/* Pure SQL Solution:
SELECT 
    s1.employee_id,
    COUNT(*) as overlap_count
FROM EmployeeShifts s1
CROSS JOIN EmployeeShifts s2
WHERE s1.employee_id = s2.employee_id
    AND s1.start_time < s2.start_time  
    AND s1.end_time > s2.start_time
GROUP BY s1.employee_id
ORDER BY s1.employee_id;
*/

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct {
    int employee_id;
    char start_time[20];
    char end_time[20];
} Shift;

typedef struct {
    int employee_id;
    int overlap_count;
} Result;

int compareShifts(const void* a, const void* b) {
    return ((Shift*)a)->employee_id - ((Shift*)b)->employee_id;
}

int countOverlapsSQLStyle(Shift* shifts, int n, Result* results) {
    // Sort by employee_id first
    qsort(shifts, n, sizeof(Shift), compareShifts);
    
    int resultCount = 0;
    int i = 0;
    
    while (i < n) {
        int currentEmp = shifts[i].employee_id;
        int j = i;
        
        // Find all shifts for current employee
        while (j < n && shifts[j].employee_id == currentEmp) {
            j++;
        }
        
        // Count overlaps for this employee
        int overlapCount = 0;
        for (int x = i; x < j; x++) {
            for (int y = x + 1; y < j; y++) {
                // Check overlap: end_time > start_time
                if (strcmp(shifts[x].end_time, shifts[y].start_time) > 0 &&
                    strcmp(shifts[y].end_time, shifts[x].start_time) > 0) {
                    overlapCount++;
                }
            }
        }
        
        results[resultCount].employee_id = currentEmp;
        results[resultCount].overlap_count = overlapCount;
        resultCount++;
        
        i = j;
    }
    
    return resultCount;
}

int main() {
    printf("SQL-style C solution implemented\n");
    return 0;
}