What is the time complexity of Diagonal Traverse II?

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

What data structures are used to solve Diagonal Traverse II?

This problem uses Array, Sorting, Heap (Priority Queue). Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Diagonal Traverse II?

Insight 1: Elements on the same diagonal have the same sum of row and column indices (r + c). Insight 2: We can group elements based on the sum of their row and column indices.

Which companies ask LeetCode 1424. Diagonal Traverse II in interviews?

This problem is frequently asked at BP, Liftoff.

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Diagonal Traverse II - LeetCode 1424 Solution

Q: How to solve Diagonal Traverse II?

The solution utilizes a hash map (or dictionary) to group elements based on the sum of their row and column indices (r + c). The keys of the hash map are the sums (r + c), and the values are lists of elements belonging to that diagonal. The elements for a given sum r+c are appended to the corresponding list. After grouping, the code iterates through the hash map in increasing order of the keys (r + c), reversing the order of elements in each diagonal's list, and then concatenating the lists into a final result list. This reversal ensures that elements are added to the final result from bottom to top along the diagonal.

Diagonal Traverse II - Complete Solution Guide

Diagonal Traverse II is LeetCode problem 1424, 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 2D integer array nums , return all elements of nums in diagonal order as shown in the below images . Example 1: Input: nums = [[1,2,3],[4,5,6],[7,8,9]] Output: [1,4,2,7,5,3,8,6,9] Example 2: Input: nums = [[1,2,3,4,5],[6,7],[8],[9,10,11],[12,13,14,15,16]] Output: [1,6,2,8,7,3,9,4,12,10,5,13,11,14,15,16] Constraints: 1 <= nums.length <= 10 5 1 <= nums[i].length <= 10 5 1 <= sum(nums[i].length) <= 10 5 1 <= nums[i][j] <= 10 5

Detailed Explanation

The problem asks us to traverse a 2D integer array `nums` in a diagonal order and return a 1D array containing all elements in that order. Diagonal order means grouping elements where the sum of their row and column indices is the same. We read the diagonals in increasing order of their sum (r + c), and within each diagonal, we traverse elements from bottom to top (increasing row and decreasing column index). The input is a 2D array (list of lists) of integers, and the output is a 1D array (list) containing all elements from the input in diagonal order.

Solution Approach

The solution utilizes a hash map (or dictionary) to group elements based on the sum of their row and column indices (r + c). The keys of the hash map are the sums (r + c), and the values are lists of elements belonging to that diagonal. The elements for a given sum r+c are appended to the corresponding list. After grouping, the code iterates through the hash map in increasing order of the keys (r + c), reversing the order of elements in each diagonal's list, and then concatenating the lists into a final result list. This reversal ensures that elements are added to the final result from bottom to top along the diagonal.

Step-by-Step Algorithm

Step 1: Create a hash map (or dictionary) called `diagonals` to store elements grouped by the sum of their row and column indices.
Step 2: Iterate through the input 2D array `nums` row by row.
Step 3: For each element `nums[r][c]`, calculate the sum `r + c`.
Step 4: Append the element `nums[r][c]` to the list associated with the key `r + c` in the `diagonals` hash map. If the key does not exist, create a new list.
Step 5: Create an empty list called `result` to store the elements in diagonal order.
Step 6: Initialize `curr_sum` to 0. This variable represents the current diagonal sum we're processing.
Step 7: While `curr_sum` exists as a key in the `diagonals` hash map:
Step 8: Get the list of elements associated with `curr_sum` from the `diagonals` hash map.
Step 9: Reverse the order of elements in the list to maintain the correct order in each diagonal.
Step 10: Extend the `result` list by appending the reversed list of elements.
Step 11: Increment `curr_sum` by 1 to move to the next diagonal.
Step 12: Return the `result` list.

Key Insights

Insight 1: Elements on the same diagonal have the same sum of row and column indices (r + c).
Insight 2: We can group elements based on the sum of their row and column indices.
Insight 3: Within each diagonal (same r+c value), we need to extract elements in reverse order of their occurrence in the input matrix during the row-wise iteration.

Complexity Analysis

Time Complexity: O(N)

Space Complexity: O(N)

Topics

This problem involves: Array, Sorting, Heap (Priority Queue).

Companies

Asked at: BP, Liftoff.

Diagonal Traverse II | LeetCode 1424

Medium

Array Sorting Heap (Priority Queue)

Time: O(N)Space: O(N)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Collections;
import java.util.Map;

class Solution {
    public int[] findDiagonalOrder(List<List<Integer>> nums) {
        Map<Integer, List<Integer>> diagonals = new HashMap<>();
        
        for (int r = 0; r < nums.size(); r++) {
            List<Integer> row = nums.get(r);
            for (int c = 0; c < row.size(); c++) {
                int val = row.get(c);
                if (!diagonals.containsKey(r + c)) {
                    diagonals.put(r + c, new ArrayList<>());
                }
                diagonals.get(r + c).add(val);
            }
        }
        
        List<Integer> resultList = new ArrayList<>();
        int curr_sum = 0;
        while (diagonals.containsKey(curr_sum)) {
            List<Integer> diagonal = diagonals.get(curr_sum);
            Collections.reverse(diagonal);
            resultList.addAll(diagonal);
            curr_sum += 1;
        }
        
        int[] result = new int[resultList.size()];
        for (int i = 0; i < resultList.size(); i++) {
            result[i] = resultList.get(i);
        }
        
        return result;
    }
}

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Problem Info

Description

Problem 1424: Diagonal Traverse II - Detailed explanation and solution approaches.

Topics

Array Sorting Heap (Priority Queue)

Companies

BPLiftoff

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1424

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

12 steps

1
Step 1: Create a hash map (or dictionary) called `diagonals` to store elements grouped by the sum of their row and column indices.
2
Step 2: Iterate through the input 2D array `nums` row by row.
3
Step 3: For each element `nums[r][c]`, calculate the sum `r + c`.
4
Step 4: Append the element `nums[r][c]` to the list associated with the key `r + c` in the `diagonals` hash map. If the key does not exist, create a new list.
5
Step 5: Create an empty list called `result` to store the elements in diagonal order.
6
Step 6: Initialize `curr_sum` to 0. This variable represents the current diagonal sum we're processing.
7
Step 7: While `curr_sum` exists as a key in the `diagonals` hash map:
8
Step 8: Get the list of elements associated with `curr_sum` from the `diagonals` hash map.
9
Step 9: Reverse the order of elements in the list to maintain the correct order in each diagonal.
10
Step 10: Extend the `result` list by appending the reversed list of elements.
11
Step 11: Increment `curr_sum` by 1 to move to the next diagonal.
12
Step 12: Return the `result` list.

💡

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)

💾

Space Complexity

O(N)

📝

Why?

The time complexity is O(N) because we visit each element of the input array a constant number of times. The space complexity is O(N) because, in the worst case, we might have to store all the elements of the input array in the hash map and the result list. Consider the case where all elements are on unique diagonals (e.g., a square matrix with elements only on the main diagonal). In that case, both hashmap and the final result list will contain all the elements of the input matrix.

📖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

Sorting Approach

Store elements as (r, c, value) tuples and sort them based on the diagonal index (r + c). Within the same diagonal, sort by row index in descending order (or column index in ascending order) to ensure correct order. This can be done using a custom comparator.

Time: O(N log N) due to sorting, where N is the total number of elements.Space: O(N) to store the tuples.

Trade-offs: The sorting approach is less efficient than the hash map approach due to the O(N log N) sorting time. It might be preferred if memory usage is very constrained and the hash map approach becomes problematic, or if the sorting approach is easier to implement for a specific use case. However, the O(N) time of the hash map approach makes it more desirable overall.

✨

Key Insights

1Insight 1: Elements on the same diagonal have the same sum of row and column indices (r + c).
2Insight 2: We can group elements based on the sum of their row and column indices.
3Insight 3: Within each diagonal (same r+c value), we need to extract elements in reverse order of their occurrence in the input matrix during the row-wise iteration.

📚

Key Terms Explained

For Students

Array

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

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty input array. The code handles this gracefully since the outer loops will not execute, and an empty result list will be returned.
Edge case 2: Input array with only one element. The code correctly identifies the single element and returns it in a list.
Edge case 3: Highly irregular array sizes (e.g., [[1,2,3,4,5],[6],[7,8,9,10,11,12,13]]). The grouping by diagonal sum correctly handles cases with varying row lengths.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not reversing the elements in each diagonal before adding them to the result. This will lead to an incorrect diagonal order.

✗Common mistake 2: Incorrectly calculating the diagonal sum (r + c).

✗Common mistake 3: Not handling the case where the input array is empty.

✗Common mistake 4: Using an inefficient sorting algorithm. The sorting approach can be competitive if an efficient algorithm such as merge sort is used.

✗Common mistake 5: In the C solution, forgetting to properly manage the memory allocated for the hashmap and result arrays. Memory leaks can occur if `free()` is not called on allocated memory blocks.

✅

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: Image processing - Diagonal traversal can be used for certain image filtering or feature extraction techniques where analyzing pixels along diagonals is relevant.

🏦Real-world scenario 2: Matrix operations - Some matrix algorithms benefit from processing elements in a specific diagonal order. Sparse matrix manipulations could leverage this pattern.

🎮Real-world scenario 3: Game development - Processing game board elements in a diagonal pattern for certain game logic or special effects. For example, a game might need to check for diagonal wins in a board game.

💡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: November 12, 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

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

@TechLeadJul 13

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

@DevExpert92Jul 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.

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 "Diagonal Traverse 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 sorting techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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