[LeetCode] 533. Lonely Pixel II 孤独的像素 II
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].
531. Lonely Pixel I 的拓展，这题多了一个整数N，行和列可以最多含有N个黑元素。
解法：思路和531题一样，还是用一个行和一个列先统计黑元素的个数，然后在遍历一遍图，是黑元素并且行和列都等于N的就加到结果，最后返回结果。这题又多出一个条件2：要保证这N行的黑元素的位置要完全一样，可以一个一个的来比较。好的做法是把行元素转换成字符串或tuple，用一个哈希表记录行元素，然后直接比较行。
0[['W', 'B', 'W', 'B', 'B', 'W'],
1 ['W', 'W', 'B', 'B', 'B', 'W'],
2 ['W', 'B', 'W', 'B', 'B', 'W'],
3 ['W', 'B', 'W', 'W', 'B', 'W']]
Java:
public int findBlackPixel(char[][] picture, int N) {
int m = picture.length;
if (m == 0) return 0;
int n = picture[0].length;
if (n == 0) return 0;
Map<String, Integer> map = new HashMap<>();
int[] colCount = new int[n];
for (int i = 0; i < m; i++) {
String key = scanRow(picture, i, N, colCount);
if (key.length() != 0) {
map.put(key, map.getOrDefault(key, 0) + 1);
}
}
int result = 0;
for (String key : map.keySet()) {
if (map.get(key) == N) {
for (int j = 0; j < n; j++) {
if (key.charAt(j) == 'B' && colCount[j] == N) {
result += N;
}
}
}
}
return result;
}
private String scanRow(char[][] picture, int row, int target, int[] colCount) {
int n = picture[0].length;
int rowCount = 0;
StringBuilder sb = new StringBuilder();
for (int j = 0; j < n; j++) {
if (picture[row][j] == 'B') {
rowCount++;
colCount[j]++;
}
sb.append(picture[row][j]);
}
if (rowCount == target) return sb.toString();
return "";
}
Java: using hashmap to store the string of rows. So that we don't need to check rule 2
public int findBlackPixel(char[][] picture, int N) {
if (picture == null || picture.length == 0 || picture[0].length == 0) return 0;
int m = picture.length, n = picture[0].length;
int[] cols = new int[n];
Map<String, Integer> map = new HashMap<>();
for (int i = 0; i < m; i++) {
int count = 0;
StringBuilder sb = new StringBuilder();
for (int j = 0; j < n; j++) {
if (picture[i][j] == 'B') {
cols[j]++;
count++;
}
sb.append(picture[i][j]);
}
if (count == N) {
String curRow = sb.toString();
map.put(curRow, map.getOrDefault(curRow, 0) + 1);
}
}
int res = 0;
for (String row : map.keySet()) {
if (map.get(row) != N) continue;
for (int i = 0; i < n; i++) {
if (row.charAt(i) == 'B' && cols[i] == N) {
res += N;
}
}
}
return res;
}
Python:
class Solution(object):
def findBlackPixel(self, picture, N):
"""
:type picture: List[List[str]]
:type N: int
:rtype: int
"""
rows, cols = [0] * len(picture), [0] * len(picture[0])
lookup = collections.defaultdict(int)
for i in xrange(len(picture)):
for j in xrange(len(picture[0])):
if picture[i][j] == 'B':
rows[i] += 1
cols[j] += 1
lookup[tuple(picture[i])] += 1
result = 0
for i in xrange(len(picture)):
if rows[i] == N and lookup[tuple(picture[i])] == N:
for j in xrange(len(picture[0])):
result += picture[i][j] == 'B' and cols[j] == N
return result
Python:
class Solution(object):
def findBlackPixel(self, picture, N):
"""
:type picture: List[List[str]]
:type N: int
:rtype: int
"""
lookup = collections.Counter(map(tuple, picture))
cols = [col.count('B') for col in zip(*picture)]
return sum(N * zip(row, cols).count(('B', N)) \
for row, cnt in lookup.iteritems() \
if cnt == N == row.count('B'))
Python:
class Solution(object):
def findBlackPixel(self, picture, N):
"""
:type picture: List[List[str]]
:type N: int
:rtype: int
"""
w, h = len(picture), len(picture[0])
rows, cols = [0] * w, [0] * h
for x in range(w):
for y in range(h):
if picture[x][y] == 'B':
rows[x] += 1
cols[y] += 1
sdict = collections.defaultdict(int)
for idx, row in enumerate(picture):
sdict[''.join(row)] += 1
ans = 0
for x in range(w):
row = ''.join(picture[x])
if sdict[row] != N:
continue
for y in range(h):
if picture[x][y] == 'B':
if rows[x] == N:
if cols[y] == N:
ans += 1
return ans
C++:
class Solution {
public:
int findBlackPixel(vector<vector<char>>& picture, int N) {
vector<int> rows = vector<int>(picture.size());
vector<int> cols = vector<int>(picture[0].size());
unordered_map<string, int> lookup;
vector<string> srows;
for (int i = 0; i < picture.size(); ++i) {
string s;
for (int j = 0; j < picture[0].size(); ++j) {
if (picture[i][j] == 'B') {
++rows[i];
++cols[j];
}
s.push_back(picture[i][j]);
}
++lookup[s];
srows.emplace_back(move(s));
}
int result = 0;
for (int i = 0; i < picture.size(); ++i) {
if (rows[i] == N && lookup[srows[i]] == N) {
for (int j = 0; j < picture[0].size(); ++j) {
result += picture[i][j] == 'B' && cols[j] == N;
}
}
}
return result;
}
};
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[LeetCode] 531. Lonely Pixel I 孤独的像素 I
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