Hard
Minimum Pair Removal to Sort Array II — C++
Full explanation · Time O(nlogn) · Space O(n)
// Time: O(nlogn)
// Space: O(n)
// simulation, doubly linked list, bst
class Solution {
public:
int minimumPairRemoval(vector<int>& nums) {
vector<int64_t> arr(cbegin(nums), cend(nums));
vector<int> left(size(arr) + 1);
iota(begin(left), end(left), -1);
vector<int> right(size(arr));
iota(begin(right), end(right), 1);
int cnt = 0;
set<pair<int64_t, int>> bst;
const auto& add = [&](int i) {
if (0 <= i && i < right[i] && right[i] < size(arr)) {
bst.emplace(arr[i] + arr[right[i]], i);
if (arr[i] > arr[right[i]]) {
++cnt;
}
}
};
const auto& remove = [&](int i) {
if (0 <= i && i < right[i] && right[i] < size(arr)) {
bst.erase({arr[i] + arr[right[i]], i});
if (arr[i] > arr[right[i]]) {
--cnt;
}
}
};
for (int i = 0; i + 1 < size(arr); ++i) {
bst.emplace(arr[i] + arr[i + 1], i);
if (arr[i] > arr[i + 1]) {
++cnt;
}
}
int result = 0;
for (; cnt; ++result) {
const auto [_, i] = *cbegin(bst);
remove(left[i]);
remove(i);
remove(right[i]);
arr[i] += arr[right[i]];
left[right[right[i]]] = i;
right[i] = right[right[i]];
add(left[i]);
add(i);
}
return result;
}
};