# 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.

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

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++ implementation to find k ` `// odd integers such that their sum is N ` ` `  `#include ` `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; ` `} `

 `// 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 `

 `# 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 `

 `// 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 `

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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