2786. Visit Array Positions to Maximize Score

You are given a 0-indexed integer array nums and a positive integer x.

You are initially at position 0 in the array and you can visit other positions according to the following rules:

• If you are currently in position i, then you can move to any position j such that i < j.
• For each position i that you visit, you get a score of nums[i].
• If you move from a position i to a position j and the parities of nums[i] and nums[j] differ, then you lose a score of x.

Return the maximum total score you can get.

Note that initially you have nums[0] points.

class Solution {
public:
    long long maxScore(vector<int>& nums, int x) {
        long long dp1 = nums[0] & 1 ? INT_MIN : nums[0], dp2 = nums[0] & 1 ? nums[0] : INT_MIN;
        for(int i = 1; i < nums.size(); i++) {
            if(nums[i] & 1) {
                dp2 = max(dp2 + nums[i], dp1 + nums[i] - x);
            } else {
                dp1 = max(dp1 + nums[i], dp2 + nums[i] - x);
            }
        }
        return max(dp1, dp2);
    }
};