Everyone knows that the sum of two even/odd numbers is even, but the sum of an even and an odd number is odd.

We iterate through the array starting from the 2nd element and check if the sum of the previous element and the current element is odd.

If the sum is odd, we increment the size of the current subarray; otherwise, we start a new subarray.

This approach is very similar to the second approach discussed earlier.

Code

#include <bits/stdc++.h>
using namespace std;

// compute and print length of the longest even-odd subarray (Kadane-like)
void evenodd_mathKadane(int arr[], int size) {
  // best length found
  int ans = 0;
  // current alternating length
  int count = 1;
  // scan from second element
  for (int i = 1; i < size; i++) {
    // extend when consecutive pair has odd sum (alternating parity)
    if ((arr[i - 1] + arr[i]) % 2 != 0) {
      // increase current length
      count++;
      // update best length
      ans = max(ans, count);
    }
    // restart when parity does not alternate
    else {
      // reset current length
      count = 1;
    }
  }
  // output result
  cout << "The length of the longest even-odd subarray is " << ans;
}

// program entry
int main() {
  // input array
  int arr[] = {1, 2, 3, 4, 5, 7, 9};
  // size of array
  int size = 7;
  // run computation
  evenodd_mathKadane(arr, size);
  // exit
  return 0;
}