What is the time complexity of Maximum Width Ramp?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Maximum Width Ramp?

This problem uses Array, Two Pointers, Stack, Monotonic Stack. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Width Ramp?

Insight 1: Constructing a monotonic stack of indices is crucial. We need to maintain a stack of indices that correspond to decreasing elements in the input array. This allows us to efficiently find potential starting points (i) for the ramp. Insight 2: Iterating from the end of the array allows us to maximize the width of the ramp. By scanning from the end, we can try to match each element with the smallest element to its left within the built stack to get the maximum possible width.

Which companies ask LeetCode 962. Maximum Width Ramp in interviews?

This problem is frequently asked at Zepto.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Maximum Width Ramp - LeetCode 962 Solution

Q: How to solve Maximum Width Ramp?

The solution utilizes a monotonic stack to efficiently find the maximum width ramp. First, it constructs a stack of indices corresponding to decreasing elements in the input array. Then, it iterates through the array from the end, attempting to form a ramp with the elements in the stack. As it iterates from the end, it checks if the current element is greater than or equal to the element at the top of the stack. If it is, a ramp is found, and the width is calculated. The algorithm keeps track of the maximum width found so far and returns it at the end.

Maximum Width Ramp - Complete Solution Guide

Maximum Width Ramp is LeetCode problem 962, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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,

Detailed Explanation

The problem asks us to find the maximum width of a 'ramp' in an array of integers. A ramp is defined as a pair of indices (i, j) such that i < j and nums[i] <= nums[j]. The width of the ramp is j - i. The goal is to find the pair (i, j) that satisfies these conditions and maximizes the width j - i. If no such ramp exists, we should return 0. The input is an array of integers, and the output is the maximum width of the ramp.

Solution Approach

The solution utilizes a monotonic stack to efficiently find the maximum width ramp. First, it constructs a stack of indices corresponding to decreasing elements in the input array. Then, it iterates through the array from the end, attempting to form a ramp with the elements in the stack. As it iterates from the end, it checks if the current element is greater than or equal to the element at the top of the stack. If it is, a ramp is found, and the width is calculated. The algorithm keeps track of the maximum width found so far and returns it at the end.

Step-by-Step Algorithm

Step 1: Initialize an empty stack `s` to store indices and a variable `max_width` to 0.
Step 2: Iterate through the input array `nums` from left to right (index `i`).
Step 3: For each element `nums[i]`, check if the stack is empty or if `nums[i]` is smaller than the element at the index on top of the stack (`nums[s[-1]]`).
Step 4: If the condition in step 3 is true, push the current index `i` onto the stack. This builds a decreasing stack of indices, ensuring the stack only holds the indices of potential starting points for the ramp.
Step 5: Iterate through the input array `nums` from right to left (index `j`).
Step 6: While the stack is not empty and the element `nums[j]` is greater than or equal to the element at the index on top of the stack (`nums[s[-1]]`), pop the index from the stack.
Step 7: Calculate the width of the ramp as `j - popped_index`. Update `max_width` to be the maximum of `max_width` and the calculated width.
Step 8: After iterating through the array from right to left, return `max_width`.

Key Insights

Insight 1: Constructing a monotonic stack of indices is crucial. We need to maintain a stack of indices that correspond to decreasing elements in the input array. This allows us to efficiently find potential starting points (i) for the ramp.
Insight 2: Iterating from the end of the array allows us to maximize the width of the ramp. By scanning from the end, we can try to match each element with the smallest element to its left within the built stack to get the maximum possible width.
Insight 3: The stack only stores potential 'left' ends of the ramp. Indices are added to the stack if the element at that index is smaller than any element already represented in the stack. This ensures that we only consider the smallest possible 'left' element for a ramp, which is optimal for maximizing width.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Two Pointers, Stack, Monotonic Stack.

Companies

Asked at: Zepto.

Maximum Width Ramp | LeetCode 962

Medium

Array Two Pointers Stack Monotonic Stack

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.Stack;

class Solution {
    public int maxWidthRamp(int[] nums) {
        int n = nums.length;
        Stack<Integer> s = new Stack<>();
        int max_width = 0;

        for (int i = 0; i < n; i++) {
            if (s.isEmpty() || nums[i] < nums[s.peek()]) {
                s.push(i);
            }
        }

        for (int j = n - 1; j >= 0; j--) {
            while (!s.isEmpty() && nums[j] >= nums[s.peek()]) {
                int i = s.pop();
                max_width = Math.max(max_width, j - i);
            }
        }

        return max_width;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 962: Maximum Width Ramp - Detailed explanation and solution approaches.

Topics

Array Two Pointers Stack Monotonic Stack

Companies

Zepto

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:962

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize an empty stack `s` to store indices and a variable `max_width` to 0.
2
Step 2: Iterate through the input array `nums` from left to right (index `i`).
3
Step 3: For each element `nums[i]`, check if the stack is empty or if `nums[i]` is smaller than the element at the index on top of the stack (`nums[s[-1]]`).
4
Step 4: If the condition in step 3 is true, push the current index `i` onto the stack. This builds a decreasing stack of indices, ensuring the stack only holds the indices of potential starting points for the ramp.
5
Step 5: Iterate through the input array `nums` from right to left (index `j`).
6
Step 6: While the stack is not empty and the element `nums[j]` is greater than or equal to the element at the index on top of the stack (`nums[s[-1]]`), pop the index from the stack.
7
Step 7: Calculate the width of the ramp as `j - popped_index`. Update `max_width` to be the maximum of `max_width` and the calculated width.
8
Step 8: After iterating through the array from right to left, return `max_width`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because we iterate through the array twice in separate loops. Each element is pushed and popped from the stack at most once. The space complexity is O(n) because the stack can grow up to the size of the input array in the worst-case scenario (decreasing array).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Iterate through all possible pairs (i, j) with i < j and check if nums[i] <= nums[j]. Keep track of the maximum width.

Time: O(n^2), where n is the length of the input array.Space: O(1).

Trade-offs: Simple to implement but inefficient for large arrays. The brute force approach is only suitable for smaller input sizes due to its quadratic time complexity. The stack-based approach is much more efficient for larger datasets.

✨

Key Insights

1Insight 1: Constructing a monotonic stack of indices is crucial. We need to maintain a stack of indices that correspond to decreasing elements in the input array. This allows us to efficiently find potential starting points (i) for the ramp.
2Insight 2: Iterating from the end of the array allows us to maximize the width of the ramp. By scanning from the end, we can try to match each element with the smallest element to its left within the built stack to get the maximum possible width.
3Insight 3: The stack only stores potential 'left' ends of the ramp. Indices are added to the stack if the element at that index is smaller than any element already represented in the stack. This ensures that we only consider the smallest possible 'left' element for a ramp, which is optimal for maximizing width.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Edge Cases

Edge case 1: Empty array. The code implicitly handles an empty input array because the loops won't execute, and the initial value of `max_width` (0) will be returned.
Edge case 2: Array with all elements the same. In this case, the stack will contain only the index of the first element. The outer loop from the end will compare each element to the first element, and `max_width` will be updated to `n - 1` if n > 0.
Edge case 3: Decreasing array. The stack will contain all indices, and the `while` loop inside the second `for` loop will find one ramp for each index j from end until j = 0, since numbers[j] can be equal to numbers[i] where i is the index at the top of the stack. It finds ramp, by comparing the smallest numbers with indices from n-1 to 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using a stack and trying to directly iterate through the array. This can lead to an inefficient solution with a higher time complexity.

✗Common mistake 2: Incorrectly implementing the stack operations or the comparison conditions. This can result in incorrect results or infinite loops.

✗Common mistake 3: Not initializing the `max_width` to 0. If there are no ramps, the function should return 0. If you don't initialize `max_width`, you might return garbage data from memory.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Stock market analysis. Finding patterns where a stock price decreases and then increases later can be similar to finding a ramp. The width can represent the time elapsed between the low and the high, and it can be used to assess the strength of the pattern.

🏦Real-world scenario 2: Signal processing. Identifying trends in data streams where values decrease and then increase. The ramp represents the period of increase after a drop.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 2, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Width Ramp"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?