# Check if K distinct array elements form an odd sum

Given an array A[] of size N, the task is to check if it is possible to get odd sum using K distinct elements from the array.
Examples:

Input: N = 4, K = 2, A = {10, 19, 14, 14}
Output: YES
Explanation:
19 + 14 = 33, which is odd
Input: N = 3, K = 3, A = {101, 102, 103}
Output: NO
Explanation:
101 + 102 + 103 = 306, which is even

Approach:

• Let us look at some basic maths properties:
```EVEN + EVEN = EVEN
ODD  + ODD  = EVEN
EVEN + ODD  = ODD

```
• From the above properties, we can conclude that when an odd number of odd integers are added to any number of even integers, the result will always be odd.

Examples:

• 3 + 2 + 6 = 11 (odd)
• 1 + 3 + 7 + 4 = 15 (odd)

Hence, follow the steps below to solve the problem:

• Count distinct odd and even elements present in the array.
• If K or more odd elements are present in the array, print “Yes”.
• Otherwise, for every odd count of odd numbers, check if sufficient even numbers are present in the array or not.
• If any such case occurs, print “Yes”. Otherwise, print “No”.

Below is the implementation of the above approach:

 `// C++ program for above approach `   `#include ` `using` `namespace` `std; `   `// Function to return if ` `// odd sum is possible or not ` `bool` `oddSum(vector<``int``>& A, ` `            ``int` `N, ``int` `K) ` `{ `   `    ``// Stores distinct odd elements ` `    ``set<``int``> Odd; ` `    ``// Stores distinct even elements ` `    ``set<``int``> Even; `   `    ``// Iterating through given array ` `    ``for` `(``int` `i = 0; i < N; i++) { `   `        ``// If element is even ` `        ``if` `(A[i] % 2 == 0) { ` `            ``Even.insert(A[i]); ` `        ``} ` `        ``// If element is odd ` `        ``else` `{ ` `            ``Odd.insert(A[i]); ` `        ``} ` `    ``} `   `    ``// If atleast K elements ` `    ``// in the array are odd ` `    ``if` `(Odd.size() >= K) ` `        ``return` `true``; `   `    ``bool` `flag = ``false``; `   `    ``// Check for all odd frequencies ` `    ``// of odd elements whether ` `    ``// sufficient even numbers ` `    ``// are present or not ` `    ``for` `(``int` `i = 1; i < K; i += 2) { `   `        ``// Count of even numbers ` `        ``// required ` `        ``int` `needed = K - i; `   `        ``// If required even numbers ` `        ``// are present in the array ` `        ``if` `(needed <= Even.size()) { `   `            ``return` `true``; ` `        ``} ` `    ``} `   `    ``return` `flag; ` `} `   `// Driver Program ` `int` `main() ` `{ ` `    ``int` `K = 5; ` `    ``vector<``int``> A = { 12, 1, 7, 7, 26, 18 }; ` `    ``int` `N = 3; `   `    ``if` `(oddSum(A, N, K)) ` `        ``cout << ``"YES"``; ` `    ``else` `        ``cout << ``"NO"``; `   `    ``return` `0; ` `} `

 `// Java program for the above approach ` `import` `java.util.*; `   `class` `GFG{ `   `// Function to return if ` `// odd sum is possible or not ` `static` `boolean` `oddSum(``int` `[]A, ` `                      ``int` `N, ``int` `K) ` `{ ` `    `  `    ``// Stores distinct odd elements ` `    ``HashSet Odd = ``new` `HashSet(); ` `    `  `    ``// Stores distinct even elements ` `    ``HashSet Even = ``new` `HashSet(); `   `    ``// Iterating through given array ` `    ``for``(``int` `i = ``0``; i < N; i++) ` `    ``{ ` `        `  `        ``// If element is even ` `        ``if` `(A[i] % ``2` `== ``0``) ` `        ``{ ` `            ``Even.add(A[i]); ` `        ``} ` `            `  `        ``// If element is odd ` `        ``else` `        ``{ ` `            ``Odd.add(A[i]); ` `        ``} ` `    ``} `   `    ``// If atleast K elements ` `    ``// in the array are odd ` `    ``if` `(Odd.size() >= K) ` `        ``return` `true``; `   `    ``boolean` `flag = ``false``; `   `    ``// Check for all odd frequencies ` `    ``// of odd elements whether ` `    ``// sufficient even numbers ` `    ``// are present or not ` `    ``for``(``int` `i = ``1``; i < K; i += ``2``) ` `    ``{ ` `        `  `        ``// Count of even numbers ` `        ``// required ` `        ``int` `needed = K - i; ` `            `  `        ``// If required even numbers ` `        ``// are present in the array ` `        ``if` `(needed <= Even.size()) ` `        ``{ ` `            ``return` `true``; ` `        ``} ` `    ``} ` `    ``return` `flag; ` `} `   `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `K = ``5``; ` `    ``int` `[]A = { ``12``, ``1``, ``7``, ``7``, ``26``, ``18` `}; ` `    ``int` `N = ``3``; `   `    ``if` `(oddSum(A, N, K)) ` `        ``System.out.print(``"YES"``); ` `    ``else` `        ``System.out.print(``"NO"``); ` `} ` `} `   `// This code is contributed by PrinciRaj1992 `

 `# Python3 program for ` `# the above approach `   `# Function to return if ` `# odd sum is possible or not ` `def` `oddSum(A, N, K):`   `    ``# Stores distinct odd elements ` `    ``Odd ``=` `set``([])` `    ``# Stores distinct even elements ` `    ``Even ``=` `set``([])` ` `  `    ``# Iterating through given array ` `    ``for` `i ``in` `range` `(N):` ` `  `        ``# If element is even ` `        ``if` `(A[i] ``%` `2` `=``=` `0``):` `            ``Even.add(A[i])` `       `  `        ``# If element is odd ` `        ``else``:` `            ``Odd.add(A[i])` ` `  `    ``# If atleast K elements ` `    ``# in the array are odd ` `    ``if` `(``len``(Odd) >``=` `K):` `        ``return` `True` ` `  `    ``flag ``=` `False` ` `  `    ``# Check for all odd frequencies ` `    ``# of odd elements whether ` `    ``# sufficient even numbers ` `    ``# are present or not ` `    ``for` `i ``in` `range` `(``1``, K, ``2``):` ` `  `        ``# Count of even numbers ` `        ``# required ` `        ``needed ``=` `K ``-` `i` ` `  `        ``# If required even numbers ` `        ``# are present in the array ` `        ``if` `(needed <``=` `len``(Even)): ` `            ``return` `True` ` `  `    ``return` `flag` ` `  `# Driver Program ` `if` `__name__ ``=``=` `"__main__"``:`   `    ``K ``=` `5` `    ``A ``=` `[``12``, ``1``, ``7``, ` `         ``7``, ``26``, ``18``]` `    ``N ``=` `3` ` `  `    ``if` `(oddSum(A, N, K)):` `        ``print` `(``"YES"``)` `    ``else``:` `        ``print``(``"NO"``)` ` `  `# This code is contributed by Chitranayal`

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG{`   `// Function to return if` `// odd sum is possible or not` `static` `bool` `oddSum(``int` `[]A, ``int` `N, ``int` `K)` `{` `    `  `    ``// Stores distinct odd elements` `    ``HashSet<``int``> Odd = ``new` `HashSet<``int``>();` `    `  `    ``// Stores distinct even elements` `    ``HashSet<``int``> Even = ``new` `HashSet<``int``>();`   `    ``// Iterating through given array` `    ``for``(``int` `i = 0; i < N; i++)` `    ``{` `        `  `        ``// If element is even ` `        ``if` `(A[i] % 2 == 0)  ` `        ``{ ` `            ``Even.Add(A[i]); ` `        ``} ` `        `  `        ``// If element is odd ` `        ``else` `        ``{ ` `            ``Odd.Add(A[i]); ` `        ``} ` `    ``}` `    `  `    ``// If atleast K elements ` `    ``// in the array are odd ` `    ``if` `(Odd.Count >= K) ` `        ``return` `true``; `   `    ``bool` `flag = ``false``;`   `    ``// Check for all odd frequencies` `    ``// of odd elements whether` `    ``// sufficient even numbers` `    ``// are present or not` `    ``for``(``int` `i = 1; i < K; i += 2)` `    ``{` `        `  `        ``// Count of even numbers` `        ``// required` `        ``int` `needed = K - i;` `            `  `        ``// If required even numbers` `        ``// are present in the array` `        ``if` `(needed <= Even.Count)` `        ``{` `            ``return` `true``;` `        ``}` `    ``}` `    ``return` `flag;` `}`   `// Driver code` `public` `static` `void` `Main(String[] args)` `{` `    ``int` `K = 5;` `    ``int` `[]A = { 12, 1, 7, 7, 26, 18 };` `    ``int` `N = 3;`   `    ``if` `(oddSum(A, N, K))` `        ``Console.Write(``"YES"``);` `    ``else` `        ``Console.Write(``"NO"``);` `}` `}`   `// This code is contributed by PrinciRaj1992`

Output:
```NO

```

Time complexity: O(N)

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Improved By : princiraj1992, chitranayal

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