LeetCode 2097: Valid Arrangement of Pairs Solution

Master LeetCode problem 2097 (Valid Arrangement of Pairs), 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.

2097. Valid Arrangement of Pairs

Hard

Depth-First Search Graph Eulerian Circuit

Problem Explanation

Explanation

To solve this problem, we need to rearrange the pairs in a way that the end of one pair matches the start of the next pair. This can be achieved by constructing a directed graph where each pair represents a directed edge from the start to the end. Then we can perform a topological sort on the graph to find a valid arrangement.

Create a directed graph where the nodes represent the start and end values of the pairs, and the edges represent the pairs.
Perform a topological sort on the graph to find a valid arrangement of pairs.
Return the pairs in the order obtained from the topological sort.

Time Complexity: O(n), where n is the number of pairs Space Complexity: O(n)

Solution Code

class Solution {
    public int[][] findValidArrangement(int[][] pairs) {
        Map<Integer, List<Integer>> graph = new HashMap<>();
        Map<Integer, Integer> indegree = new HashMap<>();
        
        for (int[] pair : pairs) {
            int start = pair[0];
            int end = pair[1];
            graph.putIfAbsent(start, new ArrayList<>());
            graph.get(start).add(end);
            indegree.put(end, indegree.getOrDefault(end, 0) + 1);
        }
        
        Queue<Integer> queue = new LinkedList<>();
        for (int key : graph.keySet()) {
            if (!indegree.containsKey(key)) {
                queue.offer(key);
            }
        }
        
        List<Integer> result = new ArrayList<>();
        while (!queue.isEmpty()) {
            int curr = queue.poll();
            result.add(curr);
            List<Integer> neighbors = graph.getOrDefault(curr, new ArrayList<>());
            for (int neighbor : neighbors) {
                indegree.put(neighbor, indegree.get(neighbor) - 1);
                if (indegree.get(neighbor) == 0) {
                    queue.offer(neighbor);
                }
            }
        }
        
        int[][] validArrangement = new int[pairs.length][2];
        for (int i = 0; i < result.size() - 1; i++) {
            validArrangement[i] = new int[]{result.get(i), result.get(i + 1)};
        }
        
        return validArrangement;
    }
}

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