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

Code

Two Pointer

Time Complexity: $O(n)$, Space Complexity: $O(1)$

Intuition：我們想要 mini 盡可能的大，且 j - i + 1 盡可能的長。因此由 index k 開始，往左右兩側延伸，取大者（因為要 mini 盡可能的大）。

這個 intuition （i, j 向左向右移動的優先順序）類似 Container With Most Water，只是在這題是由內往兩側延伸，而在 Container With Most Water 中是由兩側往中心點去找。

We start from k and track the current maximum score. We have two choices - expand left or right. To maximize the score, we expand towards the higher number.

class Solution {
public:
    int maximumScore(vector<int>& nums, int k) {
        int res = nums[k], mini = nums[k], i = k, j = k, n = nums.size();
        while(i > 0 || j < n - 1) {
            if((i > 0 ? nums[i - 1] : 0) < (j < n - 1 ? nums[j + 1] : 0)) {
                mini = min(mini, nums[++j]);
            } else {
                mini = min(mini, nums[--i]);
            }
            res = max(res, mini * (j - i + 1));
        }
 
        return res;
    }
};

Monotonic Stack

Time Complexity: $O(n)$, Space Complexity: $O(1)$

其實就是在找 Largest Rectangle in Histogram。只是多了一個 constraint： if(idx < k && i > k)。

class Solution {
public:
    int maximumScore(vector<int>& nums, int k) {
        stack<int> s;
        nums.push_back(0);
 
        int res = 0;
        for(int i = 0; i < nums.size(); i++) {
            while(!s.empty() && nums[s.top()] >= nums[i]) {
                int h = nums[s.top()];
                s.pop();
                int idx = s.empty() ? -1 : s.top();
                int width = i - idx - 1;
                int area = h * width;
 
                if(idx < k && i > k)
                    res = max(res, area);
            }
            s.push(i);
        }
 
        return res;
    }
};

Source

• Maximum Score of a Good Subarray - LeetCode