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Problem solution · Java

Minimum Operations to Remove Adjacent Ones in Matrix

Minimum Operations to Remove Adjacent Ones in Matrix: a Java solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
48 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Minimum Operations to Remove Adjacent Ones in Matrix, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 48 lines of Java from the credited upstream file 2123.java.
  • The implementation visibly relies on sequence storage.
  • 3 loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Operations to Remove Adjacent Ones in Matrix · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int minimumOperations(int[][] grid) {    final int m = grid.length;    final int n = grid[0].length;    int ans = 0;    int[][] seen = new int[m][n];    int[][] match = new int[m][n];    Arrays.stream(match).forEach(A -> Arrays.fill(A, -1));     for (int i = 0; i < m; ++i)      for (int j = 0; j < n; ++j)        if (grid[i][j] == 1 && match[i][j] == -1) {          final int sessionId = i * n + j;          seen[i][j] = sessionId;          if (dfs(grid, i, j, sessionId, seen, match))            ++ans;        }     return ans;  }   private static final int[][] DIRS = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};   private boolean dfs(final int[][] grid, int i, int j, int sessionId, int[][] seen,                      int[][] match) {    final int m = grid.length;    final int n = grid[0].length;     for (int[] dir : DIRS) {      final int x = i + dir[0];      final int y = j + dir[1];      if (x < 0 || x == m || y < 0 || y == n)        continue;      if (grid[x][y] == 0 || seen[x][y] == sessionId)        continue;      seen[x][y] = sessionId;      if (match[x][y] == -1 ||          dfs(grid, match[x][y] / n, match[x][y] % n, sessionId, seen, match)) {        match[x][y] = i * n + j;        match[i][j] = x * n + y;        return true;      }    }     return false;  }}

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