LeetCode //C - 332. Reconstruct Itinerary

332. Reconstruct Itinerary

You are given a list of airline tickets where tickets[i] = [fromi, toi] represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.

All of the tickets belong to a man who departs from "JFK", thus, the itinerary must begin with "JFK". If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string.

• For example, the itinerary ["JFK", "LGA"] has a smaller lexical order than ["JFK", "LGB"].

You may assume all tickets form at least one valid itinerary. You must use all the tickets once and only once.

Example 1:

Input: tickets = [["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]
Output: ["JFK","MUC","LHR","SFO","SJC"]

Example 2:

Input: tickets = [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
Output: ["JFK","ATL","JFK","SFO","ATL","SFO"]
Explanation: Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"] but it is larger in lexical order.

Constraints:

• 1 <= tickets.length <= 300
• tickets[i].length == 2
• f r o m i . l e n g t h = = 3 from_i.length == 3 fromi.length==3
• t o i . l e n g t h = = 3 to_i.length == 3 toi.length==3
• f r o m i a n d t o i c o n s i s t o f u p p e r c a s e E n g l i s h l e t t e r s . from_i and toi consist of uppercase English letters. fromiandtoiconsistofuppercaseEnglishletters.
• f r o m i ! = t o i from_i != to_i fromi!=toi

From: LeetCode

Link: 332. Reconstruct Itinerary

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Solution:

Ideas:

• Deferred Memory Freeing: The nodes (Node*) in the adjacency list are no longer freed immediately within the DFS loop. Instead, they're freed only after the DFS traversal is complete, preventing use-after-free errors.
• DFS Stack Handling: The stack is used to manage the DFS iteration, and nodes are correctly removed from the adjacency list as they are processed. The itinerary is built in reverse order and then freed once fully constructed.

Code:

#define MAX_TICKETS 300
#define MAX_AIRPORTS 26 * 26 * 26 // Assuming three-letter airport codes
#define AIRPORT_CODE_LEN 4

// Node structure for adjacency list
typedef struct Node {
    char airport[AIRPORT_CODE_LEN];
    struct Node* next;
} Node;

// Function to insert an edge into the adjacency list, maintaining lexicographical order
void insertEdge(Node* adjList[], char* from, char* to) {
    Node* newNode = (Node*)malloc(sizeof(Node));
    strcpy(newNode->airport, to);
    newNode->next = NULL;

    int index = (from[0] - 'A') * 26 * 26 + (from[1] - 'A') * 26 + (from[2] - 'A');

    if (adjList[index] == NULL || strcmp(adjList[index]->airport, to) > 0) {
        newNode->next = adjList[index];
        adjList[index] = newNode;
    } else {
        Node* current = adjList[index];
        while (current->next != NULL && strcmp(current->next->airport, to) < 0) {
            current = current->next;
        }
        newNode->next = current->next;
        current->next = newNode;
    }
}

// Iterative DFS using a stack to build the itinerary
void dfs(char* airport, Node* adjList[], char** itinerary, int* index) {
    char* stack[MAX_TICKETS + 1];
    int stackSize = 0;

    stack[stackSize++] = airport;

    while (stackSize > 0) {
        char* currentAirport = stack[stackSize - 1];
        int idx = (currentAirport[0] - 'A') * 26 * 26 + (currentAirport[1] - 'A') * 26 + (currentAirport[2] - 'A');

        if (adjList[idx] != NULL) {
            Node* nextNode = adjList[idx];
            stack[stackSize++] = nextNode->airport;
            adjList[idx] = nextNode->next;
            // Do not free nextNode here; defer the free operation until the end
        } else {
            itinerary[(*index)--] = stack[--stackSize];
        }
    }
}

// Function to find the itinerary from the list of tickets
char** findItinerary(char*** tickets, int ticketsSize, int* ticketsColSize, int* returnSize) {
    Node* adjList[MAX_AIRPORTS] = {NULL};

    // Populate the adjacency list with the tickets
    for (int i = 0; i < ticketsSize; i++) {
        insertEdge(adjList, tickets[i][0], tickets[i][1]);
    }

    char** itinerary = (char**)malloc((ticketsSize + 1) * sizeof(char*));
    int index = ticketsSize;
    dfs("JFK", adjList, itinerary, &index);

    // Freeing nodes after DFS traversal is complete
    for (int i = 0; i < MAX_AIRPORTS; i++) {
        Node* current = adjList[i];
        while (current) {
            Node* next = current->next;
            free(current->airport);
            free(current);
            current = next;
        }
    }

    *returnSize = ticketsSize + 1;
    return itinerary;
}