LeetCode 3401: Find Circular Gift Exchange Chains Solution

Master LeetCode problem 3401 (Find Circular Gift Exchange Chains), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3401. Find Circular Gift Exchange Chains

Hard

Database

Problem Explanation

Explanation:

To solve this problem, we can use a graph-based approach where each person is represented as a node in the graph, and the gift exchange relationships are represented as directed edges between the nodes. Then, we can perform a depth-first search (DFS) starting from each node to find circular gift exchange chains.

Here are the steps:

Build a directed graph where each node represents a person and each directed edge represents a gift exchange relationship.
Perform a DFS starting from each node to find circular gift exchange chains.
During the DFS, keep track of the visited nodes and the current path to detect cycles.
If a cycle is detected during the DFS, add the nodes in the cycle to the result.

Time complexity: O(N + E), where N is the number of nodes (people) and E is the number of edges (gift exchange relationships). Space complexity: O(N), where N is the number of nodes (people) for storing the graph and visited nodes.

Solution Code

import java.util.*;

class Solution {
    public List<List<Integer>> circularExchange(int n, int[][] giftExchange) {
        List<List<Integer>> result = new ArrayList<>();
        Map<Integer, List<Integer>> graph = new HashMap<>();
        
        for (int[] exchange : giftExchange) {
            int u = exchange[0];
            int v = exchange[1];
            graph.putIfAbsent(u, new ArrayList<>());
            graph.get(u).add(v);
        }
        
        for (int i = 1; i <= n; i++) {
            Set<Integer> visited = new HashSet<>();
            List<Integer> path = new ArrayList<>();
            dfs(i, i, graph, visited, path, result);
        }
        
        return result;
    }
    
    private void dfs(int start, int current, Map<Integer, List<Integer>> graph, Set<Integer> visited, List<Integer> path, List<List<Integer>> result) {
        if (visited.contains(current)) {
            int index = path.indexOf(current);
            if (index != -1) {
                result.add(new ArrayList<>(path.subList(index, path.size())));
            }
            return;
        }
        
        visited.add(current);
        path.add(current);
        
        if (graph.containsKey(current)) {
            for (int neighbor : graph.get(current)) {
                dfs(start, neighbor, graph, visited, path, result);
            }
        }
        
        path.remove(path.size() - 1);
    }
}

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