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

# 1793. Maximum Score of a Good Subarray

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);
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
}

• function maximumScore(nums: number[], k: number): number {
const n = nums.length;
const left: number[] = Array(n).fill(-1);
const right: number[] = Array(n).fill(n);
const stk: number[] = [];
for (let i = 0; i < n; ++i) {
while (stk.length && nums[stk.at(-1)] >= nums[i]) {
stk.pop();
}
if (stk.length) {
left[i] = stk.at(-1);
}
stk.push(i);
}
stk.length = 0;
for (let i = n - 1; ~i; --i) {
while (stk.length && nums[stk.at(-1)] > nums[i]) {
stk.pop();
}
if (stk.length) {
right[i] = stk.at(-1);
}
stk.push(i);
}
let ans = 0;
for (let 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;
}