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Formatted question description: https://leetcode.ca/all/1793.html

1793. Maximum Score of a Good Subarray

Level

Hard

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 <= 10^5
  • 1 <= nums[i] <= 2 * 10^4
  • 0 <= k < nums.length

Solution

Loop over nums and use a monotonic stack to store each element’s edges, where the elements increase from botton to top of the monotonic stack.

For each 0 <= i < nums.length, use left[i] to store the maximum j such that j < i and nums[j] < nums[i], and use right[i] to store the minimum j such that j > i and nums[j] < nums[i]. Suppose nums[-1] = nums[nums.length] = -1.

For each 0 <= i < nums.length, if right[i] > k and left[i] < k, calculate the score at index i as nums[i] * (right[i] - left[i] - 1). Maintain the maximum score and return the maximum score.

  • class Solution {
    public int maximumScore(int[] nums, int k) {
        int length = nums.length;
        int[] left = new int[length];
        int[] right = new int[length];
        Arrays.fill(right, length);
        Deque<Integer> stack = new LinkedList<Integer>();
        for (int i = 0; i < length; i++) {
            while (!stack.isEmpty() && nums[stack.peek()] >= nums[i])
                right[stack.pop()] = i;
            left[i] = stack.isEmpty() ? -1 : stack.peek();
            stack.push(i);
        }
        int score = 0;
        for (int i = 0; i < length; i++) {
            if (left[i] < k && right[i] > k)
                score = Math.max(score, nums[i] * (right[i] - left[i] - 1));
        }
        return score;
    }
}

############

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;
    }
}

  • // OJ: https://leetcode.com/contest/weekly-contest-232/problems/maximum-score-of-a-good-subarray/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int maximumScore(vector<int>& A, int k) {
        int i = k - 1, j = k + 1, N = A.size(), ans = A[k], mn = A[k];
        while (i >= 0 || j < N) {
            if (i < 0 || (j < N && A[j] > A[i])) {
                mn = min(mn, A[j]);
                ++j;
            } else {
                mn = min(mn, A[i]);
                --i;
            }
            ans = max(ans, mn * (j - i - 1));
        }
        return ans;
    }
};

  • 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

############

# 1793. Maximum Score of a Good Subarray
# https://leetcode.com/problems/maximum-score-of-a-good-subarray/

class Solution:
    def maximumScore(self, nums: List[int], k: int) -> int:
        res = mini = nums[k]
        i, j, n = k, k, len(nums)
        
        while i > 0 or j < n - 1:
            if (nums[i-1] if i > 0 else 0) < (nums[j + 1] if j < n - 1 else 0):
                j += 1
            else:
                i -= 1
            
            mini = min(mini, nums[i], nums[j])
            res = max(res, mini * (j - i + 1))
        
        return res


  • func maximumScore(nums []int, k int) (ans int) {
	n := len(nums)
	left := make([]int, n)
	right := make([]int, n)
	for i := range left {
		left[i] = -1
		right[i] = n
	}
	stk := []int{}
	for i, v := range nums {
		for len(stk) > 0 && nums[stk[len(stk)-1]] >= v {
			stk = stk[:len(stk)-1]
		}
		if len(stk) > 0 {
			left[i] = stk[len(stk)-1]
		}
		stk = append(stk, i)
	}
	stk = []int{}
	for i := n - 1; i >= 0; i-- {
		v := nums[i]
		for len(stk) > 0 && nums[stk[len(stk)-1]] > v {
			stk = stk[:len(stk)-1]
		}
		if len(stk) > 0 {
			right[i] = stk[len(stk)-1]
		}
		stk = append(stk, i)
	}
	for i, v := range nums {
		if left[i]+1 <= k && k <= right[i]-1 {
			ans = max(ans, v*(right[i]-left[i]-1))
		}
	}
	return
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

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