# 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.

• 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]) {
stk.pop();
}
if (!stk.empty()) {
right[i] = stk.top();
}
stk.push(i);
}
vector<int> ans(n);
for (int i = 0; i < n; ++i) {
ans[i] = right[i] - left[i] - 1;
}
return ans;
}
};

• 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)]

• func maximumLengthOfRanges(nums []int) []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, x := range nums {
for len(stk) > 0 && nums[stk[len(stk)-1]] <= x {
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-- {
x := nums[i]
for len(stk) > 0 && nums[stk[len(stk)-1]] <= x {
stk = stk[:len(stk)-1]
}
if len(stk) > 0 {
right[i] = stk[len(stk)-1]
}
stk = append(stk, i)
}
ans := make([]int, n)
for i := range ans {
ans[i] = right[i] - left[i] - 1
}
return ans
}

• function maximumLengthOfRanges(nums: 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 >= 0; --i) {
while (stk.length && nums[stk.at(-1)] <= nums[i]) {
stk.pop();
}
if (stk.length) {
right[i] = stk.at(-1);
}
stk.push(i);
}
return left.map((l, i) => right[i] - l - 1);
}