Hard
Find Minimum Time to Finish All Jobs — C++
Full explanation · Time O(k^n * logr) · Space O(n + k)
// Time: O(k^n * logr), the real complexity should be much less, but hard to analyze
// Space: O(n + k)
class Solution {
public:
int minimumTimeRequired(vector<int>& jobs, int k) {
sort(begin(jobs), end(jobs), greater<int>());
int left = *max_element(cbegin(jobs), cend(jobs));
int right = accumulate(cbegin(jobs), cend(jobs), 0);
while (left <= right) {
const auto& mid = left + (right - left) / 2;
vector<int> counts(k);
if (backtracking(jobs, 0, mid, &counts)) {
right = mid - 1;
} else {
left = mid + 1;
}
}
return left;
}
private:
bool backtracking(const vector<int>& jobs,
int i, int cap,
vector<int> *counts) {
if (i == size(jobs)) {
return true;
}
for (int j = 0; j < size(*counts); ++j) {
if ((*counts)[j] + jobs[i] <= cap) {
(*counts)[j] += jobs[i];
if (backtracking(jobs, i + 1, cap, counts)) {
return true;
}
(*counts)[j] -= jobs[i];
}
if ((*counts)[j] == 0) {
break;
}
}
return false;
}
};
// Time: O(k * k^n), the real complexity should be less, but hard to analyze
// Space: O(n + k)
class Solution2 {
public:
int minimumTimeRequired(vector<int>& jobs, int k) {
sort(begin(jobs), end(jobs), greater<int>());
int result = accumulate(cbegin(jobs), cend(jobs), 0);
vector<int> counts(k);
backtracking(jobs, 0, &counts, &result);
return result;
}
private:
void backtracking(const vector<int>& jobs,
int i,
vector<int> *counts,
int *result) {
if (i == size(jobs)) {
*result = min(*result, *max_element(cbegin(*counts), cend(*counts)));
return;
}
for (int j = 0; j < size(*counts); ++j) {
if ((*counts)[j] + jobs[i] <= *result) {
(*counts)[j] += jobs[i];
backtracking(jobs, i + 1, counts, result);
(*counts)[j] -= jobs[i];
}
if ((*counts)[j] == 0) {
break;
}
}
}
};