2832. Maximal Range That Each Element Is Maximum in It 🔒

Description

You are given a 0-indexed array nums of distinct integers.

Let us define a 0-indexed array ans of the same length as nums in the following way:

• ans[i] is the maximum length of a subarray nums[l..r], such that the maximum element in that subarray is equal to nums[i].

Return the array ans.

Note that a subarray is a contiguous part of the array.

 

Example 1:

Input: nums = [1,5,4,3,6]
Output: [1,4,2,1,5]
Explanation: For nums[0] the longest subarray in which 1 is the maximum is nums[0..0] so ans[0] = 1.
For nums[1] the longest subarray in which 5 is the maximum is nums[0..3] so ans[1] = 4.
For nums[2] the longest subarray in which 4 is the maximum is nums[2..3] so ans[2] = 2.
For nums[3] the longest subarray in which 3 is the maximum is nums[3..3] so ans[3] = 1.
For nums[4] the longest subarray in which 6 is the maximum is nums[0..4] so ans[4] = 5.

Example 2:

Input: nums = [1,2,3,4,5]
Output: [1,2,3,4,5]
Explanation: For nums[i] the longest subarray in which it's the maximum is nums[0..i] so ans[i] = i + 1.

 

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 105
• All elements in nums are distinct.

Solutions

Solution 1: Monotonic Stack

This problem is a template for monotonic stack. We only need to use the monotonic stack to find the position of the first element larger than $nums[i]$ on the left and right, denoted as $left[i]$ and $right[i]$. Then, the interval length with $nums[i]$ as the maximum value is $right[i] - left[i] - 1$.

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

Python3JavaC++GoTypeScript

class Solution:
    def maximumLengthOfRanges(self, nums: List[int]) -> List[int]:
        n = len(nums)
        left = [-1] * n
        right = [n] * n
        stk = []
        for i, x in enumerate(nums):
            while stk and nums[stk[-1]] <= x:
                stk.pop()
            if stk:
                left[i] = stk[-1]
            stk.append(i)
        stk = []
        for i in range(n - 1, -1, -1):
            while stk and nums[stk[-1]] <= nums[i]:
                stk.pop()
            if stk:
                right[i] = stk[-1]
            stk.append(i)
        return [r - l - 1 for l, r in zip(left, right)]

class Solution {
    public int[] maximumLengthOfRanges(int[] nums) {
        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) {
            while (!stk.isEmpty() && nums[stk.peek()] <= nums[i]) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                left[i] = stk.peek();
            }
            stk.push(i);
        }
        stk.clear();
        for (int i = n - 1; i >= 0; --i) {
            while (!stk.isEmpty() && nums[stk.peek()] <= nums[i]) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                right[i] = stk.peek();
            }
            stk.push(i);
        }
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            ans[i] = right[i] - left[i] - 1;
        }
        return ans;
    }
}

class Solution {
public:
    vector<int> maximumLengthOfRanges(vector<int>& nums) {
        int n = nums.size();
        vector<int> left(n, -1);
        vector<int> right(n, n);
        stack<int> stk;
        for (int i = 0; i < n; ++i) {
            while (!stk.empty() && nums[stk.top()] <= nums[i]) {
                stk.pop();
            }
            if (!stk.empty()) {
                left[i] = stk.top();
            }
            stk.push(i);
        }
        stk = stack<int>();
        for (int i = n - 1; ~i; --i) {
            while (!stk.empty() && nums[stk.top()] <= nums[i]) {