What is the time complexity of Length of Longest V-Shaped Diagonal Segment?

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

What is the space complexity of Length of Longest V-Shaped Diagonal Segment?

The space complexity is O(n*m).

What data structures are used to solve Length of Longest V-Shaped Diagonal Segment?

This problem uses Array, Dynamic Programming, Memoization, Matrix. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Length of Longest V-Shaped Diagonal Segment?

Insight 1: Dynamic programming is essential to efficiently compute the lengths of diagonal segments starting from each cell in all four diagonal directions. Insight 2: The alternating sequence of 2s and 0s must be maintained when extending the diagonal segments. The length of a segment depends on the previous cell's segment length and the current cell's value.

Which companies ask LeetCode 3459. Length of Longest V-Shaped Diagonal Segment in interviews?

This problem is frequently asked at Visa.

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Length of Longest V-Shaped Diagonal Segment - LeetCode 3459 Solution

Length of Longest V-Shaped Diagonal Segment - Complete Solution Guide

Length of Longest V-Shaped Diagonal Segment is LeetCode problem 3459, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a 2D integer matrix grid of size n x m , where each element is either 0 , 1 , or 2 . A V-shaped diagonal segment is defined as: The segment starts with 1 . The subsequent elements follow this infinite sequence: 2, 0, 2, 0, ... . The segment: Starts along a diagonal direction (top-left to bottom-right, bottom-right to top-left, top-right to bottom-left, or bottom-left to top-right). Continues the sequence in the same diagonal direction. Makes at most one clockwise 90-degree turn to

Detailed Explanation

The problem asks us to find the length of the longest "V-shaped diagonal segment" within a given 2D integer matrix (grid) containing only 0s, 1s, and 2s. A V-shaped diagonal segment starts with a '1' and continues with alternating '2' and '0' values along diagonal directions. The segment can turn at most once by 90 degrees clockwise while maintaining the sequence. The goal is to find the longest such segment and return its length, or 0 if no valid segment exists. The input is a 2D array (grid), and the output is an integer representing the length of the longest valid segment.

Solution Approach

The solution uses dynamic programming to calculate the lengths of diagonal segments starting from each cell in all four diagonal directions (top-left, top-right, bottom-left, bottom-right). For each cell and direction, it checks if a valid segment can be extended based on the grid value and the length of the segment from the adjacent cell in that direction. It stores these lengths in separate DP tables for each direction and value (1, 2, 0). Then, for each cell, it considers all possible 90-degree clockwise turns from one direction to another, combining the segment lengths to find the longest V-shaped segment.

Step-by-Step Algorithm

Step 1: Initialize twelve 2D arrays (DP tables) d1_tl, d2_tl, d0_tl, d1_tr, d2_tr, d0_tr, d1_bl, d2_bl, d0_bl, d1_br, d2_br, and d0_br of the same size as the input grid, initialized to 0. 'd1' indicates segment starting from '1', while 'd2' and 'd0' represent subsequent segment after '2' and '0' respectively. 'tl', 'tr', 'bl', 'br' represent the segment direction.
Step 2: Iterate through the grid from top-left to bottom-right to populate d1_tl, d2_tl, and d0_tl. If grid[r][c] is 1, set d1_tl[r][c] = 1. If grid[r][c] is 2, set d2_tl[r][c] = 1. If grid[r][c] is 0, set d0_tl[r][c] = 1. Also, check the top-left diagonal neighbor (r-1, c-1) to potentially extend existing segments. if the current value is 2 and the previous segment has length > 0 with odd length, then d1_tl[r][c] = d1_tl[r-1][c-1] + 1. if the current value is 0 and the previous segment has length > 0 with even length, then d1_tl[r][c] = d1_tl[r-1][c-1] + 1. Update d2_tl[r][c] with d0_tl[r-1][c-1] and d0_tl[r][c] with d2_tl[r-1][c-1].
Step 3: Repeat Step 2 for top-right to bottom-left and populate d1_tr, d2_tr, and d0_tr.
Step 4: Repeat Step 2 for bottom-left to top-right and populate d1_bl, d2_bl, and d0_bl.
Step 5: Repeat Step 2 for bottom-right to top-left and populate d1_br, d2_br, and d0_br.
Step 6: Initialize max_len to 1 if there is at least one element 1 in the grid or to 0 otherwise.
Step 7: Iterate through the grid, and update max_len with the maximum length of the diagonal segment starting at each cell in each direction.
Step 8: Iterate through the grid again and check all possible 90 degree turns in the direction tl->tr, tr->br, br->bl, bl->tl
Step 9: Return max_len.

Key Insights

Insight 1: Dynamic programming is essential to efficiently compute the lengths of diagonal segments starting from each cell in all four diagonal directions.
Insight 2: The alternating sequence of 2s and 0s must be maintained when extending the diagonal segments. The length of a segment depends on the previous cell's segment length and the current cell's value.
Insight 3: The 'V' shape allows for only one 90-degree clockwise turn. Therefore, we need to consider all possible combinations of diagonal directions and turning points.
Insight 4: The solution needs to handle the case where the matrix contains no 1s, returning 0 in such instance.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(n*m)

Topics

This problem involves: Array, Dynamic Programming, Memoization, Matrix.

Companies

Asked at: Visa.

Length of Longest V-Shaped Diagonal Segment | LeetCode 3459

Hard

Array Dynamic Programming Memoization Matrix

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

Solution Code

Available in:

class Solution {
    public int lenOfVDiagonal(int[][] grid) {
        int n = grid.length;
        int m = grid[0].length;

        int[][] d1_tl = new int[n][m];
        int[][] d2_tl = new int[n][m];
        int[][] d0_tl = new int[n][m];
        int[][] d1_tr = new int[n][m];
        int[][] d2_tr = new int[n][m];
        int[][] d0_tr = new int[n][m];
        int[][] d1_bl = new int[n][m];
        int[][] d2_bl = new int[n][m];
        int[][] d0_bl = new int[n][m];
        int[][] d1_br = new int[n][m];
        int[][] d2_br = new int[n][m];
        int[][] d0_br = new int[n][m];

        boolean has_one = false;
        for (int[] row : grid) {
            for (int val : row) {
                if (val == 1) {
                    has_one = true;
                    break;
                }
            }
            if(has_one) break;
        }

        // From Top-Left
        for (int r = 0; r < n; r++) {
            for (int c = 0; c < m; c++) {
                int val = grid[r][c];
                if (val == 1) d1_tl[r][c] = 1;
                if (val == 2) d2_tl[r][c] = 1;
                if (val == 0) d0_tl[r][c] = 1;
                int pr = r - 1;
                int pc = c - 1;
                if (0 <= pr && pr < n && 0 <= pc && pc < m) {
                    if (val == 2 && d1_tl[pr][pc] > 0 && d1_tl[pr][pc] % 2 == 1) d1_tl[r][c] = d1_tl[pr][pc] + 1;
                    if (val == 0 && d1_tl[pr][pc] > 0 && d1_tl[pr][pc] % 2 == 0) d1_tl[r][c] = d1_tl[pr][pc] + 1;
                    if (val == 2) d2_tl[r][c] += d0_tl[pr][pc];
                    if (val == 0) d0_tl[r][c] += d2_tl[pr][pc];
                }
            }
        }

        // From Top-Right
        for (int r = 0; r < n; r++) {
            for (int c = m - 1; c >= 0; c--) {
                int val = grid[r][c];
                if (val == 1) d1_tr[r][c] = 1;
                if (val == 2) d2_tr[r][c] = 1;
                if (val == 0) d0_tr[r][c] = 1;
                int pr = r - 1;
                int pc = c + 1;
                if (0 <= pr && pr < n && 0 <= pc && pc < m) {
                    if (val == 2 && d1_tr[pr][pc] > 0 && d1_tr[pr][pc] % 2 == 1) d1_tr[r][c] = d1_tr[pr][pc] + 1;
                    if (val == 0 && d1_tr[pr][pc] > 0 && d1_tr[pr][pc] % 2 == 0) d1_tr[r][c] = d1_tr[pr][pc] + 1;
                    if (val == 2) d2_tr[r][c] += d0_tr[pr][pc];
                    if (val == 0) d0_tr[r][c] += d2_tr[pr][pc];
                }
            }
        }

        // From Bottom-Left
        for (int r = n - 1; r >= 0; r--) {
            for (int c = 0; c < m; c++) {
                int val = grid[r][c];
                if (val == 1) d1_bl[r][c] = 1;
                if (val == 2) d2_bl[r][c] = 1;
                if (val == 0) d0_bl[r][c] = 1;
                int pr = r + 1;
                int pc = c - 1;
                if (0 <= pr && pr < n && 0 <= pc && pc < m) {
                    if (val == 2 && d1_bl[pr][pc] > 0 && d1_bl[pr][pc] % 2 == 1) d1_bl[r][c] = d1_bl[pr][pc] + 1;
                    if (val == 0 && d1_bl[pr][pc] > 0 && d1_bl[pr][pc] % 2 == 0) d1_bl[r][c] = d1_bl[pr][pc] + 1;
                    if (val == 2) d2_bl[r][c] += d0_bl[pr][pc];
                    if (val == 0) d0_bl[r][c] += d2_bl[pr][pc];
                }
            }
        }

        // From Bottom-Right
        for (int r = n - 1; r >= 0; r--) {
            for (int c = m - 1; c >= 0; c--) {
                int val = grid[r][c];
                if (val == 1) d1_br[r][c] = 1;
                if (val == 2) d2_br[r][c] = 1;
                if (val == 0) d0_br[r][c] = 1;
                int pr = r + 1;
                int pc = c + 1;
                if (0 <= pr && pr < n && 0 <= pc && pc < m) {
                    if (val == 2 && d1_br[pr][pc] > 0 && d1_br[pr][pc] % 2 == 1) d1_br[r][c] = d1_br[pr][pc] + 1;
                    if (val == 0 && d1_br[pr][pc] > 0 && d1_br[pr][pc] % 2 == 0) d1_br[r][c] = d1_br[pr][pc] + 1;
                    if (val == 2) d2_br[r][c] += d0_br[pr][pc];
                    if (val == 0) d0_br[r][c] += d2_br[pr][pc];
                }
            }
        }

        int max_len = has_one ? 1 : 0;
        for (int r = 0; r < n; r++) {
            for (int c = 0; c < m; c++) {
                max_len = Math.max(max_len, Math.max(Math.max(d1_tl[r][c], d1_tr[r][c]), Math.max(d1_bl[r][c], d1_br[r][c])));
            }
        }

        for (int r = 0; r < n; r++) {
            for (int c = 0; c < m; c++) {
                // tl(1,1) -> tr(1,-1)
                int l1 = d1_tl[r][c];
                if (l1 > 0) {
                    int nr = r + 1;
                    int nc = c - 1;
                    if (0 <= nr && nr < n && 0 <= nc && nc < m) {
                        int l2 = (l1 % 2 == 1) ? d2_bl[nr][nc] : d0_bl[nr][nc];
                        if (l2 > 0) max_len = Math.max(max_len, l1 + l2);
                    }
                }

                // tr(1,-1) -> br(-1,-1)
                l1 = d1_tr[r][c];
                if (l1 > 0) {
                    int nr = r - 1;
                    int nc = c - 1;
                    if (0 <= nr && nr < n && 0 <= nc && nc < m) {
                        int l2 = (l1 % 2 == 1) ? d2_tl[nr][nc] : d0_tl[nr][nc];
                        if (l2 > 0) max_len = Math.max(max_len, l1 + l2);
                    }
                }

                // br(-1,-1) -> bl(-1,1)
                l1 = d1_br[r][c];
                if (l1 > 0) {
                    int nr = r - 1;
                    int nc = c + 1;
                    if (0 <= nr && nr < n && 0 <= nc && nc < m) {
                        int l2 = (l1 % 2 == 1) ? d2_tr[nr][nc] : d0_tr[nr][nc];
                        if (l2 > 0) max_len = Math.max(max_len, l1 + l2);
                    }
                }

                // bl(-1,1) -> tl(1,1)
                l1 = d1_bl[r][c];
                if (l1 > 0) {
                    int nr = r + 1;
                    int nc = c + 1;
                    if (0 <= nr && nr < n && 0 <= nc && nc < m) {
                        int l2 = (l1 % 2 == 1) ? d2_br[nr][nc] : d0_br[nr][nc];
                        if (l2 > 0) max_len = Math.max(max_len, l1 + l2);
                    }
                }
            }
        }

        return max_len;
    }
}

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Problem Info

Description

Problem 3459: Length of Longest V-Shaped Diagonal Segment - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Memoization Matrix

Companies

Visa

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3459

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 twelve 2D arrays (DP tables) d1_tl, d2_tl, d0_tl, d1_tr, d2_tr, d0_tr, d1_bl, d2_bl, d0_bl, d1_br, d2_br, and d0_br of the same size as the input grid, initialized to 0. 'd1' indicates segment starting from '1', while 'd2' and 'd0' represent subsequent segment after '2' and '0' respectively. 'tl', 'tr', 'bl', 'br' represent the segment direction.
2
Step 2: Iterate through the grid from top-left to bottom-right to populate d1_tl, d2_tl, and d0_tl. If grid[r][c] is 1, set d1_tl[r][c] = 1. If grid[r][c] is 2, set d2_tl[r][c] = 1. If grid[r][c] is 0, set d0_tl[r][c] = 1. Also, check the top-left diagonal neighbor (r-1, c-1) to potentially extend existing segments. if the current value is 2 and the previous segment has length > 0 with odd length, then d1_tl[r][c] = d1_tl[r-1][c-1] + 1. if the current value is 0 and the previous segment has length > 0 with even length, then d1_tl[r][c] = d1_tl[r-1][c-1] + 1. Update d2_tl[r][c] with d0_tl[r-1][c-1] and d0_tl[r][c] with d2_tl[r-1][c-1].
3
Step 3: Repeat Step 2 for top-right to bottom-left and populate d1_tr, d2_tr, and d0_tr.
4
Step 4: Repeat Step 2 for bottom-left to top-right and populate d1_bl, d2_bl, and d0_bl.
5
Step 5: Repeat Step 2 for bottom-right to top-left and populate d1_br, d2_br, and d0_br.
6
Step 6: Initialize max_len to 1 if there is at least one element 1 in the grid or to 0 otherwise.
7
Step 7: Iterate through the grid, and update max_len with the maximum length of the diagonal segment starting at each cell in each direction.
8
Step 8: Iterate through the grid again and check all possible 90 degree turns in the direction tl->tr, tr->br, br->bl, bl->tl
9
Step 9: Return max_len.

💡

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*m)

💾

Space Complexity

O(n*m)

📝

Why?

The time complexity is derived from the multiple nested loops that iterate through the entire grid. The space complexity is determined by the size of the DP tables, which are proportional to the size of the input grid.

📖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

Recursive Approach with Memoization

A recursive function can be defined to explore each diagonal and track the length of the V-shaped segment. Memoization can be used to store the lengths of segments already computed, preventing redundant calculations.

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

Trade-offs: The recursive approach may be easier to understand for some, but can suffer from stack overflow issues for very large grids. The DP approach is generally more efficient due to its iterative nature.

✨

Key Insights

1Insight 1: Dynamic programming is essential to efficiently compute the lengths of diagonal segments starting from each cell in all four diagonal directions.
2Insight 2: The alternating sequence of 2s and 0s must be maintained when extending the diagonal segments. The length of a segment depends on the previous cell's segment length and the current cell's value.
3Insight 3: The 'V' shape allows for only one 90-degree clockwise turn. Therefore, we need to consider all possible combinations of diagonal directions and turning points.
4Insight 4: The solution needs to handle the case where the matrix contains no 1s, returning 0 in such instance.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty grid (n=0 or m=0). The code implicitly handles this case as the loops will not execute, returning 0.
Edge case 2: Grid with no '1's. The code initializes 'max_len' to 0 if there are no 1's in the grid, ensuring that the correct value is returned in this case.
Edge case 3: Grid with only one element '1'. The code initializes the corresponding cell in the DP table to 1, and since there will be no valid V shape in this case, it returns the correct result of 1.
Edge case 4: Grid with a '1' that cannot be extended. The code will properly find length 1 for this situation, since no other 0 or 2 values form a diagonal after it.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly updating the DP tables, leading to inaccurate segment lengths. This can happen if the sequence of 2s and 0s is not properly maintained during extension.

✗Common mistake 2: Not considering all possible 90-degree turns, resulting in missing the longest V-shaped segment. It's crucial to check all direction combinations.

✗Common mistake 3: Failing to initialize or handle the case where there are no '1's in the grid. The 'max_len' should be initialized to 0 in that case.

✅

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: Pattern recognition in image processing. Identifying V-shaped patterns along diagonals in a pixel grid could be relevant in object detection or feature extraction.

🏦Real-world scenario 2: Analysis of data sequences with alternating characteristics. Finding the longest alternating sequence with a single turn point could be used in time series analysis or financial modeling.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: July 10, 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

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

@AlgoMasterJun 29

❓

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.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Length of Longest V-Shaped Diagonal Segment"?

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

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