Hard
Count Good Triplets in an Array — C++
Full explanation · Time O(nlogn) · Space O(n)
// Time: O(nlogn)
// Space: O(n)
// bit, fenwick tree, combinatorics
class Solution {
public:
long long goodTriplets(vector<int>& nums1, vector<int>& nums2) {
unordered_map<int, int> lookup;
for (int i = 0; i < size(nums1); ++i) {
lookup[nums1[i]] = i;
}
int64_t result = 0;
BIT bit(size(nums1));
for (int i = 0; i < size(nums2); ++i) {
const int64_t smaller = bit.query(lookup[nums2[i]] - 1);
const int64_t larger = (size(nums1) - (lookup[nums2[i]] + 1)) - (i - smaller);
result += smaller * larger;
bit.add(lookup[nums2[i]], 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_;
};
};