533. Lonely Pixel II
Explanation
The problem "Lonely Pixel II" asks us to determine the number of "black" pixels in a given picture that meet specific criteria. A black pixel is represented as 'B' and a white pixel is represented as 'W'. The criteria for a black pixel to be considered "lonely" are:
- There must be exactly one black pixel in the same row as the given black pixel.
- There must be exactly one black pixel in the same column as the given black pixel.
To solve this problem efficiently, we can follow these steps:
- Count the number of black pixels in each row and each column.
- For each black pixel, check if it is lonely based on the above criteria.
- Count the total number of lonely black pixels.
Time complexity: O(mn) where m is the number of rows and n is the number of columns in the picture. Space complexity: O(m + n) to store the counts of black pixels in each row and each column.
class Solution {
public int findBlackPixel(char[][] picture, int N) {
int m = picture.length;
int n = picture[0].length;
int[] rowCount = new int[m];
int[] colCount = new int[n];
Map<String, Integer> map = new HashMap<>();
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (picture[i][j] == 'B') {
rowCount[i]++;
colCount[j]++;
}
}
}
for (int i = 0; i < m; i++) {
if (rowCount[i] == N) {
String key = Arrays.toString(picture[i]);
map.put(key, map.getOrDefault(key, 0) + 1);
}
}
int count = 0;
for (int j = 0; j < n; j++) {
if (colCount[j] == N) {
for (Map.Entry<String, Integer> entry : map.entrySet()) {
if (entry.getValue() == N && entry.getKey().charAt(j) == 'B') {
count += N;
}
}
}
}
return count;
}
}Code Editor (Testing phase)
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