3387. Maximize Amount After Two Days of Conversions

You are given a string initialCurrency, and you start with 1.0 of initialCurrency.

You are also given four arrays with currency pairs (strings) and rates (real numbers):

• pairs1[i] = [startCurrencyi, targetCurrencyi] denotes that you can convert from startCurrencyi to targetCurrencyi at a rate of rates1[i] on day 1.
• pairs2[i] = [startCurrencyi, targetCurrencyi] denotes that you can convert from startCurrencyi to targetCurrencyi at a rate of rates2[i] on day 2.
• Also, each targetCurrency can be converted back to its corresponding startCurrency at a rate of 1 / rate.

You can perform any number of conversions, including zero, using rates1 on day 1, followed by any number of additional conversions, including zero, using rates2 on day 2.

Return the maximum amount of initialCurrency you can have after performing any number of conversions on both days in order.

Note: Conversion rates are valid, and there will be no contradictions in the rates for either day. The rates for the days are independent of each other.

class Solution {
public:
    double maxAmount(string initialCurrency, vector<vector<string>>& currencyPairs1, vector<double>& rates1, vector<vector<string>>& currencyPairs2, vector<double>& rates2) {
        unordered_map<string, int> ord;
        int idx = 0;
        auto apply = [&](string& s) {
            if(ord.count(s)) return;
            ord[s] = idx++;
        };
        apply(initialCurrency);
        for(auto& p : currencyPairs1) for(auto& c : p) apply(c);
        for(auto& p : currencyPairs2) for(auto& c : p) apply(c);
        vector<double> dp(idx);
        dp[0] = 1.;
        auto run = [&](vector<vector<string>>& p, vector<double> r) {
            vector<vector<pair<int,double>>> adj(idx);
            priority_queue<pair<double,int>> q;
            for(int i = 0; i < p.size(); i++) {
                int u = ord[p[i][0]], v = ord[p[i][1]];
                assert(u != v);
                adj[u].push_back({v,r[i]});
                adj[v].push_back({u,1. / r[i]});
            }
            for(int i = 0; i < idx; i++) q.push({dp[i], i});
            while(q.size()) {
                auto [val, u] = q.top(); q.pop();
                
                if(dp[u] != val) continue;
                for(auto& [v,r] : adj[u]) {
                    double x = val * r;
                    if(dp[v] >= x) continue;
                    dp[v] = x;
                    q.push({dp[v],v});
                }
            }
        };
        run(currencyPairs1,rates1);
        run(currencyPairs2,rates2);
        return dp[0];
    }
};