Medium
Minimum Domino Rotations For Equal Row — C++
Full explanation · Time O(n) · Space O(1)
// Time: O(n)
// Space: O(1)
class Solution {
public:
int minDominoRotations(vector<int>& A, vector<int>& B) {
set<int> intersect{A[0], B[0]};
for (int i = 1; i < A.size() && !intersect.empty(); ++i) {
const auto s1 = move(intersect);
const set<int> s2{A[i], B[i]};
set_intersection(s1.cbegin(), s1.cend(),
s2.cbegin(), s2.cend(),
inserter(intersect, intersect.begin()));
}
if (intersect.empty()) {
return -1;
}
const auto& x = *intersect.cbegin();
return min(A.size() - count(A.cbegin(), A.cend(), x),
B.size() - count(B.cbegin(), B.cend(), x));
}
};
// Time: O(n)
// Space: O(1)
class Solution2 {
public:
int minDominoRotations(vector<int>& A, vector<int>& B) {
unordered_set<int> intersect{A[0], B[0]};
for (int i = 1; i < A.size() && !intersect.empty(); ++i) {
const auto s1 = move(intersect);
const unordered_set<int> s2{A[i], B[i]};
copy_if(s1.cbegin(), s1.cend(),
inserter(intersect, intersect.begin()),
[&s2](const int x) {
return s2.count(x) > 0;
});
}
if (intersect.empty()) {
return -1;
}
const auto& x = *intersect.cbegin();
return min(A.size() - count(A.cbegin(), A.cend(), x),
B.size() - count(B.cbegin(), B.cend(), x));
}
};