-- Optimal SQL Solution using RANK()
SELECT transaction_id
FROM (
    SELECT transaction_id,
           RANK() OVER (PARTITION BY DATE(day) ORDER BY amount DESC) as rnk
    FROM Transactions
) ranked
WHERE rnk = 1
ORDER BY transaction_id;

/* C implementation */
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int* solution(Transaction* transactions, int n, int* returnSize) {
    // Sort transactions by day first
    qsort(transactions, n, sizeof(Transaction), compareByDay);
    
    int* result = malloc(n * sizeof(int));
    int count = 0;
    int i = 0;
    
    while (i < n) {
        int dayStart = i;
        char currentDay[11];
        strcpy(currentDay, transactions[i].day);
        int maxAmount = transactions[i].amount;
        
        // Find end of current day and maximum amount
        while (i < n && strcmp(transactions[i].day, currentDay) == 0) {
            if (transactions[i].amount > maxAmount) {
                maxAmount = transactions[i].amount;
            }
            i++;
        }
        
        // Add all transactions with maximum amount for this day
        for (int j = dayStart; j < i; j++) {
            if (transactions[j].amount == maxAmount) {
                result[count++] = transactions[j].transaction_id;
            }
        }
    }
    
    // Sort result by transaction_id
    qsort(result, count, sizeof(int), compare);
    *returnSize = count;
    return result;
}

int compareByDay(const void* a, const void* b) {
    Transaction* ta = (Transaction*)a;
    Transaction* tb = (Transaction*)b;
    return strcmp(ta->day, tb->day);
}