531

531. Lonely Pixel I (Medium)

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","W"],["B","B","B"]]
Output: 0

 

Constraints:

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

Companies: Amazon, Google

Related Topics:
Array, Hash Table, Matrix

Similar Questions:

• Lonely Pixel II (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/lonely-pixel-i
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(M + N)
class Solution {
public:
    int findLonelyPixel(vector<vector<char>>& A) {
        int M = A.size(), N = A[0].size(), ans = 0;
        vector<int> row(N), col(N);
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (A[i][j] == 'W') continue;
                row[i]++;
                col[j]++;
            }
        }
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (A[i][j] == 'W') continue;
                if (row[i] == 1 && col[j] == 1) ++ans;
            }
        }
        return ans;
    }
};