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Merge pull request #543 from suzzzal/main
Salary Queries solution
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/*
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Problem: Salary Queries
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Approach:
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- Use coordinate compression because salaries can be up to 1e9.
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- Maintain frequencies of salaries using a Fenwick Tree (Binary Indexed Tree).
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- For update queries, decrease the old salary count and increase the new one.
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- For range queries, use prefix sums from the Fenwick Tree.
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Time Complexity:
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O((n + q) log (n + q))
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Space Complexity:
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O(n + q)
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Example:
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Input:
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5 3
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3 7 2 2 5
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? 2 3
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! 3 6
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? 2 3
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Output:
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3
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2
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Submission Link:
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https://cses.fi/paste/129b78ff774776c0f188fe/
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*/
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#include <bits/stdc++.h>
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using namespace std;
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struct Fenwick {
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int n;
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vector<int> bit;
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Fenwick(int n) : n(n), bit(n + 1, 0) {}
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void update(int i, int v) {
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for (; i <= n; i += i & -i)
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bit[i] += v;
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}
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int query(int i) {
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int s = 0;
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for (; i > 0; i -= i & -i)
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s += bit[i];
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return s;
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}
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};
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int main() {
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int n, q;
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cin >> n >> q;
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vector<int> salary(n);
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for (int i = 0; i < n; i++)
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cin >> salary[i];
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vector<int> all = salary;
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vector<pair<char, pair<int, int>>> queries;
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for (int i = 0; i < q; i++) {
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char type;
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cin >> type;
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if (type == '!') {
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int k, x;
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cin >> k >> x;
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queries.push_back({type, {k - 1, x}});
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all.push_back(x);
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} else {
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int a, b;
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cin >> a >> b;
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queries.push_back({type, {a, b}});
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all.push_back(a);
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all.push_back(b);
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}
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}
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// Coordinate compression
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sort(all.begin(), all.end());
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all.erase(unique(all.begin(), all.end()), all.end());
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auto get_id = [&](int x) {
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return int(lower_bound(all.begin(), all.end(), x) - all.begin()) + 1;
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};
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Fenwick fw(all.size());
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for (int x : salary)
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fw.update(get_id(x), 1);
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for (auto &qr : queries) {
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if (qr.first == '!') {
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int idx = qr.second.first;
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int x = qr.second.second;
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fw.update(get_id(salary[idx]), -1);
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fw.update(get_id(x), 1);
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salary[idx] = x;
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} else {
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int a = qr.second.first;
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int b = qr.second.second;
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cout << fw.query(get_id(b)) - fw.query(get_id(a) - 1) << "\n";
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}
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}
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return 0;
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}
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