SELECT 
    c.name,
    ROUND(SUM(CASE WHEN EXTRACT(QUARTER FROM t.dt) = 1 THEN t.amount ELSE 0 END), 2) AS q1_amount,
    ROUND(SUM(CASE WHEN EXTRACT(QUARTER FROM t.dt) = 2 THEN t.amount ELSE 0 END), 2) AS q2_amount,
    ROUND(SUM(CASE WHEN EXTRACT(QUARTER FROM t.dt) = 3 THEN t.amount ELSE 0 END), 2) AS q3_amount,
    ROUND(SUM(CASE WHEN EXTRACT(QUARTER FROM t.dt) = 4 THEN t.amount ELSE 0 END), 2) AS q4_amount,
    COUNT(t.coin_id) AS total_transactions,
    ROUND(SUM(t.amount), 2) AS total_amount
FROM 
    coins c
LEFT JOIN 
    transactions t ON c.id = t.coin_id AND EXTRACT(YEAR FROM t.dt) = 2023
GROUP BY 
    c.name
ORDER BY 
    c.name ASC;

#include <bits/stdc++.h>
using namespace std;
vector<string> solve(vector<string>& chain, vector<string>& elements) {
    unordered_map<string, int> required, window;
    for (const auto& elem : elements) {
        required[elem]++;
    }
    int requiredSize = required.size();
    int have = 0;
    int left = 0, right = 0;
    int minLength = INT_MAX;
    int start = -1;
    while (right < chain.size()) {
        string rightElem = chain[right];
        if (required.count(rightElem)) {
            window[rightElem]++;
            if (window[rightElem] == required[rightElem]) {
                have++;
            }
        }
        while (have == requiredSize) {
            if (right - left + 1 < minLength) {
                minLength = right - left + 1;
                start = left;
            }
            string leftElem = chain[left];
            if (required.count(leftElem)) {
                window[leftElem]--;
                if (window[leftElem] < required[leftElem]) {
                    have--;
                }
            }
            left++;
        }
        right++;
    }
    if (start == -1) {
        return {};
    }
    return vector<string>(chain.begin() + start, chain.begin() + start + minLength);
}