# My hacky straightforward hashtable solution

• ``````void setZeroes(vector<vector<int> > &matrix) {
unordered_set<int> zeroR, zeroC;
for (int i = 0; i < matrix.size(); i++) {
for (int j = 0; j < matrix[i].size(); j++) {
if (matrix[i][j] == 0) {
if (zeroR.find(i) == zeroR.end())
zeroR.insert(i);
if (zeroC.find(j) == zeroC.end())
zeroC.insert(j);
}
}
}

for (auto it = zeroR.begin(); it != zeroR.end(); it++) {
int i = *it;
for (int j = 0; j < matrix[i].size(); j++) {
matrix[i][j] = 0;
}
}

for (auto it = zeroC.begin(); it != zeroC.end(); it++) {
int j = *it;
for (int i = 0; i < matrix.size(); i++) {
matrix[i][j] = 0;
}
}
}
``````

I think this might be too tricky so in real interview my solution may not be accepted... but I think this is quite straightforward to me although in worst case I need O(m+n) space for storing the hash table itself. The size of the hash table is completely depending on how many zeros.

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