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Maximize count of odd-sum pairs in given Array with at most one conversion

Last Updated : 23 Dec, 2023
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Given an array of integers size n, find maximum number of pairs from array such that their sum is odd, by changing at most one number.

Example:

Input: N = 6, arr = [1, 5, 3, 6, 8, 0]
Output: 3
Explanation: There are 3 even and 3 odd numbers

Input: N = 8, arr = [1, 5, 3, 6, 8, 0, 2, 4]
Output: 4
Explanation: There are 5 even and 3 odd numbers, and an even number can be converted to an odd number. Hence final output will be 4.

Approach: For sum of element to be odd, one number should be odd and other number should be even. Below steps can be followed:

  • Calculate x which is equal to number of elements which are odd
  • Calculate Y which is equal to number of element which are even.
  • initialize the answer variable by minimum of x and y
  • if |x – y| >= 2, then increment answer by 1 (i.e., if y – x >= 2 then we can convert one odd number into even and if x – y >= 2 then we can convert one even number into odd number)
  • resultant answer variable is our required answer.

Below is the implementation of the above approach:

C++




#include <iostream>
 
// To find the maximum number of pairs in
// array with conversion of at most one element
int maximumNumberofpairs(int n, int arr[]) {
    // Initialize count of even elements
    int x = 0;
    // Initialize count of odd elements
    int y = 0;
    for (int i = 0; i < n; i++) {
        // If the current number is even, increment x by 1
        if (arr[i] % 2 == 0) {
            x++;
        }
        // If the current number is odd, increment y by 1
        else {
            y++;
        }
    }
    // Initialize the answer by min(x, y)
    int answer = std::min(x, y);
    // If the difference in count of odd and even
    // is more than 2, then increment the answer
    if (std::abs(x - y) >= 2) {
        answer++;
    }
    // Return the final answer
    return answer;
}
 
// Driver code
int main() {
    // Given array
    int arr[] = {1, 2, 4, 6, 5, 10, 12};
    int n = sizeof(arr) / sizeof(arr[0]);
    std::cout << maximumNumberofpairs(n, arr) << std::endl;
    return 0;
}


Java




import java.io.*;
 
class GFG {
    // To find the maximum number of pairs in
    // array with conversion of at most one element
    static int maximumNumberofpairs(int n, int arr[]) {
        // Initialize count of even elements
        int x = 0;
        // Initialize count of odd elements
        int y = 0;
        for (int i = 0; i < n; i++) {
            // If the current number is even, increment x by 1
            if (arr[i] % 2 == 0) {
                x++;
            }
            // If the current number is odd, increment y by 1
            else {
                y++;
            }
        }
        // Initialize the answer by min(x, y)
        int answer = Math.min(x, y);
        // If the difference in count of odd and even
        // is more than 2, then increment the answer
        if (Math.abs(x - y) >= 2) {
            answer++;
        }
        // Return the final answer
        return answer;
    }
 
    // Driver code
    public static void main(String[] args) {
        // Given array
        int arr[] = {1, 2, 4, 6, 5, 10, 12};
        int n = arr.length;
        System.out.println(maximumNumberofpairs(n, arr));
    }
}


Python3




# To find the maximum number of pairs in
# array with conversion of at most one element
def maximumNumberofpairs(n, arr):
    # Initialize count of even elements
    x = 0
    # Initialize count of odd elements
    y = 0
    for i in range(n):
        # If the current number is even, increment x by 1
        if arr[i] % 2 == 0:
            x += 1
        # If the current number is odd, increment y by 1
        else:
            y += 1
    # Initialize the answer by min(x, y)
    answer = min(x, y)
    # If the difference in count of odd and even
    # is more than 2, then increment the answer
    if abs(x - y) >= 2:
        answer += 1
    # Return the final answer
    return answer
 
# Driver code
if __name__ == '__main__':
    # Given array
    arr = [1, 2, 4, 6, 5, 10, 12]
    n = len(arr)
    print(maximumNumberofpairs(n, arr))


C#




using System;
 
class GFG {
    // To find the maximum number of pairs in
    // array with conversion of at most one element
    static int maximumNumberofpairs(int n, int[] arr) {
        // Initialize count of even elements
        int x = 0;
        // Initialize count of odd elements
        int y = 0;
        for (int i = 0; i < n; i++) {
            // If the current number is even, increment x by 1
            if (arr[i] % 2 == 0) {
                x++;
            }
            // If the current number is odd, increment y by 1
            else {
                y++;
            }
        }
        // Initialize the answer by min(x, y)
        int answer = Math.Min(x, y);
        // If the difference in count of odd and even
        // is more than 2, then increment the answer
        if (Math.Abs(x - y) >= 2) {
            answer++;
        }
        // Return the final answer
        return answer;
    }
 
    // Driver code
    public static void Main(string[] args) {
        // Given array
        int[] arr = {1, 2, 4, 6, 5, 10, 12};
        int n = arr.Length;
        Console.WriteLine(maximumNumberofpairs(n, arr));
    }
}


Javascript




// To find the maximum number of pairs in
// array with conversion of at most one element
function maximumNumberofpairs(n, arr) {
    // Initialize count of even elements
    var x = 0;
    // Initialize count of odd elements
    var y = 0;
    for (var i = 0; i < n; i++) {
        // If the current number is even, increment x by 1
        if (arr[i] % 2 == 0) {
            x++;
        }
        // If the current number is odd, increment y by 1
        else {
            y++;
        }
    }
    // Initialize the answer by min(x, y)
    var answer = Math.min(x, y);
    // If the difference in count of odd and even
    // is more than 2, then increment the answer
    if (Math.abs(x - y) >= 2) {
        answer++;
    }
    // Return the final answer
    console.log(answer);
}
 
// Driver code
// Given array
var arr = [1, 2, 4, 6, 5, 10, 12];
var n = arr.length;
maximumNumberofpairs(n, arr)


Output

3

Time complexity: O(N)
Auxiliary Space: O(1)



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