533. Lonely Pixel II

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LeetCode

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Problem

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

A black lonely pixel is a character ‘B’ that located at a specific position (r, c) where:

• Row r and column c both contain exactly target black pixels.
• For all rows that have a black pixel at column c, they should be exactly the same as row r.

Example 1:

Input: picture = [["W","B","W","B","B","W"],["W","B","W","B","B","W"],["W","B","W","B","B","W"],["W","W","B","W","B","W"]], target = 3
Output: 6
Explanation: All the green 'B' are the black pixels we need (all 'B's at column 1 and 3).
Take 'B' at row r = 0 and column c = 1 as an example:
 - Rule 1, row r = 0 and column c = 1 both have exactly target = 3 black pixels. 
 - Rule 2, the rows have black pixel at column c = 1 are row 0, row 1 and row 2. They are exactly the same as row r = 0.

Example 2:

Input: picture = [["W","W","B"],["W","W","B"],["W","W","B"]], target = 1
Output: 0

Constraints:

• m == picture.length
• n == picture[i].length
• 1 <= m, n <= 200
• picture[i][j] is ‘W’ or ‘B’.
• 1 <= target <= min(m, n)

Code

[
    ["W","B","W","B","B","W"],
    ["B","W","B","W","W","B"],
    ["W","B","W","B","B","W"],
    ["B","W","B","W","W","B"],
    ["W","W","W","B","B","W"],
    ["B","W","B","W","W","B"]
]

target = 3

result is 9, not 13

class Solution {
    public int findBlackPixel(char[][] picture, int target) {
        int m = picture.length;
        int n = picture[0].length;
        
        Map<String, Integer> map = new HashMap<>();
        // 每一列有多少B
        int[] colCount = new int[n];
        
        for (int i = 0; i < m; i++) {
            int rowCount = 0;
            StringBuilder sb = new StringBuilder();
            
            for(int j = 0; j < n; j++) {
                if (picture[i][j] == 'B') {
                    rowCount++;
                    colCount[j]++;
                }

                sb.append(picture[i][j]);
            }
            
            // 本行符合要求
            if (rowCount == target) {
                String str = sb.toString();
                map.put(str, map.getOrDefault(str, 0) + 1);
            }
        }
        
        int res = 0;
        
        for (String rowStr : map.keySet()) {
            // 所有有target个B的行 正好有target个
            if (map.get(rowStr) == target) {
                for (int j = 0; j < n; j++) {
                    if (rowStr.charAt(j) == 'B' && colCount[j] == target) {
                        res += target;
                    }
                }
            }
        }
        
        return res;
    }
}