531. Lonely Pixel I

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Problem

Given an m x n picture consisting of black ‘B’ and white ‘W’ pixels, return the number of black lonely pixels.

A black lonely pixel is a character ‘B’ that located at a specific position where the same row and same column don’t have any other black pixels.

Example 1:

Input: picture = [["W","W","B"],["W","B","W"],["B","W","W"]]
Output: 3
Explanation: All the three 'B's are black lonely pixels.

Example 2:

Input: picture = [["B","B","B"],["B","B","B"],["B","B","B"]]
Output: 0

Constraints:

• m == picture.length
• n == picture[i].length
• 1 <= m, n <= 500
• picture[i][j] is ‘W’ or ‘B’.

Code

73. Set Matrix Zeroes

class Solution {
    public int findLonelyPixel(char[][] picture) {
        int m = picture.length;
        int n = picture[0].length;
        
        int res = 0;
        // check col 0 and row 0
        for (int i = 0; i < m; i++) {
            if(checkLonelyPixel(picture, i, 0)) res++;
        }

        for (int j = 1; j < n; j++) {
            if(checkLonelyPixel(picture, 0, j)) res++;
        }

        // B -> 1, W -> 0        
        for (int i = 0; i < m; i++) {
            picture[i][0] = (picture[i][0] == 'B' ? '1' : '0');
        }
        
        for (int j = 0; j < n; j++) {
            picture[0][j] = (picture[0][j] == 'B' ? '1' : '0');
        }

        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (picture[i][j] == 'B') {
                    picture[i][0]++;
                    picture[0][j]++;
                }
            }
        }
        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (picture[i][j] == 'B') {
                    if (picture[0][j] == '1' && picture[i][0] == '1') {
                        res++;
                    }
                }
            }
        }
        
        return res;
    }
    
    private boolean checkLonelyPixel(char[][] picture, int x, int y) {
        if(picture[x][y] != 'B') return false;
        
        int res = 1;
        
        // check y column
        for (int i = 0; i < picture.length; i++) {
            if(i != x && picture[i][y] == 'B') res++;
        }
        
        // check x row
        for (int j = 0; j < picture[0].length; j++) {
            if(j != y && picture[x][j] == 'B') res++;
        }
        
        return res == 1;
    }
}