README_EN.md

comments	true
difficulty	Medium
edit_url	https://github.com/doocs/leetcode/edit/main/solution/0900-0999/0962.Maximum%20Width%20Ramp/README_EN.md
tags	Stack Array Two Pointers Monotonic Stack

962. Maximum Width Ramp

中文文档

Description

A ramp in an integer array nums is a pair (i, j) for which i < j and nums[i] <= nums[j]. The width of such a ramp is j - i.

Given an integer array nums, return the maximum width of a ramp in nums. If there is no ramp in nums, return 0.

 

Example 1:

Input: nums = [6,0,8,2,1,5]
Output: 4
Explanation: The maximum width ramp is achieved at (i, j) = (1, 5): nums[1] = 0 and nums[5] = 5.

Example 2:

Input: nums = [9,8,1,0,1,9,4,0,4,1]
Output: 7
Explanation: The maximum width ramp is achieved at (i, j) = (2, 9): nums[2] = 1 and nums[9] = 1.

 

Constraints:

• 2 <= nums.length <= 5 * 104
• 0 <= nums[i] <= 5 * 104

Solutions

Solution 1: Monotonic Stack

According to the problem, we can find that the subsequence formed by all possible $\textit{nums}[i]$ must be monotonically decreasing. Why is that? Let's prove it by contradiction.

Suppose there exist $i_1<i_2$ and $\textit{nums}[i_1]\leq\textit{nums}[i_2]$, then actually $\textit{nums}[i_2]$ cannot be a candidate value, because $\textit{nums}[i_1]$ is more to the left and would be a better value. Therefore, the subsequence formed by $\textit{nums}[i]$ must be monotonically decreasing, and $i$ must start from 0.

We use a monotonically decreasing stack $\textit{stk}$ (from bottom to top) to store all possible $\textit{nums}[i]$. Then we traverse $j$ starting from the right boundary. If we encounter $\textit{nums}[\textit{stk.top()}]\leq\textit{nums}[j]$, it means a ramp is formed at this point. We cyclically pop the top elements from the stack and update $\textit{ans}$.

The time complexity is $O(n)$ and the space complexity is $O(n)$, where $n$ represents the length of $\textit{nums}$.

Python3

class Solution:
    def maxWidthRamp(self, nums: List[int]) -> int:
        stk = []
        for i, v in enumerate(nums):
            if not stk or nums[stk[-1]] > v:
                stk.append(i)
        ans = 0
        for i in range(len(nums) - 1, -1, -1):
            while stk and nums[stk[-1]] <= nums[i]:
                ans = max(ans, i - stk.pop())
            if not stk:
                break
        return ans

Java

class Solution {
    public int maxWidthRamp(int[] nums) {
        int n = nums.length;
        Deque<Integer> stk = new ArrayDeque<>();
        for (int i = 0; i < n; ++i) {
            if (stk.isEmpty() || nums[stk.peek()] > nums[i]) {
                stk.push(i);
            }
        }
        int ans = 0;
        for (int i = n - 1; i >= 0; --i) {
            while (!stk.isEmpty() && nums[stk.peek()] <= nums[i]) {
                ans = Math.max(ans, i - stk.pop());
            }
            if (stk.isEmpty()) {
                break;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int maxWidthRamp(vector<int>& nums) {
        int n = nums.size();
        stack<int> stk;
        for (int i = 0; i < n; ++i) {
            if (stk.empty() || nums[stk.top()] > nums[i]) stk.push(i);
        }
        int ans = 0;
        for (int i = n - 1; i; --i) {
            while (!stk.empty() && nums[stk.top()] <= nums[i]) {
                ans = max(ans, i - stk.top());
                stk.pop();
            }
            if (stk.empty()) break;
        }
        return ans;
    }
};

Go

func maxWidthRamp(nums []int) int {
	n := len(nums)
	stk := []int{}
	for i, v := range nums {
		if len(stk) == 0 || nums[stk[len(stk)-1]] > v {
			stk = append(stk, i)
		}
	}
	ans := 0
	for i := n - 1; i >= 0; i-- {
		for len(stk) > 0 && nums[stk[len(stk)-1]] <= nums[i] {
			ans = max(ans, i-stk[len(stk)-1])
			stk = stk[:len(stk)-1]
		}
		if len(stk) == 0 {
			break
		}
	}
	return ans
}

TypeScript

function maxWidthRamp(nums: number[]): number {
    let [ans, n] = [0, nums.length];
    const stk: number[] = [];

    for (let i = 0; i < n - 1; i++) {
        if (stk.length === 0 || nums[stk.at(-1)!] > nums[i]) {
            stk.push(i);
        }
    }

    for (let i = n - 1; i >= 0; i--) {
        while (stk.length && nums[stk.at(-1)!] <= nums[i]) {
            ans = Math.max(ans, i - stk.pop()!);
        }
        if (stk.length === 0) break;
    }

    return ans;
}

JavaScript

function maxWidthRamp(nums) {
    let [ans, n] = [0, nums.length];
    const stk = [];

    for (let i = 0; i < n - 1; i++) {
        if (stk.length === 0 || nums[stk.at(-1)] > nums[i]) {
            stk.push(i);
        }
    }

    for (let i = n - 1; i >= 0; i--) {
        while (stk.length && nums[stk.at(-1)] <= nums[i]) {
            ans = Math.max(ans, i - stk.pop());
        }
        if (stk.length === 0) break;
    }

    return ans;
}