Skip to content

Capture Regions On Board

Logical Breakdown

  • [ ] Subproblem Identification:
  • [ ] Optimal Substructure:
  • [ ] Constraint Handling: (e.g., Modulo \(10^9 + 7\))
  • [ ] Optimization: (Matrix Exponentiation if 'A' is huge)

Mathematical Rigor

State Definition

Let \(dp[i]\) be the state for the \(i\)-th subproblem.

Recurrence Relation

\[ dp[i] = ... \]

Visualization

graph TD A["Current State"] --> B["Previous State 1"] A --> C["Previous State 2"]

Complexity Analysis

Approach Time Complexity Space Complexity
Iterative DP \(O(N)\) \(O(1)\)

Code Reference

package com.dsa.graphs;

import java.util.*;

/ * ============================================================================================ * PROBLEM: Capture Regions on Board * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * Given an N x M board containing 'X' and 'O'. A region of 'O's is "captured" (flipped to 'X') * if it is completely surrounded by 'X's. * * Boundary Rule: * Any 'O' on the boundary, or connected to an 'O' on the boundary, cannot be captured. * * --------------------- * 2. Logical Breakdown & Mathematical Formulation * --------------------- * Instead of finding surrounded regions directly, it is mathematically simpler to find * "unsurroundable" regions using a Flood Fill approach starting from the boundaries. * * Step 1: Boundary DFS/BFS * - Traverse the four boundaries of the matrix: * - Top row, Bottom row, Left column, Right column. * - For every 'O' found on the boundary, initiate a DFS/BFS to find all connected 'O's. * - Mark these "safe" 'O's with a temporary placeholder (e.g., ''). * * Step 2: Final Transformation * - Iterate through the entire matrix: * - If an element is 'O', it must be surrounded (since it wasn't reached from the boundary). * Flip it to 'X'. * - If an element is our placeholder '', it is safe. Flip it back to 'O'. * - If an element is 'X', leave it as is. * * --------------------- * 3. Constraints & Complexity Analysis * --------------------- * Constraints: * 1 <= N, M <= 1000 * * Time Complexity: * O(N * M) - Every cell is visited at most twice (once for boundary check/DFS, once for final flip). * * Space Complexity: * O(N * M) - In the worst case, the recursion stack (DFS) or queue (BFS) can grow to the size * of the board. * ============================================================================================ */ public class CaptureRegionsOnBoard { public int solve(ArrayList> A) { // TODO: Implement solution return 0; } }