What is the time complexity of Next Greater Element II?

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

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

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

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

Insight 1: The circular nature of the array requires a way to effectively 'wrap around' when searching for the next greater element. Insight 2: A monotonic stack is an efficient data structure for tracking potential next greater elements as we iterate through the array.

Which companies ask LeetCode 503. Next Greater Element II in interviews?

This problem is frequently asked at Intuit, Zeta.

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Next Greater Element II - LeetCode 503 Solution

Next Greater Element II - Complete Solution Guide

Next Greater Element II is LeetCode problem 503, 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

Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0] ), return the next greater number for every element in nums . The next greater number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn't exist, return -1 for this number. Example 1: Input: nums = [1,2,1] Output: [2,-1,2] Explanation: The first 1's next greater number is 2; The numbe

Detailed Explanation

The problem requires us to find the next greater element for each number in a given circular integer array. A circular array means that the element after the last element is the first element. The 'next greater element' for a number `x` is the first number greater than `x` encountered while traversing the array in a circular manner. If no greater element exists, we should return -1 for that number. The input is an array of integers (`nums`), and the output is an array of integers representing the next greater elements for each corresponding element in the input array.

Solution Approach

The provided solutions use a monotonic stack to efficiently find the next greater element for each number in the array. The algorithm iterates through the array twice (simulating the circular nature) in reverse order. For each element, it compares the current element to the elements at the top of the stack. If the top of the stack is less than or equal to the current element, it's popped from the stack because it cannot be the next greater element for any preceding element. The stack maintains a decreasing order of elements. If, after popping smaller elements, the stack is not empty, the top of the stack contains the next greater element. The current element's index is then pushed onto the stack.

Step-by-Step Algorithm

Step 1: Initialize an output array `res` of the same size as `nums` with all elements set to -1. This array will store the next greater elements.
Step 2: Initialize an empty stack `stack` to store the indices of the elements in `nums`.
Step 3: Iterate through the array from `2 * n - 1` down to 0, where `n` is the length of `nums`. This effectively iterates through the array twice, simulating the circular nature.
Step 4: In each iteration, calculate the actual index `current_index` by taking the modulo of the loop variable `i` with `n` (`current_index = i % n`).
Step 5: While the stack is not empty and the element at the top of the stack is less than or equal to the element at `current_index`, pop the element from the stack.
Step 6: If the stack is not empty after popping, then the element at the top of the stack is the next greater element for the element at `current_index`. Assign this value to `res[current_index]`.
Step 7: Push the `current_index` onto the stack.
Step 8: After the loop completes, return the `res` array.

Key Insights

Insight 1: The circular nature of the array requires a way to effectively 'wrap around' when searching for the next greater element.
Insight 2: A monotonic stack is an efficient data structure for tracking potential next greater elements as we iterate through the array.
Insight 3: Duplicating the array's iteration allows the algorithm to simulate the circular search without complex indexing or modulo operations everywhere.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Stack, Monotonic Stack.

Companies

Asked at: Intuit, Zeta.

Next Greater Element II | LeetCode 503

Medium

Array Stack Monotonic Stack

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

Solution Code

Available in:

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int n = nums.length;
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = -1;
        }
        java.util.Stack<Integer> stack = new java.util.Stack<>();

        for (int i = 2 * n - 1; i >= 0; i--) {
            int current_index = i % n;
            while (!stack.isEmpty() && nums[stack.peek()] <= nums[current_index]) {
                stack.pop();
            }

            if (!stack.isEmpty()) {
                res[current_index] = nums[stack.peek()];
            }

            stack.push(current_index);
        }

        return res;
    }
}

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Problem Info

Description

Problem 503: Next Greater Element II - Detailed explanation and solution approaches.

Topics

Array Stack Monotonic Stack

Companies

IntuitZeta

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:503

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 output array `res` of the same size as `nums` with all elements set to -1. This array will store the next greater elements.
2
Step 2: Initialize an empty stack `stack` to store the indices of the elements in `nums`.
3
Step 3: Iterate through the array from `2 * n - 1` down to 0, where `n` is the length of `nums`. This effectively iterates through the array twice, simulating the circular nature.
4
Step 4: In each iteration, calculate the actual index `current_index` by taking the modulo of the loop variable `i` with `n` (`current_index = i % n`).
5
Step 5: While the stack is not empty and the element at the top of the stack is less than or equal to the element at `current_index`, pop the element from the stack.
6
Step 6: If the stack is not empty after popping, then the element at the top of the stack is the next greater element for the element at `current_index`. Assign this value to `res[current_index]`.
7
Step 7: Push the `current_index` onto the stack.
8
Step 8: After the loop completes, return the `res` 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(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 at most twice, and each element is pushed onto and popped from the stack at most once. The space complexity is O(n) because the stack can potentially store all the indices of the array elements in the worst-case scenario.

📖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, iterate through the array (wrapping around) to find the next greater element.

Time: O(n^2)Space: O(1)

Trade-offs: Simple to understand, but inefficient for larger arrays due to the nested loops. Avoid for sizable inputs.

✨

Key Insights

1Insight 1: The circular nature of the array requires a way to effectively 'wrap around' when searching for the next greater element.
2Insight 2: A monotonic stack is an efficient data structure for tracking potential next greater elements as we iterate through the array.
3Insight 3: Duplicating the array's iteration allows the algorithm to simulate the circular search without complex indexing or modulo operations everywhere.

📚

Key Terms Explained

For Students

Array

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

Stack

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

Edge Cases

Edge case 1: Array with all identical elements (e.g., [5, 5, 5]). In this case, the output will be [-1, -1, -1] because no element has a greater element.
Edge case 2: Array sorted in descending order (e.g., [5, 4, 3, 2, 1]). In this case, the algorithm will correctly assign -1 to all elements because there are no greater elements.
Edge case 3: Array sorted in ascending order (e.g., [1, 2, 3, 4, 5]). In this case the last element will be assigned -1 and rest of the elements will have their next greater element correctly calculated.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not understanding the circular nature of the array, leading to incorrect results when the next greater element requires wrapping around.

✗Common mistake 2: Not using a monotonic stack, leading to a less efficient solution with higher time complexity. For instance, not popping smaller elements from the stack can make the algorithm slow.

✅

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: Analyzing stock prices to identify potential buying opportunities when a stock price is about to increase (next greater element).

🏦Real-world scenario 2: Analyzing sensor data in a circular buffer to detect anomalies or patterns where a certain value exceeds subsequent values.

💡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: October 22, 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

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

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What is a monotonic stack and when should I use it?

A monotonic stack maintains elements in sorted order (increasing or decreasing). Use it for "next greater element," "largest rectangle," or "stock span" problems. It achieves O(n) time by avoiding redundant comparisons through maintaining order invariants.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Next Greater Element II"?

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 stack techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?