ` `        ``int` `sum = 0; ` `        ``for``(``int` `i = 1; i < k; i++){ ` `            ``cout << val << ``" "``; ` `            ``sum += val; ` `            ``val += 2; ` `        ``} ` `        ``cout << n - sum << endl; ` `    ``} ` `    ``else` `        ``cout << ``"NO \n"``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``ll n = 100; ` `    ``ll k = 4; ` `    ``findDistinctOddSum(n, k); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation to find K  ` `// odd positive integers such that ` `// their sum is equal to given number ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `  `  `// Function to find K odd positive ` `// integers such that their sum is N ` `static` `void` `findDistinctOddSum(``int` `n, ``int` `k) ` `{ ` `    ``// Condition to check if there ` `    ``// are enough values to check ` `    ``if` `((k * k) <= n &&  ` `        ``(n + k) % ``2` `== ``0``){ ` `        ``int` `val = ``1``; ` `        ``int` `sum = ``0``; ` `        ``for``(``int` `i = ``1``; i < k; i++){ ` `            ``System.out.print(val+ ``" "``); ` `            ``sum += val; ` `            ``val += ``2``; ` `        ``} ` `        ``System.out.print(n - sum +``"\n"``); ` `    ``} ` `    ``else` `        ``System.out.print(``"NO \n"``); ` `} ` `  `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `n = ``100``; ` `    ``int` `k = ``4``; ` `    ``findDistinctOddSum(n, k); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python3 implementation to find K  ` `# odd positive integers such that ` `# their summ is equal to given number ` ` `  `# Function to find K odd positive ` `# integers such that their summ is N ` `def` `findDistinctOddsumm(n, k): ` `     `  `    ``# Condition to check if there ` `    ``# are enough values to check ` `    ``if` `((k ``*` `k) <``=` `n ``and` `(n ``+` `k) ``%` `2` `=``=` `0``): ` `        ``val ``=` `1` `        ``summ ``=` `0` `        ``for` `i ``in` `range``(``1``, k): ` `            ``print``(val, end ``=`  `" "``) ` `            ``summ ``+``=` `val ` `            ``val ``+``=` `2` `        ``print``(n ``-` `summ) ` `    ``else``: ` `        ``print``(``"NO"``) ` ` `  `# Driver Code ` `n ``=` `100` `k ``=` `4` `findDistinctOddsumm(n, k) ` ` `  `# This code is contributed by shubhamsingh10 `

## C#

 `// C# implementation to find K  ` `// odd positive integers such that ` `// their sum is equal to given number ` `using` `System; ` ` `  `public` `class` `GFG{ ` `   `  `// Function to find K odd positive ` `// integers such that their sum is N ` `static` `void` `findDistinctOddSum(``int` `n, ``int` `k) ` `{ ` `    ``// Condition to check if there ` `    ``// are enough values to check ` `    ``if` `((k * k) <= n &&  ` `        ``(n + k) % 2 == 0){ ` `        ``int` `val = 1; ` `        ``int` `sum = 0; ` `        ``for``(``int` `i = 1; i < k; i++){ ` `            ``Console.Write(val+ ``" "``); ` `            ``sum += val; ` `            ``val += 2; ` `        ``} ` `        ``Console.Write(n - sum +``"\n"``); ` `    ``} ` `    ``else` `        ``Console.Write(``"NO \n"``); ` `} ` `   `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `n = 100; ` `    ``int` `k = 4; ` `    ``findDistinctOddSum(n, k); ` `} ` `} ` `// This code is contributed by 29AjayKumar `

Output:

```1 3 5 91
```

Performance Analysis:

• Time Complexity: In the above-given approach, there is one loop which takes O(K) time in the worst case. Therefore, the time complexity for this approach will be O(K).
• Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Practice until my ideals becomes my rivals

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.