Matrix Questions

Maximal Square

Question

Original Question: Leetcode 221. Maximal Square

Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.

Example:

Input: 

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Output: 4

Solution

Java

class Solution {
    public int maximalSquare(char[][] matrix) {
        if(matrix.length == 0) return 0;
        int[][] dp = new int[matrix.length+1][matrix[0].length+1];
        int maxLen = 0;
        for(int j = 1; j <= matrix.length; j++){
            for(int i = 1; i <= matrix[0].length; i++){
                if(matrix[j-1][i-1] == '1'){
                    dp[j][i] = Math.min(Math.min(dp[j][i-1], dp[j-1][i]), dp[j-1][i-1]) + 1;
                    maxLen = Math.max(maxLen, dp[j][i]);
                }
            }
        }
        return maxLen * maxLen;
    }
}

Python

class Solution:
    def maximalSquare(self, matrix: List[List[str]]) -> int:
        if not matrix or not matrix[0]: 
            return 0 

        m,n, maxL, prev = len(matrix), len(matrix[0]), 0, 0 
        sqLength = [0] *(n+1) 
        for i in range(1,m+1):
            for j in range(1,n+1):
                temp = sqLength[j]
                if matrix[i-1][j-1] == '1':
                    sqLength[j] = min(sqLength[j], prev, sqLength[j-1]) + 1 
                    maxL = max(maxL, sqLength[j])
                else:
                    sqLength[j] = 0 
                prev = temp 

        return maxL*maxL

Count Square Submatrices with All Ones

Question

Original Question: Leetcode 1277. Count Square Submatrices with All Ones

Given a m * n matrix of ones and zeros, return how many square submatrices have all ones.

Example 1:

Input: matrix =
[
  [0,1,1,1],
  [1,1,1,1],
  [0,1,1,1]
]
Output: 15
Explanation: 
There are 10 squares of side 1.
There are 4 squares of side 2.
There is  1 square of side 3.
Total number of squares = 10 + 4 + 1 = 15.

Example 2:

Input: matrix = 
[
  [1,0,1],
  [1,1,0],
  [1,1,0]
]
Output: 7
Explanation: 
There are 6 squares of side 1.  
There is 1 square of side 2. 
Total number of squares = 6 + 1 = 7.

Solution

Java

class Solution {
    public int countSquares(int[][] A) {
        int res = 0;
        for (int i = 0; i < A.length; ++i) {
            for (int j = 0; j < A[0].length; ++j) {
                if (A[i][j] > 0 && i > 0 && j > 0) {
                    A[i][j] = Math.min(A[i - 1][j - 1], Math.min(A[i - 1][j], A[i][j - 1])) + 1;
                }
                res += A[i][j];
            }
        }
        return res;
    }
}

Bomb Enemy

;

Question

Original Question: Leetcode 361. Bomb Enemy

Given a 2D grid, each cell is either a wall 'W', an enemy 'E' or empty '0' (the number zero), return the maximum enemies you can kill using one bomb. The bomb kills all the enemies in the same row and column from the planted point until it hits the wall since the wall is too strong to be destroyed. Note: You can only put the bomb at an empty cell.

Example:

Input: [["0","E","0","0"],["E","0","W","E"],["0","E","0","0"]]
Output: 3 
Explanation: For the given grid,

0 E 0 0 
E 0 W E 
0 E 0 0

Placing a bomb at (1,1) kills 3 enemies.

Solution

Java

class Solution {
    public int maxKilledEnemies(char[][] grid) {
        if(grid.length == 0 || grid[0].length == 0) return 0;
        int m = grid.length, n=grid[0].length;
        int[][] up = new int[m][n];
        int[][] down = new int[m][n];
        int[][] left = new int[m][n];
        int[][] right = new int[m][n];
        
        // fill up;
        for(int i = 0; i < m; i++){
           for(int j = 0; j < n; j++){
               if(grid[i][j] == 'W'){
                   up[i][j] = 0;
                   continue;
               }
               
               up[i][j] = grid[i][j] == 'E' ? 1 : 0;
               if(i>0){
                   up[i][j] += up[i-1][j];
               }
           } 
        }
        // fill left;
        for(int i = 0; i < m; i++){
           for(int j = 0; j < n; j++){
               if(grid[i][j] == 'W'){
                   left[i][j] = 0;
                   continue;
               }
               
               left[i][j] = grid[i][j] == 'E' ? 1 : 0;
               if(j>0){
                   left[i][j] += left[i][j-1];
               }
           } 
        }
        
        // fill down;
        for(int i = m-1; i >= 0; i--){
           for(int j = n-1; j >= 0; j--){
               if(grid[i][j] == 'W'){
                   down[i][j] = 0;
                   continue;
               }
               
               down[i][j] = grid[i][j] == 'E' ? 1 : 0;
               if(i < m-1){
                   down[i][j] += down[i+1][j];
               }
           } 
        }
        
        // fill right;
        for(int i = m-1; i >= 0; i--){
           for(int j = n-1; j >= 0; j--){
               if(grid[i][j] == 'W'){
                   right[i][j] = 0;
                   continue;
               }
               
               right[i][j] = grid[i][j] == 'E' ? 1 : 0;
               if(j < n-1){
                   right[i][j] += right[i][j+1];
               }
           } 
        }
        
        int res = 0;
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == '0'){
                    res = Math.max(res, up[i][j] + down[i][j] + right[i][j] + left[i][j]);
                }
            }
        }
        return res;
    }
}