2786. Visit Array Positions to Maximize Score

Description

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.

Example 1:

Input: nums = [2,3,6,1,9,2], x = 5
Output: 13
Explanation: We can visit the following positions in the array: 0 -> 2 -> 3 -> 4.
The corresponding values are 2, 6, 1 and 9. Since the integers 6 and 1 have different parities, the move 2 -> 3 will make you lose a score of x = 5.
The total score will be: 2 + 6 + 1 + 9 - 5 = 13.

Example 2:

Input: nums = [2,4,6,8], x = 3
Output: 20
Explanation: All the integers in the array have the same parities, so we can visit all of them without losing any score.
The total score is: 2 + 4 + 6 + 8 = 20.

Constraints:

2 <= nums.length <= 10⁵
1 <= nums[i], x <= 10⁶

Solutions

Solution 1: Dynamic Programming

Based on the problem description, we can draw the following conclusions:

Moving from position $i$ to position $j$, if $nums[i]$ and $nums[j]$ have different parities, then $x$ points will be lost;
Moving from position $i$ to position $j$, if $nums[i]$ and $nums[j]$ have the same parity, then no points will be lost.

Therefore, we can use an array $f$ of length $2$ to represent the maximum score when the current position's parity is $0$ and $1$. Initially, the values of $f$ are $-\infty$, and then we initialize $f[nums[0] \& 1] = nums[0]$, indicating the score at the initial position.

Next, we start traversing the array $nums$ from position $1$. For each position $i$ corresponding to the value $v$, we update the value of $f[v \& 1]$ to be the larger value between $f[v \& 1]$ and $f[v \& 1 \oplus 1] - x$ plus $v$, i.e., $f[v \& 1] = \max(f[v \& 1], f[v \& 1 \oplus 1] - x) + v$.

The answer is the larger value between $f[0]$ and $f[1]$.

The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.

Python3JavaC++GoTypeScript

class Solution:
    def maxScore(self, nums: List[int], x: int) -> int:
        f = [-inf] * 2
        f[nums[0] & 1] = nums[0]
        for v in nums[1:]:
            f[v & 1] = max(f[v & 1], f[v & 1 ^ 1] - x) + v
        return max(f)

class Solution {
    public long maxScore(int[] nums, int x) {
        long[] f = new long[2];
        Arrays.fill(f, -(1L << 60));
        f[nums[0] & 1] = nums[0];
        for (int i = 1; i < nums.length; ++i) {
            int v = nums[i];
            f[v & 1] = Math.max(f[v & 1], f[v & 1 ^ 1] - x) + v;
        }
        return Math.max(f[0], f[1]);
    }
}

class Solution {
public:
    long long maxScore(vector<int>& nums, int x) {
        const long long inf = 1LL << 60;
        vector<long long> f(2, -inf);
        f[nums[0] & 1] = nums[0];
        int n = nums.size();
        for (int i = 1; i < n; ++i) {
            int v = nums[i];
            f[v & 1] = max(f[v & 1], f[v & 1 ^ 1] - x) + v;
        }
        return max(f[0], f[1]);
    }
};

func maxScore(nums []int, x int) int64 {
    const inf int = 1 << 40
    f := [2]int{-inf, -inf}
    f[nums[0]&1] = nums[0]
    for _, v := range nums[1:] {
        f[v&1] = max(f[v&1], f[v&1^1]-x) + v
    }
    return int64(max(f[0], f[1]))
}

function maxScore(nums: number[], x: number): number {
    const f: number[] = Array(2).fill(-Infinity);
    f[nums[0] & 1] = nums[0];
    for (let i = 1; i < nums.length; ++i) {
        const v = nums[i];
        f[v & 1] = Math.max(f[v & 1], f[(v & 1) ^ 1] - x) + v;
    }
    return Math.max(...f);
}

2786. Visit Array Positions to Maximize Score

Description

Solutions

Solution 1: Dynamic Programming

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