- Difficulty: Medium
- Tags: LeetCode, Medium, Array, Hash Table, Matrix, leetcode-533, O(m * n), 🔒
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
rand columncboth contain exactlytargetblack pixels. - For all rows that have a black pixel at column
c, they should be exactly the same as rowr.
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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
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Constraints:
m ==Â picture.lengthn ==Â picture[i].length1 <= m, n <= 200picture[i][j]is'W'or'B'.1 <= target <= min(m, n)