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1793. Maximum Score of a Good Subarray

Description

You are given an array of integers nums (0-indexed) and an integer k.

The score of a subarray (i, j) is defined as min(nums[i], nums[i+1], ..., nums[j]) * (j - i + 1). A good subarray is a subarray where i <= k <= j.

Return the maximum possible score of a good subarray.

 

Example 1:

Input: nums = [1,4,3,7,4,5], k = 3
Output: 15
Explanation: The optimal subarray is (1, 5) with a score of min(4,3,7,4,5) * (5-1+1) = 3 * 5 = 15.

Example 2:

Input: nums = [5,5,4,5,4,1,1,1], k = 0
Output: 20
Explanation: The optimal subarray is (0, 4) with a score of min(5,5,4,5,4) * (4-0+1) = 4 * 5 = 20.

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 2 * 104
  • 0 <= k < nums.length

Solutions

Solution 1: Monotonic Stack

We can enumerate each element $nums[i]$ in $nums$ as the minimum value of the subarray, and use a monotonic stack to find the first position $left[i]$ on the left that is less than $nums[i]$ and the first position $right[i]$ on the right that is less than or equal to $nums[i]$. Then, the score of the subarray with $nums[i]$ as the minimum value is $nums[i] \times (right[i] - left[i] - 1)$.

It should be noted that the answer can only be updated when the left and right boundaries $left[i]$ and $right[i]$ satisfy $left[i]+1 \leq k \leq right[i]-1$.

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

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class Solution:
    def maximumScore(self, nums: List[int], k: int) -> int:
        n = len(nums)
        left = [-1] * n
        right = [n] * n
        stk = []
        for i, v in enumerate(nums):
            while stk and nums[stk[-1]] >= v:
                stk.pop()
            if stk:
                left[i] = stk[-1]
            stk.append(i)
        stk = []
        for i in range(n - 1, -1, -1):
            v = nums[i]
            while stk and nums[stk[-1]] > v:
                stk.pop()
            if stk:
                right[i] = stk[-1]
            stk.append(i)
        ans = 0
        for i, v in enumerate(nums):
            if left[i] + 1 <= k <= right[i] - 1:
                ans = max(ans, v * (right[i] - left[i] - 1))
        return ans

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class Solution {
    public int maximumScore(int[] nums, int k) {
        int n = nums.length;
        int[] left = new int[n];
        int[] right = new int[n];
        Arrays.fill(left, -1);
        Arrays.fill(right, n);
        Deque<Integer> stk = new ArrayDeque<>();
        for (int i = 0; i < n; ++i) {
            int v = nums[i];
            while (!stk.isEmpty() && nums[stk.peek()] >= v) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                left[i] = stk.peek();
            }
            stk.push(i);
        }
        stk.clear();
        for (int i = n - 1; i >= 0; --i) {
            int v = nums[i];
            while (!stk.isEmpty() && nums[stk.peek()] > v) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                right[i] = stk.peek();
            }
            stk.push(i);
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (left[i] + 1 <= k && k <= right[i] - 1) {
                ans = Math.max(ans, nums[i] * (right[i] - left[i] - 1));
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int maximumScore(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> left(n, -1);
        vector<int> right(n, n);
        stack<int> stk;
        for (int i = 0; i < n; ++i)