533. Lonely Pixel II
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 r and column c both contain exactly target black pixels.
- For all rows that have a black pixel at column c, they should be exactly the same as row r.
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| class Solution {
string toString(vector<char>& A) {
string res = "";
for(auto& a : A) res.push_back(a);
return res;
}
public:
int findBlackPixel(vector<vector<char>>& picture, int target) {
vector<string> P;
vector<int> rsum, csum;
for(auto& pic : picture) {
P.push_back(toString(pic));
int s = 0;
for(auto& pp : pic) s += pp == 'B';
rsum.push_back(s);
}
int n = picture.size(), m = picture[0].size(), res = 0;
unordered_map<int, vector<int>> mp;
for(int i = 0; i < m; i++) {
int s = 0;
for(int j = 0; j < n; j++) {
s += picture[j][i] == 'B';
if(picture[j][i] == 'B') mp[i].push_back(j);
}
csum.push_back(s);
}
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
if(picture[i][j] == 'W') continue;
if(rsum[i] != target or csum[j] != target) continue;
bool pass = true;
for(auto& v : mp[j]) {
if(P[v] == P[i]) continue;
pass = false;
break;
}
res += pass;
}
}
return res;
}
};
|
