What is the time complexity of Next Greater Element I?

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

What data structures are used to solve Next Greater Element I?

This problem uses Array, Hash Table, Stack, Monotonic Stack. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Next Greater Element I?

Insight 1: The key observation is that we need to efficiently find the next greater element for each number in `nums2` to avoid redundant searches. We can precompute these and then use them to answer queries for `nums1` quickly. Insight 2: A stack can be used to keep track of potential next greater elements. The stack maintains a decreasing order, and when we encounter a larger element, we can pop elements from the stack and map them to the larger element as their next greater element.

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Next Greater Element I - LeetCode 496 Solution

Q: How to solve Next Greater Element I?

The optimal solution utilizes a monotonic stack and a hash map. The monotonic stack helps efficiently determine the next greater element for each number in `nums2`. We iterate through `nums2`. While the stack is not empty and the current element in `nums2` is greater than the top of the stack, we pop the top element and store the current element as its next greater element in the hash map. After processing `nums2`, we iterate through `nums1` and look up the next greater element for each number in the hash map. If a number is not found in the hash map, its next greater element is -1.

Q: Which companies ask LeetCode 496. Next Greater Element I in interviews?

This problem is frequently asked at Accenture, Agoda, Flipkart, Goldman Sachs, Morgan Stanley and 2 more companies.

Next Greater Element I - Complete Solution Guide

Next Greater Element I is LeetCode problem 496, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

The next greater element of some element x in an array is the first greater element that is to the right of x in the same array. You are given two distinct 0-indexed integer arrays nums1 and nums2 , where nums1 is a subset of nums2 . For each 0 <= i < nums1.length , find the index j such that nums1[i] == nums2[j] and determine the next greater element of nums2[j] in nums2 . If there is no next greater element, then the answer for this query is -1 . Return an array ans of length nums1.length such

Detailed Explanation

The problem asks us to find the 'next greater element' for each number in `nums1` within `nums2`, where `nums1` is a subset of `nums2`. The next greater element for a number `x` is the first element to the right of `x` in `nums2` that is strictly greater than `x`. If no such element exists, we return -1. For each element in `nums1`, we need to find its corresponding position in `nums2`, then search for its next greater element to the right. The inputs are two integer arrays, `nums1` and `nums2`, where `nums1` is a subset of `nums2`, and all elements are unique. The output is an array of the same length as `nums1`, containing the next greater element for each corresponding element in `nums1` within `nums2`.

Solution Approach

The optimal solution utilizes a monotonic stack and a hash map. The monotonic stack helps efficiently determine the next greater element for each number in `nums2`. We iterate through `nums2`. While the stack is not empty and the current element in `nums2` is greater than the top of the stack, we pop the top element and store the current element as its next greater element in the hash map. After processing `nums2`, we iterate through `nums1` and look up the next greater element for each number in the hash map. If a number is not found in the hash map, its next greater element is -1.

Step-by-Step Algorithm

Step 1: Initialize an empty stack and a hash map to store the next greater element for each number.
Step 2: Iterate through `nums2`.
Step 3: While the stack is not empty and the current element in `nums2` is greater than the element at the top of the stack:
Step 4: Pop the element from the stack and store the current element of `nums2` as its next greater element in the hash map.
Step 5: Push the current element of `nums2` onto the stack.
Step 6: After iterating through `nums2`, initialize an array to store the results for `nums1`.
Step 7: Iterate through `nums1`.
Step 8: For each element in `nums1`, look up its next greater element in the hash map. If it exists, add it to the result array; otherwise, add -1.
Step 9: Return the result array.

Key Insights

Insight 1: The key observation is that we need to efficiently find the next greater element for each number in `nums2` to avoid redundant searches. We can precompute these and then use them to answer queries for `nums1` quickly.
Insight 2: A stack can be used to keep track of potential next greater elements. The stack maintains a decreasing order, and when we encounter a larger element, we can pop elements from the stack and map them to the larger element as their next greater element.
Insight 3: Using a hash map (or unordered_map in C++) to store the mapping between elements and their next greater elements allows for quick lookups.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m)

Topics

This problem involves: Array, Hash Table, Stack, Monotonic Stack.

Companies

Asked at: Accenture, Agoda, Flipkart, Goldman Sachs, Morgan Stanley, Swiggy, Tencent.

Next Greater Element I | LeetCode 496

Easy

Array Hash Table Stack Monotonic Stack

Time: O(m*n)Space: O(m)

Solution Code

Available in:

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        Map<Integer, Integer> map = new HashMap<>();
        Stack<Integer> stack = new Stack<>();
        for (int num : nums2) {
            while (!stack.isEmpty() && stack.peek() < num) {
                map.put(stack.pop(), num);
            }
            stack.push(num);
        }
        int[] result = new int[nums1.length];
        for (int i = 0; i < nums1.length; i++) {
            result[i] = map.getOrDefault(nums1[i], -1);
        }
        return result;
    }
}

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Problem Info

Description

Problem 496: Next Greater Element I - Detailed explanation and solution approaches.

Topics

Array Hash Table Stack Monotonic Stack

Companies

AccentureAgodaFlipkartGoldman SachsMorgan StanleySwiggyTencent

Stats

Difficulty:Easy

Languages:4

Companies:7

Problem #:496

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

9 steps

1
Step 1: Initialize an empty stack and a hash map to store the next greater element for each number.
2
Step 2: Iterate through `nums2`.
3
Step 3: While the stack is not empty and the current element in `nums2` is greater than the element at the top of the stack:
4
Step 4: Pop the element from the stack and store the current element of `nums2` as its next greater element in the hash map.
5
Step 5: Push the current element of `nums2` onto the stack.
6
Step 6: After iterating through `nums2`, initialize an array to store the results for `nums1`.
7
Step 7: Iterate through `nums1`.
8
Step 8: For each element in `nums1`, look up its next greater element in the hash map. If it exists, add it to the result array; otherwise, add -1.
9
Step 9: Return the result array.

💡

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(m*n)

💾

Space Complexity

O(m)

📝

Why?

The time complexity is dominated by the two loops iterating through `nums1` and `nums2`. The stack operations and hash map operations are performed within those loops but have an amortized constant time complexity. The space complexity is determined by the stack and hash map, which can store at most all elements of `nums2`.

📖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

For each element in `nums1`, find its index in `nums2` and then iterate through the rest of `nums2` to find the next greater element.

Time: O(m * n), where n is the length of `nums1` and m is the length of `nums2`.Space: O(1)

Trade-offs: The brute force approach is simpler to implement but has a higher time complexity. It is suitable for small input sizes but becomes inefficient for larger inputs. The optimal approach using the stack and hashmap is much faster, especially as array sizes increase.

✨

Key Insights

1Insight 1: The key observation is that we need to efficiently find the next greater element for each number in `nums2` to avoid redundant searches. We can precompute these and then use them to answer queries for `nums1` quickly.
2Insight 2: A stack can be used to keep track of potential next greater elements. The stack maintains a decreasing order, and when we encounter a larger element, we can pop elements from the stack and map them to the larger element as their next greater element.
3Insight 3: Using a hash map (or unordered_map in C++) to store the mapping between elements and their next greater elements allows for quick lookups.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Stack

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

Edge Cases

Edge case 1: `nums1` is empty. The algorithm will return an empty array, which is correct.
Edge case 2: No next greater element exists for an element in `nums1`. The algorithm correctly stores -1 as the next greater element for that number, by using `map.getOrDefault(nums1[i], -1)` in java and similar approaches in other languages. The hashmap lookup will return default value of -1 if number does not exist in the map.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where no next greater element exists. It's important to explicitly set the next greater element to -1 in this scenario.

✗Common mistake 2: Failing to use a stack or a hash map, which leads to a brute-force solution with significantly higher time complexity.

✗Common mistake 3: Incorrectly implementing the stack logic, such as pushing the wrong element or not popping elements when a greater element is found.

✅

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 price analysis. Finding the next day a stock price is higher than the current day's price is an application of finding the next greater element.

🏦Real-world scenario 2: Signal processing. Identifying peaks in a signal can be done by finding elements that are greater than their neighbors, effectively using a similar concept to find local maxima.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ 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: October 15, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

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

@InterviewProJul 7

❓

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.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Next Greater Element I"?

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

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