Length of Longest V-Shaped Diagonal Segment
Time O(n * m) · Space O(n * m) · Official statement on LeetCode
Solutions
// Time: O(n * m)
// Space: O(n * m)
// dp
class Solution {
public:
int lenOfVDiagonal(vector<vector<int>>& grid) {
const int n = size(grid);
const int m = size(grid[0]);
int result = 0;
vector<vector<int>> down_right(n, vector<int>(m, 1));
vector<vector<int>> down_left(n, vector<int>(m, 1));
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
const int x = grid[i][j];
if (x == 1) {
result = 1;
continue;
}
if (i - 1 >= 0 && j - 1 >= 0 && grid[i - 1][j - 1] == 2 - x) {
down_right[i][j] = down_right[i - 1][j - 1] + 1;
}
if (i - 1 >= 0 && j + 1 < m && grid[i - 1][j + 1] == 2 - x) {
down_left[i][j] = down_left[i - 1][j + 1] + 1;
}
}
}
vector<vector<int>> up_right(n, vector<int>(m, 1));
vector<vector<int>> up_left(n, vector<int>(m, 1));
for (int i = n - 1; i >= 0; --i) {
for (int j = 0; j < m; ++j) {
const int x = grid[i][j];
if (x == 1) {
continue;
}
if (i + 1 < n && j - 1 >= 0 && grid[i + 1][j - 1] == 2 - x) {
up_right[i][j] = up_right[i + 1][j - 1] + 1;
}
if (i + 1 < n && j + 1 < m && grid[i + 1][j + 1] == 2 - x) {
up_left[i][j] = up_left[i + 1][j + 1] + 1;
}
}
}
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
const int x = grid[i][j];
if (x == 1) {
continue;
}
if ((down_right[i][j] % 2 == 0 && x == 0) || (down_right[i][j] % 2 == 1 && x == 2)) {
const int ni = i - down_right[i][j];
const int nj = j - down_right[i][j];
if (0 <= ni && ni < n && 0 <= nj && nj < m && grid[ni][nj] == 1) {
result = max(result, down_right[i][j] + up_right[i][j]); // >
}
}
if ((down_left[i][j] % 2 == 0 && x == 0) || (down_left[i][j] % 2 == 1 && x == 2)) {
const int ni = i - down_left[i][j];
const int nj = j + down_left[i][j];
if (0 <= ni && ni < n && 0 <= nj && nj < m && grid[ni][nj] == 1) {
result = max(result, down_left[i][j] + down_right[i][j]); // v
}
}
if ((up_left[i][j] % 2 == 0 && x == 0) || (up_left[i][j] % 2 == 1 && x == 2)) {
const int ni = i + up_left[i][j];
const int nj = j + up_left[i][j];
if (0 <= ni && ni < n && 0 <= nj && nj < m && grid[ni][nj] == 1) {
result = max(result, up_left[i][j] + down_left[i][j]); // <
}
}
if ((up_right[i][j] % 2 == 0 && x == 0) || (up_right[i][j] % 2 == 1 && x == 2)) {
const int ni = i + up_right[i][j];
const int nj = j - up_right[i][j];
if (0 <= ni && ni < n && 0 <= nj && nj < m && grid[ni][nj] == 1) {
result = max(result, up_right[i][j] + up_left[i][j]); // ^
}
}
}
}
return result;
}
};
// Time: O(n * m)
// Space: O(n * m)
// memoization
class Solution2 {
public:
int lenOfVDiagonal(vector<vector<int>>& grid) {
const int n = size(grid);
const int m = size(grid[0]);
vector<pair<int, int>> directions = {{1, 1}, {1, -1}, {-1, -1}, {-1, 1}};
// be careful with the order, going from smaller dimensions to larger dimensions
vector<vector<vector<vector<vector<int>>>>> lookup(2, vector<vector<vector<vector<int>>>>(3, vector<vector<vector<int>>>(4, vector<vector<int>>(n, vector<int>(m)))));
const function<int (int, int, int, int, int)> memoization = [&](int i, int j, int x, int d, int k) {
if (!(0 <= i && i < n && 0 <= j && j < m)) {
return 0;
}
if (grid[i][j] != x) {
return 0;
}
if (lookup[k][x][d][i][j] == 0) {
const int ni = i + directions[d].first;
const int nj = j + directions[d].second;
const int nx = x == 2 ? 0 : 2;
int result = memoization(ni, nj, nx, d, k) + 1;
if (k != 1) {
const int nd = (d + 1) % 4;
result = max(result, memoization(ni, nj, nx, nd, k + 1) + 1);
}
lookup[k][x][d][i][j] = result;
}
return lookup[k][x][d][i][j];
};
int result = 0;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
if (grid[i][j] == 1) {
for (int d = 0; d < 4; ++d) {
result = max(result, memoization(i, j, 1, d, 0));
}
}
}
}
return result;
}
};
Beginner Explanation
What is Length of Longest V-Shaped Diagonal Segment?
Length of Longest V-Shaped Diagonal Segment (LeetCode #3459) is a Hard problem that primarily trains dynamic programming.
How to think about it
- Restate the goal in your own words before coding.
- Work a tiny example by hand so the invariant becomes obvious.
- Identify the pattern — this problem aligns with dynamic programming.
- Only then translate the idea into code.
Why this problem matters
Hard problems force you to combine patterns and prove complexity carefully — interview gold. Official solution notes mention: Memoization, DP.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Length of Longest V-Shaped Diagonal Segment
Opening (30–60 seconds)
- Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
- State a brute force so the interviewer knows you can solve it naively.
- Propose the optimal direction tied to dynamic programming.
Core solution narrative
- Define the state you track (pointers, DP cell, set membership, stack top, etc.).
- Explain the transition when you process the next element.
- Call out time (O(n * m)) and space (O(n * m)) before coding.
- Code cleanly; narrate variable names.
What interviewers listen for
- Correctness on edge cases
- Complexity honesty
- Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)
Follow-up questions they may ask
- Can you solve it with less memory?
- What if the input stream is infinite / doesn't fit in RAM?
- How would tests look for adversarial inputs?
Optimized Approach
Optimized solution notes
The reference solutions on AlgoForge target O(n * m) time and O(n * m) space.
Pattern focus: dynamic programming
Use the pattern as a checklist:
- dynamic programming — confirm the invariant holds after each step
Multiple methods appear in the source solutions — compare them and explain when each is preferable.
Implementation tips
- Prefer readable names over micro-optimizations in interviews.
- Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
- After AC-level logic, re-scan for off-by-one and null checks.
Complexity Analysis
Complexity
| Measure | Bound |
|---|---|
| Time | O(n * m) |
| Space | O(n * m) |
How to justify this in an interview
- Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
- Space: include hash maps, recursion stack, and output allocation when the problem asks for it.
If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.
Common Mistakes
Common mistakes on Length of Longest V-Shaped Diagonal Segment
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for dynamic programming — updating state too early or too late.
- Mutating input unexpectedly when the problem forbids it.
- Off-by-one in windows, ranges, or binary search bounds.
- Ignoring overflow / precision for integer arithmetic problems.
- Overengineering — jumping to an advanced structure when a simpler approach works.
Alternative Approaches
Alternatives
The source file includes more than one method. Compare:
- Primary optimized path — best complexity for typical interviews.
- Secondary approach — often brute force, sorting-based, or space-optimized variant.
Practice articulating when you would pick each (constraints, readability, follow-ups).
Edge Cases
Edge cases checklist
- Minimum input size
- Maximum input size / time limits
- Duplicates and already-sorted input
- Negative numbers / zeros (if applicable)
- Disconnected structures (graphs/trees)
- Single path vs branching recursion depth
Pattern Recognition
Spotting this pattern
Signal phrases that point to dynamic programming:
- Sorted input or ability to sort without changing the answer class
- Need for contiguous subarray / substring → consider sliding window
- Need for O(1) membership → hash set/map
- Optimal substructure + overlapping subproblems → DP
- Connectivity / components → graph DFS/BFS or Union-Find
Primary topics: dynamic programming.
Follow-up Interview Questions
Follow-ups
- How does the solution change if the input is a stream?
- Can you solve it in-place?
- What if duplicates must be handled differently?
- How would you parallelize the approach?
- Design tests that would break a buggy implementation.
Practice Recommendations
What to practice next
- Re-solve Length of Longest V-Shaped Diagonal Segment in a second language (cpp, python).
- Drill 3–5 more problems tagged dynamic programming.
- Teach the solution out loud in under 5 minutes.
- Add this problem to your revision calendar in 3 days and 14 days.
Visualization
Study checklist
- Read the official problem statement on LeetCode
- Solve on paper / whiteboard first
- Implement the dynamic programming approach
- Verify edge cases from the checklist
- State time and space complexity aloud
- Compare with the AlgoForge reference solution
- Schedule a revision session
Revision notes
Length of Longest V-Shaped Diagonal Segment (#3459) — Hard. Pattern: dynamic programming. Complexity: O(n * m) time / O(n * m) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Length of Longest V-Shaped Diagonal Segment?+
The reference solutions aim for O(n * m) time and O(n * m) space. Always re-derive complexity from the code you write in the interview.
What pattern does Length of Longest V-Shaped Diagonal Segment use?+
It primarily maps to dynamic programming, within the broader topic of dynamic programming.
Is Length of Longest V-Shaped Diagonal Segment good for interviews?+
Yes — as a Hard problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.
Where can I read the official statement?+
Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/length-of-longest-v-shaped-diagonal-segment/