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3459. Length of Longest V-Shaped Diagonal Segment

HardMemoizationArrayDynamic ProgrammingMatrix
Leetcode Link

Problem Description

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 another diagonal direction while maintaining the sequence.

You need to return the length of the longest V-shaped diagonal segment. If no valid segment exists, return 0.

Intuition

The problem requires finding the longest diagonal sequence in a grid that follows a specific pattern and allows a single change of direction.

To solve the problem, we can take a dynamic programming approach and utilize memoization to avoid redundant calculations:

  1. Identify Starting Points:

    • Traverse the grid and focus on elements with the value 1, as these are the only valid starting points for any potential V-shaped diagonal segment.

  2. Directional Exploration:

    • From each start point, explore four possible diagonal directions:
      • Top-left to bottom-right
      • Bottom-right to top-left
      • Top-right to bottom-left
      • Bottom-left to top-right

  3. Recursive Exploration with Memoization:

    • For each direction, recursively attempt to extend the diagonal segment following the sequence 2, 0, 2, 0, ....
    • Keep track of whether a direction change has occurred. Only one change is allowed to form a V-shape.
    • Use a cached function to store and reuse results for overlapping subproblems to enhance efficiency.

  4. Determine the Maximum Length:

    • On successfully evaluating all possible V-shaped diagonal segments starting from each 1, we keep track of the maximum length obtained.

This approach ensures that all potential segments are explored efficiently, leveraging recursion and memoization to manage computational complexity.

Learn more about Memoization and Dynamic Programming patterns.

Solution Approach

The solution leverages recursive exploration, memoization, and diagonal traversal to find the longest V-shaped diagonal segment in the grid. Here's a breakdown of the implementation:

  1. Define Helper Functions and Data Structures:

    • A dictionary next_digit is used to ensure that as you progress in the diagonal segment, the next number in the sequence is correctly selected (i.e., after 1 must come 2, followed by 0, and then 2 again).

  2. Bounds Checking:

    • A helper function within_bounds ensures we stay within the grid's limits during diagonal exploration.

  3. Recursive Function with Memoization:

    • The function f(i, j, di, dj, turned) is defined using Python's @cache to memoize results, avoiding redundant calculations.
      • i, j: Current position in the grid.
      • di, dj: Current direction of traversal.
      • turned: A boolean indicating if a direction change has occurred.
    • The function returns the length of the longest segment starting from position (i, j) moving in direction (di, dj).

  4. Diagonal Exploration Logic:

    • For every starting point (i, j) with a value 1, it explores in four diagonal directions using directions = ((1, 1), (-1, 1), (1, -1), (-1, -1)).
    • For each position, check if moving in the current direction logically follows the sequence (2, 0, 2, ...).

  5. Allowing One Direction Change:

    • If a direction change hasn’t been made yet (turned is False), attempt a 90-degree clockwise direction change by swapping direction vectors (di, dj = dj, -di) while maintaining the sequence.

  6. Determine and Return Maximum Length:

    • Traverse the grid, initiating traversal from all 1 values.
    • Keep track of and return the maximum length of all evaluated V-shaped diagonal segments.

The recursive and memoized approach efficiently evaluates all potential segments while avoiding redundant work, leveraging symmetries in the problem and caching results for reuse.

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Example Walkthrough

Let's consider a small grid to illustrate the solution approach:

Grid:
[
    [1, 2, 0, 2],
    [2, 1, 2, 0],
    [0, 2, 1, 2],
    [2, 0, 2, 0]
]

Step-by-Step Execution:

  1. Identify Starting Points:

    • Traverse the grid to find elements with the value 1. These are potential starting points for V-shaped segments:
      • (0, 0), (1, 1), and (2, 2).

  2. Directional Exploration:

    • Starting from each 1, explore in the four diagonal directions:
      • Top-left to bottom-right (↘)
      • Bottom-right to top-left (↖)
      • Top-right to bottom-left (↙)
      • Bottom-left to top-right (↗)

  3. Recursive Exploration with Memoization:

    • For each starting point, use a recursive function with memoization to follow the sequence 2, 0, 2, 0, ....
    • Continuation is only allowed within grid bounds and if the sequence 2, 0, 2, ... holds.

  4. Exploration Example:

    • Start at (1, 1) with value 1.
    • Move ↘ to (2, 2) with value 2.
    • Move ↘ to (3, 3) with value 0.
    • At this stage, if within bounds, would move ↘ to the next 2.
    • Now, try allowing one change in direction (say to ↙) to continue the sequence while maintaining the correctness.

  5. Determine Maximum Length:

    • Calculate the length of V-shaped segments starting from each 1.
    • Track the maximum length over all possible start points and directions.

Final Result:

  • The longest detected V-shaped diagonal segment in the example has a length of 4, starting from (1, 1) and following the sequence 1, 2, 0, 2 with one change allowed.

Solution Implementation

1from typing import List
2from functools import cache
3
4class Solution:
5    def lenOfVDiagonal(self, grid: List[List[int]]) -> int:
6        # Get the dimensions of the grid
7        m, n = len(grid), len(grid[0])
8      
9        # Define what the next digit should be for each current digit
10        next_digit = {1: 2, 2: 0, 0: 2}
11
12        # Check if the given coordinates are within the grid bounds
13        def within_bounds(i, j):
14            return 0 <= i < m and 0 <= j < n
15
16        # Memoized recursive function to calculate the length of the diagonal
17        @cache
18        def f(i, j, di, dj, turned):
19            result = 1
20            successor = next_digit[grid[i][j]]
21
22            # Move in the current direction if the next cell has the expected successor digit
23            if within_bounds(i + di, j + dj) and grid[i + di][j + dj] == successor:
24                result = 1 + f(i + di, j + dj, di, dj, turned)
25
26            # If not already turned, try turning and continue in the new direction, checking if allowed
27            if not turned:
28                new_di, new_dj = dj, -di
29                if within_bounds(i + new_di, j + new_dj) and grid[i + new_di][j + new_dj] == successor:
30                    result = max(result, 1 + f(i + new_di, j + new_dj, new_di, new_dj, True))
31
32            return result
33
34        # Define possible diagonal directions
35        directions = ((1, 1), (-1, 1), (1, -1), (-1, -1))
36        result = 0
37
38        # Iterate over each cell in the grid
39        for i in range(m):
40            for j in range(n):
41                # Start exploring only if the cell contains a 1
42                if grid[i][j] != 1:
43                    continue
44                # Try each diagonal direction from the current 1 cell
45                for di, dj in directions:
46                    result = max(result, f(i, j, di, dj, False))
47
48        return result
49
1import java.util.List;
2import java.util.HashMap;
3import java.util.Map;
4
5public class Solution {
6  
7    // Define a recursive function with memoization
8    private int f(int[][] grid, int i, int j, int di, int dj, boolean turned, Map<String, Integer> memo, Map<Integer, Integer> nextDigit) {
9        String key = i + "," + j + "," + di + "," + dj + "," + turned;
10        if (memo.containsKey(key)) {
11            return memo.get(key);
12        }
13      
14        int m = grid.length;
15        int n = grid[0].length;
16      
17        int result = 1; // Start with length 1
18        int successor = nextDigit.get(grid[i][j]);
19      
20        // Check if we can move in the current direction
21        if (withinBounds(i + di, j + dj, m, n) && grid[i + di][j + dj] == successor) {
22            result = 1 + f(grid, i + di, j + dj, di, dj, turned, memo, nextDigit);
23        }
24      
25        // If not yet turned, try turning to a new direction
26        if (!turned) {
27            int newDi = dj, newDj = -di;
28            if (withinBounds(i + newDi, j + newDj, m, n) && grid[i + newDi][j + newDj] == successor) {
29                result = Math.max(result, 1 + f(grid, i + newDi, j + newDj, newDi, newDj, true, memo, nextDigit));
30            }
31        }
32      
33        memo.put(key, result);
34        return result;
35    }
36  
37    // Check if the coordinates are within grid bounds
38    private boolean withinBounds(int i, int j, int m, int n) {
39        return 0 <= i && i < m && 0 <= j && j < n;
40    }
41  
42    public int lenOfVDiagonal(List<List<Integer>> grid) {
43        int m = grid.size();
44        int n = grid.get(0).size();
45      
46        // Define the transition for each digit
47        Map<Integer, Integer> nextDigit = new HashMap<>();
48        nextDigit.put(1, 2);
49        nextDigit.put(2, 0);
50        nextDigit.put(0, 2);
51      
52        int[][] gridArray = new int[m][n];
53        for (int i = 0; i < m; i++) {
54            for (int j = 0; j < n; j++) {
55                gridArray[i][j] = grid.get(i).get(j);
56            }
57        }
58      
59        // Possible diagonal directions
60        int[][] directions = {{1, 1}, {-1, 1}, {1, -1}, {-1, -1}};
61        int result = 0;
62        Map<String, Integer> memo = new HashMap<>();
63      
64        // Iterate through each cell in the grid
65        for (int i = 0; i < m; i++) {
66            for (int j = 0; j < n; j++) {
67                // Start exploring only if the cell contains a 1
68                if (gridArray[i][j] != 1) {
69                    continue;
70                }
71                // Try each diagonal direction from the current cell
72                for (int[] direction : directions) {
73                    result = Math.max(result, f(gridArray, i, j, direction[0], direction[1], false, memo, nextDigit));
74                }
75            }
76        }
77      
78        return result;
79    }
80}
81
1#include <vector>
2#include <unordered_map>
3#include <algorithm>
4using namespace std;
5
6class Solution {
7public:
8    // Function to find the length of the longest "V" shaped diagonal in the grid
9    int lenOfVDiagonal(vector<vector<int>>& grid) {
10        int m = grid.size();    // Number of rows
11        int n = grid[0].size(); // Number of columns
12
13        // Mapping to determine the expected next digit for a given digit
14        unordered_map<int, int> nextDigit = {{1, 2}, {2, 0}, {0, 2}};
15
16        // Check if the given coordinates are within the grid bounds
17        auto withinBounds = [&](int i, int j) {
18            return 0 <= i && i < m && 0 <= j && j < n;
19        };
20
21        // Memoized recursive function to calculate diagonal length
22        auto f = [&](auto&& self, int i, int j, int di, int dj, bool turned) -> int {
23            int result = 1;
24            int successor = nextDigit[grid[i][j]];
25
26            // Move in the current direction if the next cell contains the expected successor digit
27            if (withinBounds(i + di, j + dj) && grid[i + di][j + dj] == successor) {
28                result = 1 + self(self, i + di, j + dj, di, dj, turned);
29            }
30
31            // If not already turned, try turning and continue in the new direction, checking if allowed
32            if (!turned) {
33                int new_di = dj, new_dj = -di;
34                if (withinBounds(i + new_di, j + new_dj) && grid[i + new_di][j + new_dj] == successor) {
35                    result = max(result, 1 + self(self, i + new_di, j + new_dj, new_di, new_dj, true));
36                }
37            }
38
39            return result;
40        };
41
42        // Define possible diagonal directions
43        vector<pair<int, int>> directions = {{1, 1}, {-1, 1}, {1, -1}, {-1, -1}};
44        int result = 0;
45
46        // Iterate over each cell in the grid
47        for (int i = 0; i < m; ++i) {
48            for (int j = 0; j < n; ++j) {
49                // Start exploring only if the cell contains a 1
50                if (grid[i][j] != 1) continue;
51              
52                // Try each diagonal direction from the current 1 cell
53                for (auto [di, dj] : directions) {
54                    result = max(result, f(f, i, j, di, dj, false));
55                }
56            }
57        }
58      
59        return result;
60    }
61};
62
1// Import necessary types from the standard library
2type Grid = number[][];
3
4// Define the directions in which to search for diagonals (right-down, right-up, left-down, left-up)
5const directions = [
6    [1, 1],  // right-down
7    [-1, 1], // right-up
8    [1, -1], // left-down
9    [-1, -1] // left-up
10];
11
12// Define a mapping for the next expected digit in the diagonal sequence
13const nextDigit: { [key: number]: number } = { 1: 2, 2: 0, 0: 2 };
14
15// Check if the given coordinates are within the grid bounds
16const withinBounds = (m: number, n: number, i: number, j: number) =>
17    0 <= i && i < m && 0 <= j && j < n;
18
19// Memoization cache to store results of previously computed diagonal lengths
20const memo = new Map<string, number>();
21
22// Recursive function to calculate the length of the diagonal
23const f = (grid: Grid, i: number, j: number, di: number, dj: number, turned: boolean): number => {
24    const m = grid.length;
25    const n = grid[0].length;
26    const cacheKey = `${i}-${j}-${di}-${dj}-${turned}`;
27    if (memo.has(cacheKey)) {
28        return memo.get(cacheKey)!;
29    }
30
31    let result = 1;
32    const successor = nextDigit[grid[i][j]];
33
34    // Move in the current direction if the next cell has the expected successor digit
35    if (withinBounds(m, n, i + di, j + dj) && grid[i + di][j + dj] === successor) {
36        result = 1 + f(grid, i + di, j + dj, di, dj, turned);
37    }
38
39    // If not already turned, attempt to turn and continue in the new direction
40    if (!turned) {
41        const new_di = dj;
42        const new_dj = -di;
43        if (withinBounds(m, n, i + new_di, j + new_dj) && grid[i + new_di][j + new_dj] === successor) {
44            result = Math.max(result, 1 + f(grid, i + new_di, j + new_dj, new_di, new_dj, true));
45        }
46    }
47
48    memo.set(cacheKey, result);
49    return result;
50};
51
52// Main function to calculate the longest 'V' diagonal of '1','2','0' sequence
53const lenOfVDiagonal = (grid: Grid): number => {
54    const m = grid.length;
55    const n = grid[0].length;
56    let result = 0;
57
58    // Iterate over each cell in the grid to start possible diagonals
59    for (let i = 0; i < m; i++) {
60        for (let j = 0; j < n; j++) {
61            // Start exploring only if the cell contains a 1
62            if (grid[i][j] !== 1) {
63                continue;
64            }
65            // Try each diagonal direction from the current 1 cell
66            for (const [di, dj] of directions) {
67                result = Math.max(result, f(grid, i, j, di, dj, false));
68            }
69        }
70    }
71    return result;
72};
73

Time and Space Complexity

The time complexity of the code is O(m * n * 4), where m and n are the dimensions of the grid. This is because for each cell in the grid, it attempts to explore four directions, and the f function uses memoization to solve subproblems only once.

The space complexity of the code is O(m * n), primarily due to the caching of intermediate results in the recursive calls through the use of the @cache decorator.

Learn more about how to find time and space complexity quickly.

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