Hard
Create Sorted Array through Instructions — C++
Full explanation · Time O(nlogn) · Space O(n)
// Time: O(nlogm)
// Space: O(m)
class Solution {
public:
int createSortedArray(vector<int>& instructions) {
static const int MOD = 1e9 + 7;
BIT bit(*max_element(cbegin(instructions), cend(instructions)));
int result = 0;
for (int i = 0; i < size(instructions); ++i) {
const auto inst = instructions[i] - 1;
auto cost = min(bit.query(inst - 1), i - bit.query(inst));
if (MOD - result <= cost) {
cost -= MOD;
}
result += cost;
bit.add(inst, 1);
}
return result;
}
private:
class BIT {
public:
BIT(int n) : bit_(n + 1) { // 0-indexed
}
void add(int i, int val) {
++i;
for (; i < size(bit_); i += lower_bit(i)) {
bit_[i] += val;
}
}
int query(int i) const {
++i;
int total = 0;
for (; i > 0; i -= lower_bit(i)) {
total += bit_[i];
}
return total;
}
private:
int lower_bit(int i) const {
return i & -i;
}
vector<int> bit_;
};
};
// Time: O(nlogn)
// Space: O(n)
// in fact, we could use a raw array instead of vector to avoid TLE
class Solution_TLE {
public:
int createSortedArray(vector<int>& instructions) {
static const int MOD = 1e9 + 7;
vector<int> smaller_counts(size(instructions)), larger_counts(size(instructions));
vector<pair<int, int>> smaller_idxs, larger_idxs;
for (int i = 0; i < size(instructions); ++i) {
smaller_idxs.emplace_back(instructions[i], i);
larger_idxs.emplace_back(instructions[i], i);
}
smallerMergeSort(&smaller_idxs, 0, size(smaller_idxs) - 1, &smaller_counts);
largerMergeSort(&larger_idxs, 0, size(larger_idxs) - 1, &larger_counts);
int result = 0;
for (int i = 0; i < size(instructions); ++i) {
auto cost = min(smaller_counts[i], larger_counts[i]);
if (MOD - result <= cost) {
cost -= MOD;
}
result += cost;
}
return result;
}
private:
void smallerMergeSort(vector<pair<int, int>> *idxs, int start, int end, vector<int> *counts) {
if (end - start <= 0) { // The number of range [start, end] of which size is less than 2 doesn't need sort.
return;
}
int mid = start + (end - start) / 2;
smallerMergeSort(idxs, start, mid, counts);
smallerMergeSort(idxs, mid + 1, end, counts);
int r = start;
vector<pair<int, int>> tmp;
for (int i = mid + 1; i <= end; ++i) {
// Merge the two sorted arrays into tmp.
while (r <= mid && (*idxs)[r].first < (*idxs)[i].first) {
tmp.emplace_back((*idxs)[r++]);
}
tmp.emplace_back((*idxs)[i]);
(*counts)[(*idxs)[i].second] += r - start;
}
while (r <= mid) {
tmp.emplace_back((*idxs)[r++]);
}
// Copy tmp back to num_idxs.
copy(tmp.begin(), tmp.end(), idxs->begin() + start);
}
void largerMergeSort(vector<pair<int, int>> *idxs, int start, int end, vector<int> *counts) {
if (end - start <= 0) { // The number of range [start, end] of which size is less than 2 doesn't need sort.
return;
}
int mid = start + (end - start) / 2;
largerMergeSort(idxs, start, mid, counts);
largerMergeSort(idxs, mid + 1, end, counts);
int r = start;
vector<pair<int, int>> tmp;
for (int i = mid + 1; i <= end; ++i) {
// Merge the two sorted arrays into tmp.
while (r <= mid && (*idxs)[r].first <= (*idxs)[i].first) {
tmp.emplace_back((*idxs)[r++]);
}
if (r <= mid) {
tmp.emplace_back((*idxs)[i]);
}
(*counts)[(*idxs)[i].second] += mid - r + 1;
}
while (r <= mid) {
tmp.emplace_back((*idxs)[r++]);
}
// Copy tmp back to num_idxs.
copy(tmp.begin(), tmp.end(), idxs->begin() + start);
}
};