Medium
Minimize the Maximum of Two Arrays — C++
Full explanation · Time O(log(min(d1, d2))) · Space O(1)
// Time: O(log(min(d1, d2)))
// Space: O(1)
// number theory
class Solution1 {
public:
int minimizeSet(int divisor1, int divisor2, int uniqueCnt1, int uniqueCnt2) {
const auto& count = [](int cnt, int64_t d1, int64_t d2) {
const int64_t l = lcm(d1, d2);
return cnt + cnt / (l - 1) - static_cast<int>(cnt % (l - 1) == 0);
};
return max({count(uniqueCnt1, divisor1, 1),
count(uniqueCnt2, divisor2, 1),
count(uniqueCnt1 + uniqueCnt2, divisor1, divisor2)});
}
};
// Time: O(log(min(d1, d2)) + logr)
// Space: O(1)
// binary search
class Solution2 {
public:
int minimizeSet(int divisor1, int divisor2, int uniqueCnt1, int uniqueCnt2) {
const int64_t l = lcm(static_cast<int64_t>(divisor1), static_cast<int64_t>(divisor2));
const auto& check = [&](int cnt) {
return cnt - cnt / divisor1 >= uniqueCnt1 &&
cnt - cnt / divisor2 >= uniqueCnt2 &&
cnt - cnt / l >= uniqueCnt1 + uniqueCnt2;
};
int left = 2, right = numeric_limits<int>::max();
while (left <= right) {
const int mid = left + (right - left) / 2;
if (check(mid)) {
right = mid - 1;
} else {
left = mid + 1;
}
}
return left;
}
};