What is the time complexity of Diagonal Traverse?

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

What data structures are used to solve Diagonal Traverse?

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

What are the key insights for solving Diagonal Traverse?

Insight 1: The core idea is to simulate the diagonal traversal using row and column indices and a direction indicator (going_up/going_down). Insight 2: The boundary conditions (edges of the matrix) are crucial to determine when to change the direction of traversal and update row/column indices accordingly.

Which companies ask LeetCode 498. Diagonal Traverse in interviews?

This problem is frequently asked at Liftoff, Nike, Nykaa, Roblox, Zoho.

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Diagonal Traverse - LeetCode 498 Solution

Q: How to solve Diagonal Traverse?

The provided solution employs a simulation approach to traverse the matrix diagonally. It maintains two variables, 'row' and 'col', to track the current position in the matrix. A boolean variable 'going_up' indicates the current direction of traversal. The algorithm iterates through the matrix, adding elements to the result array until all elements are visited. Inside the loop, it checks for boundary conditions to determine when to change the direction and update the row and column accordingly. This approach systematically visits each element following the diagonal pattern.

Q: What is the space complexity of Diagonal Traverse?

The space complexity is O(m*n).

Diagonal Traverse - Complete Solution Guide

Diagonal Traverse is LeetCode problem 498, 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 an m x n matrix mat , return an array of all the elements of the array in a diagonal order . Example 1: Input: mat = [[1,2,3],[4,5,6],[7,8,9]] Output: [1,2,4,7,5,3,6,8,9] Example 2: Input: mat = [[1,2],[3,4]] Output: [1,2,3,4] Constraints: m == mat.length n == mat[i].length 1 <= m, n <= 10 4 1 <= m * n <= 10 4 -10 5 <= mat[i][j] <= 10 5

Detailed Explanation

The problem asks us to traverse a given m x n matrix in a diagonal order. That means we need to extract elements from the matrix following a specific pattern. We start from the top-left corner (0,0), then move diagonally upwards to the right until we hit a boundary. When we reach a boundary, we change direction and move diagonally downwards to the left. The input is a 2D array (matrix), and the output is a 1D array containing all the matrix elements in the specified diagonal order. The constraints specify the size of the matrix and the range of values it contains.

Solution Approach

The provided solution employs a simulation approach to traverse the matrix diagonally. It maintains two variables, 'row' and 'col', to track the current position in the matrix. A boolean variable 'going_up' indicates the current direction of traversal. The algorithm iterates through the matrix, adding elements to the result array until all elements are visited. Inside the loop, it checks for boundary conditions to determine when to change the direction and update the row and column accordingly. This approach systematically visits each element following the diagonal pattern.

Step-by-Step Algorithm

Step 1: Initialize an empty result array (list in Python, array in Java/C/C++) to store the diagonal traversal order.
Step 2: Initialize row and col to 0, representing the starting position (top-left corner) of the matrix.
Step 3: Initialize a boolean variable 'going_up' to True, indicating the initial direction of traversal (upwards-right).
Step 4: Iterate until all m * n elements have been added to the result array.
Step 5: In each iteration, add the element at mat[row][col] to the result array.
Step 6: Check if 'going_up' is True:
a. If col is at the rightmost edge (n-1), move down (row++) and set 'going_up' to False.
b. Else if row is at the top edge (0), move right (col++) and set 'going_up' to False.
c. Else, move diagonally upwards (row--, col++).
Step 7: If 'going_up' is False:
a. If row is at the bottom edge (m-1), move right (col++) and set 'going_up' to True.
b. Else if col is at the leftmost edge (0), move down (row++) and set 'going_up' to True.
c. Else, move diagonally downwards (row++, col--).
Step 8: After the loop finishes, return the result array.

Key Insights

Insight 1: The core idea is to simulate the diagonal traversal using row and column indices and a direction indicator (going_up/going_down).
Insight 2: The boundary conditions (edges of the matrix) are crucial to determine when to change the direction of traversal and update row/column indices accordingly.
Insight 3: The time complexity is inherently O(m*n) because we need to visit each element in the matrix exactly once.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Matrix, Simulation.

Companies

Asked at: Liftoff, Nike, Nykaa, Roblox, Zoho.

Diagonal Traverse | LeetCode 498

Medium

Array Matrix Simulation

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

Solution Code

Available in:

class Solution {
    public int[] findDiagonalOrder(int[][] mat) {
        if (mat == null || mat.length == 0 || mat[0].length == 0) {
            return new int[0];
        }

        int m = mat.length;
        int n = mat[0].length;
        int[] result = new int[m * n];
        int row = 0;
        int col = 0;
        boolean going_up = true;

        for (int i = 0; i < m * n; i++) {
            result[i] = mat[row][col];

            if (going_up) {
                if (col == n - 1) {
                    row += 1;
                    going_up = false;
                } else if (row == 0) {
                    col += 1;
                    going_up = false;
                } else {
                    row -= 1;
                    col += 1;
                }
            } else {
                if (row == m - 1) {
                    col += 1;
                    going_up = true;
                } else if (col == 0) {
                    row += 1;
                    going_up = true;
                } else {
                    row += 1;
                    col -= 1;
                }
            }
        }

        return result;
    }
}

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Problem Info

Description

Problem 498: Diagonal Traverse - Detailed explanation and solution approaches.

Topics

Array Matrix Simulation

Companies

LiftoffNikeNykaaRobloxZoho

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:498

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

14 steps

1
Step 1: Initialize an empty result array (list in Python, array in Java/C/C++) to store the diagonal traversal order.
2
Step 2: Initialize row and col to 0, representing the starting position (top-left corner) of the matrix.
3
Step 3: Initialize a boolean variable 'going_up' to True, indicating the initial direction of traversal (upwards-right).
4
Step 4: Iterate until all m * n elements have been added to the result array.
5
Step 5: In each iteration, add the element at mat[row][col] to the result array.
6
Step 6: Check if 'going_up' is True:
7
a. If col is at the rightmost edge (n-1), move down (row++) and set 'going_up' to False.
8
b. Else if row is at the top edge (0), move right (col++) and set 'going_up' to False.
9
c. Else, move diagonally upwards (row--, col++).
10
Step 7: If 'going_up' is False:
11
a. If row is at the bottom edge (m-1), move right (col++) and set 'going_up' to True.
12
b. Else if col is at the leftmost edge (0), move down (row++) and set 'going_up' to True.
13
c. Else, move diagonally downwards (row++, col--).
14
Step 8: After the loop finishes, 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*n)

📝

Why?

Time Complexity: The algorithm visits each element of the m x n matrix exactly once. The operations performed in each visit (adding to the result, checking boundary conditions, updating row/col) take constant time, O(1). Therefore, the overall time complexity is O(m*n). Space Complexity: The algorithm creates a result array to store the diagonal traversal order. The size of this array is equal to the number of elements in the matrix, which is m*n. Therefore, the space complexity is O(m*n).

📖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

Using a HashMap to group diagonals

This approach involves using a hash map where the keys are the sums of the row and column indices (i+j), which represent the diagonals. Values are lists of elements on that diagonal. Then, we iterate through the hash map and add elements to the result array, reversing the order for even-numbered diagonals.

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

Trade-offs: This approach is more complex to implement but could be beneficial if the matrix is very sparse, and only a subset of diagonals needs to be accessed. However, for dense matrices, the simulation approach is generally simpler and more efficient due to the overhead of the HashMap.

✨

Key Insights

1Insight 1: The core idea is to simulate the diagonal traversal using row and column indices and a direction indicator (going_up/going_down).
2Insight 2: The boundary conditions (edges of the matrix) are crucial to determine when to change the direction of traversal and update row/column indices accordingly.
3Insight 3: The time complexity is inherently O(m*n) because we need to visit each element in the matrix exactly once.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: Empty matrix (m=0 or n=0). The solution correctly handles this by returning an empty array.
Edge case 2: Single row matrix (m=1). The algorithm correctly traverses the row from left to right.
Edge case 3: Single column matrix (n=1). The algorithm correctly traverses the column from top to bottom.
Edge case 4: Square matrix (m=n). The algorithm functions correctly, demonstrating that it handles symmetric inputs properly.
Edge case 5: Rectangular matrix (m != n). The solution correctly handles cases where the number of rows and columns are different.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling boundary conditions, leading to out-of-bounds errors or incorrect traversal order. The code should carefully check the row and column indices before accessing matrix elements.

✗Common mistake 2: Failing to switch the direction ('going_up' flag) at the correct boundaries, which results in traversing elements incorrectly. The logic for updating this flag must be precise.

✗Common mistake 3: Not handling the edge case of empty input matrices.

✅

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 specific image processing algorithms, such as diagonal edge detection or feature extraction.

🏦Real-world scenario 2: Data compression: Certain compression algorithms may utilize diagonal patterns to identify and encode repeating patterns or redundancies in data.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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 17, 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 13

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

@TechLeadJul 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"?

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

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