# Check if its possible to make sum of the array odd with given Operations

Given an array arr[], the task is to check if it is possible to make the sum of array odd such that any two indices i and j can be chosen and arr[i] can be set equal to arr[j] given that i != j.

Examples:

Input: arr[] = { 5, 4, 4, 5, 1, 3 }
Output: Yes
Explanation:
The sum of the array is 22. Put arr = arr where i=0 and j=1. The sum of this new array is 21 which is odd.

Input: arr[] = { 2, 2, 8, 8 }
Output: No
Explanation:
Sum of the array is 20.
The sum of the array cannot change to odd as all the elements are even. Adding even number to an even number always gives an even result.

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

Approach: The idea is to find the sum all the elements in the array and simultaneously checking the number of even and odd numbers present in the array. If the sum is even, then one of the even numbers can be replaced with an odd number thereby making the sum of the array odd. If there are no odd elements in the array, then the sum of the array cannot be made odd.

Below is the implementation of the above approach:

## C++

 `// C++ program to check if the array ` `// with odd sum is possible ` `#include ` `using` `namespace` `std; ` ` `  `// Function to check if the ` `// sum of the array can be made odd. ` `bool` `isOdd(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `l, r, flag = 0, flag1 = 0, sum = 0; ` ` `  `    ``// Find sum of all elements and increment ` `    ``// check for odd or even elements in the array ` `    ``// so that by changing ai=aj, ` `    ``// the sum of the array can be made odd ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``sum += arr[i]; ` `        ``if` `(arr[i] % 2 == 0 && flag == 0) { ` `            ``flag = 1; ` `            ``l = arr[i]; ` `        ``} ` `        ``if` `(arr[i] % 2 != 0 && flag1 == 0) { ` `            ``r = arr[i]; ` `            ``flag1 = 1; ` `        ``} ` `    ``} ` ` `  `    ``// If the sum is already odd ` `    ``if` `(sum % 2 != 0) { ` `        ``return` `true``; ` `    ``} ` ` `  `    ``// Else, then both the flags should be checked. ` `    ``// Here, flag1 and flag represent if there is ` `    ``// an odd-even pair which can be replaced. ` `    ``else` `{ ` `        ``if` `(flag1 == 1 && flag == 1) ` `            ``return` `true``; ` `        ``else` `            ``return` `false``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `ar[] = { 5, 4, 4, 5, 1, 3 }; ` `    ``int` `n = ``sizeof``(ar) / ``sizeof``(ar); ` `    ``bool` `res = isOdd(ar, n); ` ` `  `    ``if` `(res) ` `        ``cout << ``"Yes"` `<< endl; ` `    ``else` `        ``cout << ``"No"` `<< endl; ` `    ``return` `0; ` `} `

## Java

 `// Java program to check if the array ` `// with odd sum is possible ` `class` `GFG { ` `     `  `    ``// Function to check if the ` `    ``// sum of the array can be made odd. ` `    ``static` `boolean` `isOdd(``int` `[]arr, ``int` `n) ` `    ``{ ` `        ``int` `l, r, flag = ``0``, flag1 = ``0``, sum = ``0``; ` `     `  `        ``// Find sum of all elements and increment ` `        ``// check for odd or even elements in the array ` `        ``// so that by changing ai=aj, ` `        ``// the sum of the array can be made odd ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` `            ``sum += arr[i]; ` `            ``if` `(arr[i] % ``2` `== ``0` `&& flag == ``0``) { ` `                ``flag = ``1``; ` `                ``l = arr[i]; ` `            ``} ` `            ``if` `(arr[i] % ``2` `!= ``0` `&& flag1 == ``0``) { ` `                ``r = arr[i]; ` `                ``flag1 = ``1``; ` `            ``} ` `        ``} ` `     `  `        ``// If the sum is already odd ` `        ``if` `(sum % ``2` `!= ``0``) { ` `            ``return` `true``; ` `        ``} ` `     `  `        ``// Else, then both the flags should be checked. ` `        ``// Here, flag1 and flag represent if there is ` `        ``// an odd-even pair which can be replaced. ` `        ``else` `{ ` `            ``if` `(flag1 == ``1` `&& flag == ``1``) ` `                ``return` `true``; ` `            ``else` `                ``return` `false``; ` `        ``} ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `ar[] = { ``5``, ``4``, ``4``, ``5``, ``1``, ``3` `}; ` `        ``int` `n = ar.length; ` `        ``boolean` `res = isOdd(ar, n); ` `     `  `        ``if` `(res == ``true``) ` `            ``System.out.println(``"Yes"``); ` `        ``else` `            ``System.out.println(``"No"``); ` ` `  `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 program to check if the array ` `# with odd sum is possible ` ` `  `# Function to check if the ` `# sum of the array can be made odd. ` `def` `isOdd(arr,  n) : ` `    ``flag ``=` `0``; flag1 ``=` `0``; ``sum` `=` `0``; ` ` `  `    ``# Find sum of all elements and increment ` `    ``# check for odd or even elements in the array ` `    ``# so that by changing ai=aj, ` `    ``# the sum of the array can be made odd ` `    ``for` `i ``in` `range``(n) : ` `        ``sum` `+``=` `arr[i]; ` `        ``if` `(arr[i] ``%` `2` `=``=` `0` `and` `flag ``=``=` `0``) : ` `            ``flag ``=` `1``; ` `            ``l ``=` `arr[i]; ` `         `  `        ``if` `(arr[i] ``%` `2` `!``=` `0` `and` `flag1 ``=``=` `0``) : ` `            ``r ``=` `arr[i]; ` `            ``flag1 ``=` `1``; ` ` `  `    ``# If the sum is already odd ` `    ``if` `(``sum` `%` `2` `!``=` `0``) : ` `        ``return` `True``; ` ` `  `    ``# Else, then both the flags should be checked. ` `    ``# Here, flag1 and flag represent if there is ` `    ``# an odd-even pair which can be replaced. ` `    ``else` `: ` `        ``if` `(flag1 ``=``=` `1` `and` `flag ``=``=` `1``) : ` `            ``return` `True``; ` `        ``else` `: ` `            ``return` `False``; ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``arr ``=` `[ ``5``, ``4``, ``4``, ``5``, ``1``, ``3` `]; ` `    ``n ``=` `len``(arr); ` `     `  `    ``res ``=` `isOdd(arr, n); ` ` `  `    ``if` `(res) : ` `        ``print``(``"Yes"``); ` `    ``else` `: ` `        ``print``(``"No"``); ` `    `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# program to check if the array ` `// with odd sum is possible ` `using` `System; ` ` `  `class` `GFG{ ` `    ``// Function to check if the ` `    ``// sum of the array can be made odd. ` `    ``static` `bool` `isOdd(``int``[] arr, ``int` `n) ` `    ``{ ` `        ``int` `flag = 0, flag1 = 0, sum = 0; ` `     `  `        ``// Find sum of all elements and increment ` `        ``// check for odd or even elements in the array ` `        ``// so that by changing ai=aj, ` `        ``// the sum of the array can be made odd ` `        ``for` `(``int` `i = 0; i < n; i++) { ` `            ``sum += arr[i]; ` `            ``if` `(arr[i] % 2 == 0 && flag == 0) { ` `                ``flag = 1; ` `            ``} ` `            ``if` `(arr[i] % 2 != 0 && flag1 == 0) { ` `                ``flag1 = 1; ` `            ``} ` `        ``} ` `     `  `        ``// If the sum is already odd ` `        ``if` `(sum % 2 != 0) { ` `            ``return` `true``; ` `        ``} ` `     `  `        ``// Else, then both the flags should be checked. ` `        ``// Here, flag1 and flag represent if there is ` `        ``// an odd-even pair which can be replaced. ` `        ``else` `{ ` `            ``if` `(flag1 == 1 && flag == 1) ` `                ``return` `true``; ` `            ``else` `                ``return` `false``; ` `        ``} ` `    ``} ` `     `  `    ``// Driver code    ` `    ``static` `public` `void` `Main () ` `    ``{ ` `        ``int``[] ar = { 5, 4, 4, 5, 1, 3 }; ` `        ``int` `n = ar.Length; ` `        ``bool` `res = isOdd(ar, n); ` `     `  `        ``if` `(res) ` `            ``Console.WriteLine(``"Yes"``); ` `        ``else` `            ``Console.WriteLine(``"No"``); ` `    ``} ` `} ` ` `  `// This code is contributed by shivanisingh `

Output:

```Yes
```

Time Complexity: O(N)

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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.