Given a picture consisting of black and white pixels, and a positive integer N, find the number of black pixels located at some specific row R and column C that align with all the following rules:
- Row R and column C both contain exactly N black pixels.
- For all rows that have a black pixel at column C, they should be exactly the same as row R
The picture is represented by a 2D char array consisting of 'B' and 'W', which means black and white pixels respectively.
Example:
Input:
[['W', 'B', 'W', 'B', 'B', 'W'],
['W', 'B', 'W', 'B', 'B', 'W'],
['W', 'B', 'W', 'B', 'B', 'W'],
['W', 'W', 'B', 'W', 'B', 'W']]
N = 3
Output: 6
Explanation: All the bold 'B' are the black pixels we need (all 'B's at column 1 and 3).
0 1 2 3 4 5 column index
0 [['W', 'B', 'W', 'B', 'B', 'W'],
1 ['W', 'B', 'W', 'B', 'B', 'W'],
2 ['W', 'B', 'W', 'B', 'B', 'W'],
3 ['W', 'W', 'B', 'W', 'B', 'W']]
row index
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 N = 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.
Note:
- The range of width and height of the input 2D array is [1,200].
class Solution:
def findBlackPixel(self, picture: List[List[str]], target: int) -> int:
m, n = len(picture), len(picture[0])
rows = [0] * m
cols = defaultdict(list)
for i in range(m):
for j in range(n):
if picture[i][j] == 'B':
rows[i] += 1
cols[j].append(i)
t = [[False] * m for _ in range(m)]
for i in range(m):
for k in range(i, m):
if i == k:
t[i][k] = True
else:
t[i][k] = all([picture[i][j] == picture[k][j] for j in range(n)])
t[k][i] = t[i][k]
res = 0
for i in range(m):
if rows[i] == target:
for j in range(n):
if len(cols[j]) == target and all([t[i][k] for k in cols[j]]):
res += 1
return resclass Solution {
public int findBlackPixel(char[][] picture, int target) {
int m = picture.length, n = picture[0].length;
int[] rows = new int[m];
Map<Integer, List<Integer>> cols = new HashMap<>();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (picture[i][j] == 'B') {
++rows[i];
cols.computeIfAbsent(j, k -> new ArrayList<>()).add(i);
}
}
}
boolean[][] t = new boolean[m][m];
for (int i = 0; i < m; ++i) {
for (int k = i; k < m; ++k) {
t[i][k] = i == k || all(picture[i], picture[k]);
t[k][i] = t[i][k];
}
}
int res = 0;
for (int i = 0; i < m; ++i) {
if (rows[i] == target) {
for (int j = 0; j < n; ++j) {
List<Integer> col = cols.get(j);
if (col != null && col.size() == target) {
boolean check = true;
for (int k : col) {
check = check && t[i][k];
}
if (check) {
++res;
}
}
}
}
}
return res;
}
private boolean all(char[] row1, char[] row2) {
int n = row1.length;
for (int j = 0; j < n; ++j) {
if (row1[j] != row2[j]) {
return false;
}
}
return true;
}
}class Solution {
public:
int findBlackPixel(vector<vector<char>>& picture, int target) {
int m = picture.size(), n = picture[0].size();
vector<int> rows(m);
unordered_map<int, vector<int>> cols;
for (int i = 0; i < m; ++i)
{
for (int j = 0; j < n; ++j)
{
if (picture[i][j] == 'B')
{
++rows[i];
cols[j].push_back(i);
}
}
}
vector<vector<bool>> t(m, vector<bool>(m, false));
for (int i = 0; i < m; ++i)
{
for (int k = i; k < m; ++k)
{
t[i][k] = i == k || all(picture[i], picture[k]);
t[k][i] = t[i][k];
}
}
int res = 0;
for (int i = 0; i < m; ++i)
{
if (rows[i] == target)
{
for (int j = 0; j < n; ++j)
{
if (cols[j].size() == target)
{
bool check = true;
for (int k : cols[j]) check = check && t[i][k];
if (check) ++res;
}
}
}
}
return res;
}
bool all(vector<char>& row1, vector<char>& row2) {
int n = row1.size();
for (int j = 0; j < n; ++j)
if (row1[j] != row2[j]) return false;
return true;
}
};func findBlackPixel(picture [][]byte, target int) int {
m, n := len(picture), len(picture[0])
rows := make([]int, m)
cols := make(map[int][]int)
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
if picture[i][j] == 'B' {
rows[i]++
cols[j] = append(cols[j], i)
}
}
}
t := make([][]bool, m)
for i := 0; i < m; i++ {
t[i] = make([]bool, m)
}
for i := 0; i < m; i++ {
for k := i; k < m; k++ {
if i == k {
t[i][k] = true
} else {
t[i][k] = all(picture[i], picture[k])
}
t[k][i] = t[i][k]
}
}
res := 0
for i := 0; i < m; i++ {
if rows[i] == target {
for j := 0; j < n; j++ {
col, ok := cols[j]
if ok && len(col) == target {
check := true
for _, k := range col {
check = check && t[i][k]
}
if check {
res++
}
}
}
}
}
return res
}
func all(row1, row2 []byte) bool {
n := len(row1)
for i := 0; i < n; i++ {
if row1[i] != row2[i] {
return false
}
}
return true
}