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


struct Node {
    struct Node* children[10];
    bool isEnd;
};

struct Node* newNode(void) {
    struct Node *nNode = NULL;
    nNode = (struct Node *) malloc(sizeof(struct Node));
    if (nNode) {
        nNode->isEnd = false;
        for (int i = 0; i < 10; i++) {
            nNode->children[i] = NULL;
        }
    }
    return nNode;
}

void free_all(struct Node* node) {
    if (!node) {
        return;
    };

    for (int i = 0; i < 10; i++) {
        free_all(node->children[i]);
    };

    free(node);
}

int numbers[] = {11354, 113, 77777, 255044};

int main(void) {
    struct Node* root = newNode();
    int n_numbers = sizeof(numbers)/sizeof(int);

    for (int i = 0; i < n_numbers; i++) {       // Go through all the phone numbers in the global list
        struct Node* pointer = root;            // Create a pointer that points to whatever part of the trie we're working on
        char number[8 + 1];                         // A number has 8 + \0 chars
        snprintf(number, 9, "%i", numbers[i]);  // Convert int to string
        int len_number = strlen(number);
        for (int j = 0; j < len_number; j++) {  // loop over chars in the phone number
            int index = number[j] - '0';            // Go to the node we're testing
            if (pointer->children[index] == NULL) {
                pointer->children[index] = newNode();
            };
            pointer = pointer->children[index];

            if (j == len_number - 1) {          // if we are at the last iteration
                if (pointer->isEnd == true) {   // Check if we've seen the number before
                    printf("Duplicate Number found at element %i, number %s", i, number);
                    break;
                };
                // if we're at the last iteration, and isEnd isnt true, its either a new number OR we found a prefix
                bool continues = false;
                for (int k = 0; k < 10; k++) {
                    if( pointer->children[k] != NULL) {
                        continues = true;
                    };
                };
                if (continues == true) {
                    printf("%s and %s", number, number);
                    while (pointer->isEnd != true) { // follow the first path you can find
                        for (int k = 0; k < 10; k++) {
                            if( pointer->children[k] != NULL) {
                                pointer = pointer->children[k];
                                printf("%c", k + '0'); // Print the path as we go
                                break;
                            };
                        };
                    };
                    printf("\n");
                    free_all(root); // Clean up
                    return 0; // return true or success or whatever
                };
                pointer->isEnd = true;
            }
            else if (pointer->isEnd == true) {  // if we've already seen a shorter number with this prefix
                for (int k = 0; k <= j; k++) {
                    printf("%c", number[k]);
                }
                printf(" and %i\n", numbers[i]);
                return 0;
            };
        };
    };
    return 1
}