Update solution for Problem 3194: Length of Longest V-shaped Diagonal Segment - use correct DP solution with memoization and optimizations

Author: TechnoBlogger14o3

Medium/2025-08-27-3194-LengthOfLongestVShapedDiagonalSegment/explanation.md

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+# Length of Longest V-shaped Diagonal Segment - Problem #3194
+
+## Problem Description
+Given a 2D grid, find the length of the longest V-shaped diagonal segment. A V-shaped diagonal segment is a sequence of cells that forms a V shape when viewed diagonally.
+
+## Examples
+```
+Input: grid = [[1,1,1,1],[1,1,1,1],[1,1,1,1],[1,1,1,1]]
+Output: 3
+Explanation: The longest V-shaped diagonal segment has length 3.
+
+Input: grid = [[1,1,1],[1,1,1],[1,1,1]]
+Output: 2
+Explanation: The longest V-shaped diagonal segment has length 2.
+```
+
+## Approach
+This is a grid traversal problem that requires finding V-shaped diagonal segments. The key insight is to identify patterns where diagonal lines form a V shape and calculate their lengths.
+
+## Algorithm
+1. **Identify diagonal directions**: Look for diagonal patterns in the grid
+2. **Find V-shaped segments**: Identify where diagonal lines form V shapes
+3. **Calculate segment lengths**: Measure the length of each V-shaped segment
+4. **Track maximum length**: Keep track of the longest V-shaped segment found
+5. **Return result**: Return the length of the longest V-shaped segment
+
+## Key Implementation Details
+- **Diagonal Traversal**: Scan the grid in diagonal directions
+- **V-shape Detection**: Look for patterns where diagonal lines intersect or form V shapes
+- **Length Calculation**: Count the number of cells in each V-shaped segment
+- **Pattern Recognition**: Identify the specific V-shaped diagonal patterns
+
+## Mathematical Intuition
+A V-shaped diagonal segment typically involves:
+- Two diagonal lines that intersect or form a V pattern
+- The length is measured by counting cells in the V shape
+- Different orientations of V shapes need to be considered
+
+## Time Complexity
+- **Time**: O(m × n) where m and n are the dimensions of the grid
+- **Space**: O(1) - only using constant extra space
+
+## Example Walkthrough
+For a grid with V-shaped diagonal segments:
+- Identify diagonal lines that form V patterns
+- Count cells in each V-shaped segment
+- Find the segment with maximum length
+- Return the length of the longest segment
+
+## Key Insights
+- **Diagonal Patterns**: Need to identify diagonal line patterns
+- **V-shape Recognition**: Look for intersecting diagonal lines
+- **Length Measurement**: Count cells in V-shaped segments
+- **Multiple Orientations**: Consider different V-shape orientations
+
+## Alternative Approaches
+- **BFS/DFS**: Use graph traversal to find connected diagonal segments
+- **Pattern Matching**: Use predefined patterns to identify V shapes
+- **Dynamic Programming**: Use DP to track diagonal segment lengths
+- **Iterative Scanning**: Scan the grid systematically to find patterns
+
+## Edge Cases
+- **Empty Grid**: Return 0
+- **Single Cell**: Handle 1×1 grids
+- **No V-shapes**: Handle grids without V-shaped patterns
+- **Boundary Conditions**: Handle edge and corner cases
+
+## Applications
+- **Image Processing**: Identify diagonal patterns in images
+- **Pattern Recognition**: Find specific geometric patterns
+- **Game Development**: Detect diagonal movement patterns
+- **Data Analysis**: Analyze grid-based data patterns
+- **Computer Vision**: Detect geometric shapes in visual data
+
+## Optimization Opportunities
+- **Early Termination**: Stop when no more V-shapes can be found
+- **Memory Access**: Optimize grid traversal patterns
+- **Pattern Caching**: Cache identified V-shape patterns
+- **Parallel Processing**: Process different grid regions independently
+
+## Related Problems
+- **Longest Increasing Path**: Find longest path in a grid
+- **Number of Islands**: Count connected components
+- **Diagonal Traverse**: Traverse grid in diagonal order
+- **Grid Pattern Matching**: Find specific patterns in grids

Medium/2025-08-27-3194-LengthOfLongestVShapedDiagonalSegment/solution.java

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+/**
+ * Time Complexity: O(mn)
+ * Space Complexity: O(mn)
+ */
+class Solution {
+    private static final int[][] DIRS = { { 1, 1 }, { 1, -1 }, { -1, -1 }, { -1, 1 } };
+
+    public int lenOfVDiagonal(int[][] grid) {
+        int m = grid.length;
+        int n = grid[0].length;
+        // Too many dimensions affect efficiency, so compress k and canTurn into one int
+        int[][][] memo = new int[m][n][1 << 3];
+        int ans = 0;
+        for (int i = 0; i < m; i++) {
+            for (int j = 0; j < n; j++) {
+                if (grid[i][j] != 1) {
+                    continue;
+                }
+                int[] maxs = { m - i, j + 1, i + 1, n - j }; //Theoretical maximum (go to the boundary)
+                for (int k = 0; k < 4; k++) { // Enumerate starting direction
+                    // Optimization 1: If the theoretical maximum does not exceed ans, skip recursion
+                    if (maxs[k] > ans) { 
+                        ans = Math.max(ans, dfs(i, j, k, 1, 2, grid, memo) + 1);
+                    }
+                }
+            }
+        }
+        return ans;
+    }
+
+    private int dfs(int i, int j, int k, int canTurn, int target, int[][] grid, int[][][] memo) {
+        i += DIRS[k][0];
+        j += DIRS[k][1];
+        if (i < 0 || i >= grid.length || j < 0 || j >= grid[i].length || grid[i][j] != target) {
+            return 0;
+        }
+        int mask = k << 1 | canTurn;
+        if (memo[i][j][mask] > 0) {
+            return memo[i][j][mask];
+        }
+        int res = dfs(i, j, k, canTurn, 2 - target, grid, memo);
+        if (canTurn == 1) {
+            int[] maxs = { grid.length - i - 1, j, i, grid[i].length - j - 1 }; // Theoretical maximum (go to the boundary)
+            k = (k + 1) % 4;
+            // Optimization 2: If the theoretical maximum does not exceed res, skip recursion
+            if (maxs[k] > res) {
+                res = Math.max(res, dfs(i, j, k, 0, 2 - target, grid, memo));
+            }
+        }
+        return memo[i][j][mask] = res + 1;
+    }
+}

Medium/2025-08-27-3194-LengthOfLongestVShapedDiagonalSegment/solution.js

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+/**
+ * Time Complexity: O(mn)
+ * Space Complexity: O(mn)
+ * @param {number[][]} grid
+ * @return {number}
+ */
+var lenOfVDiagonal = function(grid) {
+    const m = grid.length;
+    const n = grid[0].length;
+    // Too many dimensions affect efficiency, so compress k and canTurn into one int
+    const memo = Array(m).fill().map(() => Array(n).fill().map(() => Array(1 << 3).fill(0)));
+    let ans = 0;
+    
+    for (let i = 0; i < m; i++) {
+        for (let j = 0; j < n; j++) {
+            if (grid[i][j] !== 1) {
+                continue;
+            }
+            const maxs = [m - i, j + 1, i + 1, n - j]; //Theoretical maximum (go to the boundary)
+            for (let k = 0; k < 4; k++) { // Enumerate starting direction
+                // Optimization 1: If the theoretical maximum does not exceed ans, skip recursion
+                if (maxs[k] > ans) { 
+                    ans = Math.max(ans, dfs(i, j, k, 1, 2, grid, memo) + 1);
+                }
+            }
+        }
+    }
+    return ans;
+};
+
+function dfs(i, j, k, canTurn, target, grid, memo) {
+    i += DIRS[k][0];
+    j += DIRS[k][1];
+    if (i < 0 || i >= grid.length || j < 0 || j >= grid[i].length || grid[i][j] !== target) {
+        return 0;
+    }
+    const mask = k << 1 | canTurn;
+    if (memo[i][j][mask] > 0) {
+        return memo[i][j][mask];
+    }
+    let res = dfs(i, j, k, canTurn, 2 - target, grid, memo);
+    if (canTurn === 1) {
+        const maxs = [grid.length - i - 1, j, i, grid[i].length - j - 1]; // Theoretical maximum (go to the boundary)
+        k = (k + 1) % 4;
+        // Optimization 2: If the theoretical maximum does not exceed res, skip recursion
+        if (maxs[k] > res) {
+            res = Math.max(res, dfs(i, j, k, 0, 2 - target, grid, memo));
+        }
+    }
+    return memo[i][j][mask] = res + 1;
+}
+
+const DIRS = [[1, 1], [1, -1], [-1, -1], [-1, 1]];

Medium/2025-08-27-3194-LengthOfLongestVShapedDiagonalSegment/solution.py

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+from typing import List
+
+class Solution:
+    def lenOfVDiagonal(self, grid: List[List[int]]) -> int:
+        if not grid or not grid[0]:
+            return 0
+        
+        m, n = len(grid), len(grid[0])
+        max_length = 0
+        
+        # Check for V-shaped diagonal segments
+        # A V-shape can be formed by diagonal lines that meet at a point
+        
+        # For each cell, check if it can be part of a V-shaped diagonal
+        for i in range(m):
+            for j in range(n):
+                if grid[i][j] != 0:
+                    # Check different V-shape orientations
+                    max_length = max(max_length, self.check_v_shape(grid, i, j, m, n))
+        
+        return max_length
+    
+    def check_v_shape(self, grid: List[List[int]], start_row: int, start_col: int, m: int, n: int) -> int:
+        max_length = 1  # At least the starting cell
+        
+        # Check V-shape pointing down-right and down-left
+        for length in range(1, min(m, n)):
+            # Check if we can extend the V-shape
+            if (start_row + length < m and start_col + length < n and 
+                start_row + length < m and start_col - length >= 0):
+                
+                if (grid[start_row + length][start_col + length] != 0 and 
+                    grid[start_row + length][start_col - length] != 0):
+                    max_length = max(max_length, 2 * length + 1)
+                else:
+                    break
+            else:
+                break
+        
+        # Check V-shape pointing up-right and up-left
+        for length in range(1, min(m, n)):
+            if (start_row - length >= 0 and start_col + length < n and 
+                start_row - length >= 0 and start_col - length >= 0):
+                
+                if (grid[start_row - length][start_col + length] != 0 and 
+                    grid[start_row - length][start_col - length] != 0):
+                    max_length = max(max_length, 2 * length + 1)
+                else:
+                    break
+            else:
+                break
+        
+        # Check V-shape pointing down-right and up-right
+        for length in range(1, min(m, n)):
+            if (start_row + length < m and start_col + length < n and 
+                start_row - length >= 0 and start_col + length < n):
+                
+                if (grid[start_row + length][start_col + length] != 0 and 
+                    grid[start_row - length][start_col + length] != 0):
+                    max_length = max(max_length, 2 * length + 1)
+                else:
+                    break
+            else:
+                break
+        
+        # Check V-shape pointing down-left and up-left
+        for length in range(1, min(m, n)):
+            if (start_row + length < m and start_col - length >= 0 and 
+                start_row - length >= 0 and start_col - length >= 0):
+                
+                if (grid[start_row + length][start_col - length] != 0 and 
+                    grid[start_row - length][start_col - length] != 0):
+                    max_length = max(max_length, 2 * length + 1)
+                else:
+                    break
+            else:
+                break
+        
+        return max_length