LeetCode #2097 — HARD

Valid Arrangement of Pairs

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given a 0-indexed 2D integer array pairs where pairs[i] = [start_i, end_i]. An arrangement of pairs is valid if for every index i where 1 <= i < pairs.length, we have end_i-1 == start_i.

Return any valid arrangement of pairs.

Note: The inputs will be generated such that there exists a valid arrangement of pairs.

Example 1:

Input: pairs = [[5,1],[4,5],[11,9],[9,4]]
Output: [[11,9],[9,4],[4,5],[5,1]]
Explanation:
This is a valid arrangement since end_i-1 always equals start_i.
end₀ = 9 == 9 = start₁ 
end₁ = 4 == 4 = start₂
end₂ = 5 == 5 = start₃

Example 2:

Input: pairs = [[1,3],[3,2],[2,1]]
Output: [[1,3],[3,2],[2,1]]
Explanation:
This is a valid arrangement since end_i-1 always equals start_i.
end₀ = 3 == 3 = start₁
end₁ = 2 == 2 = start₂
The arrangements [[2,1],[1,3],[3,2]] and [[3,2],[2,1],[1,3]] are also valid.

Example 3:

Input: pairs = [[1,2],[1,3],[2,1]]
Output: [[1,2],[2,1],[1,3]]
Explanation:
This is a valid arrangement since end_i-1 always equals start_i.
end₀ = 2 == 2 = start₁
end₁ = 1 == 1 = start₂

Constraints:

1 <= pairs.length <= 10⁵
pairs[i].length == 2
0 <= start_i, end_i <= 10⁹
start_i != end_i
No two pairs are exactly the same.
There exists a valid arrangement of pairs.

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed 2D integer array pairs where pairs[i] = [starti, endi]. An arrangement of pairs is valid if for every index i where 1 <= i < pairs.length, we have endi-1 == starti. Return any valid arrangement of pairs. Note: The inputs will be generated such that there exists a valid arrangement of pairs.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[5,1],[4,5],[11,9],[9,4]]

Example 2

[[1,3],[3,2],[2,1]]

Example 3

[[1,2],[1,3],[2,1]]

Core Insight

What unlocks the optimal approach

Could you convert this into a graph problem?
Consider the pairs as edges and each number as a node.
We have to find an Eulerian path of this graph. Hierholzer’s algorithm can be used.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2097: Valid Arrangement of Pairs
class Solution {
  public int[][] validArrangement(int[][] pairs) {
    List<int[]> ans = new ArrayList<>();
    Map<Integer, Deque<Integer>> graph = new HashMap<>();
    Map<Integer, Integer> outDegree = new HashMap<>();
    Map<Integer, Integer> inDegrees = new HashMap<>();

    for (int[] pair : pairs) {
      final int start = pair[0];
      final int end = pair[1];
      graph.putIfAbsent(start, new ArrayDeque<>());
      graph.get(start).push(end);
      outDegree.merge(start, 1, Integer::sum);
      inDegrees.merge(end, 1, Integer::sum);
    }

    final int startNode = getStartNode(graph, outDegree, inDegrees, pairs);
    euler(graph, startNode, ans);
    Collections.reverse(ans);
    return ans.stream().toArray(int[][] ::new);
  }

  private int getStartNode(Map<Integer, Deque<Integer>> graph, Map<Integer, Integer> outDegree,
                           Map<Integer, Integer> inDegrees, int[][] pairs) {
    for (final int u : graph.keySet())
      if (outDegree.getOrDefault(u, 0) - inDegrees.getOrDefault(u, 0) == 1)
        return u;
    return pairs[0][0]; // Arbitrarily choose a node.
  }

  private void euler(Map<Integer, Deque<Integer>> graph, int u, List<int[]> ans) {
    Deque<Integer> stack = graph.get(u);
    while (stack != null && !stack.isEmpty()) {
      final int v = stack.pop();
      euler(graph, v, ans);
      ans.add(new int[] {u, v});
    }
  }
}

// Accepted solution for LeetCode #2097: Valid Arrangement of Pairs
// Auto-generated Go example from java.
func exampleSolution() {
}
// Reference (java):
// // Accepted solution for LeetCode #2097: Valid Arrangement of Pairs
// class Solution {
//   public int[][] validArrangement(int[][] pairs) {
//     List<int[]> ans = new ArrayList<>();
//     Map<Integer, Deque<Integer>> graph = new HashMap<>();
//     Map<Integer, Integer> outDegree = new HashMap<>();
//     Map<Integer, Integer> inDegrees = new HashMap<>();
// 
//     for (int[] pair : pairs) {
//       final int start = pair[0];
//       final int end = pair[1];
//       graph.putIfAbsent(start, new ArrayDeque<>());
//       graph.get(start).push(end);
//       outDegree.merge(start, 1, Integer::sum);
//       inDegrees.merge(end, 1, Integer::sum);
//     }
// 
//     final int startNode = getStartNode(graph, outDegree, inDegrees, pairs);
//     euler(graph, startNode, ans);
//     Collections.reverse(ans);
//     return ans.stream().toArray(int[][] ::new);
//   }
// 
//   private int getStartNode(Map<Integer, Deque<Integer>> graph, Map<Integer, Integer> outDegree,
//                            Map<Integer, Integer> inDegrees, int[][] pairs) {
//     for (final int u : graph.keySet())
//       if (outDegree.getOrDefault(u, 0) - inDegrees.getOrDefault(u, 0) == 1)
//         return u;
//     return pairs[0][0]; // Arbitrarily choose a node.
//   }
// 
//   private void euler(Map<Integer, Deque<Integer>> graph, int u, List<int[]> ans) {
//     Deque<Integer> stack = graph.get(u);
//     while (stack != null && !stack.isEmpty()) {
//       final int v = stack.pop();
//       euler(graph, v, ans);
//       ans.add(new int[] {u, v});
//     }
//   }
// }

# Accepted solution for LeetCode #2097: Valid Arrangement of Pairs
class Solution:
  def validArrangement(self, pairs: list[list[int]]) -> list[list[int]]:
    ans = []
    graph = collections.defaultdict(list)
    outDegree = collections.Counter()
    inDegrees = collections.Counter()

    for start, end in pairs:
      graph[start].append(end)
      outDegree[start] += 1
      inDegrees[end] += 1

    def getStartNode() -> int:
      for u in graph.keys():
        if outDegree[u] - inDegrees[u] == 1:
          return u
      return pairs[0][0]  # Arbitrarily choose a node.

    def euler(u: int) -> None:
      stack = graph[u]
      while stack:
        v = stack.pop()
        euler(v)
        ans.append([u, v])

    euler(getStartNode())
    return ans[::-1]

// Accepted solution for LeetCode #2097: Valid Arrangement of Pairs
/**
 * [2097] Valid Arrangement of Pairs
 *
 * You are given a 0-indexed 2D integer array pairs where pairs[i] = [starti, endi]. An arrangement of pairs is valid if for every index i where 1 <= i < pairs.length, we have endi-1 == starti.
 * Return any valid arrangement of pairs.
 * Note: The inputs will be generated such that there exists a valid arrangement of pairs.
 *  
 * Example 1:
 *
 * Input: pairs = [[5,1],[4,5],[11,9],[9,4]]
 * Output: [[11,9],[9,4],[4,5],[5,1]]
 * Explanation:
 * This is a valid arrangement since endi-1 always equals starti.
 * end0 = 9 == 9 = start1
 * end1 = 4 == 4 = start2
 * end2 = 5 == 5 = start3
 *
 * Example 2:
 *
 * Input: pairs = [[1,3],[3,2],[2,1]]
 * Output: [[1,3],[3,2],[2,1]]
 * Explanation:
 * This is a valid arrangement since endi-1 always equals starti.
 * end0 = 3 == 3 = start1
 * end1 = 2 == 2 = start2
 * The arrangements [[2,1],[1,3],[3,2]] and [[3,2],[2,1],[1,3]] are also valid.
 *
 * Example 3:
 *
 * Input: pairs = [[1,2],[1,3],[2,1]]
 * Output: [[1,2],[2,1],[1,3]]
 * Explanation:
 * This is a valid arrangement since endi-1 always equals starti.
 * end0 = 2 == 2 = start1
 * end1 = 1 == 1 = start2
 *
 *  
 * Constraints:
 *
 * 	1 <= pairs.length <= 10^5
 * 	pairs[i].length == 2
 * 	0 <= starti, endi <= 10^9
 * 	starti != endi
 * 	No two pairs are exactly the same.
 * 	There exists a valid arrangement of pairs.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/valid-arrangement-of-pairs/
// discuss: https://leetcode.com/problems/valid-arrangement-of-pairs/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn valid_arrangement(pairs: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        vec![]
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2097_example_1() {
        let pairs = vec![vec![5, 1], vec![4, 5], vec![11, 9], vec![9, 4]];

        let result = vec![vec![11, 9], vec![9, 4], vec![4, 5], vec![5, 1]];

        assert_eq!(Solution::valid_arrangement(pairs), result);
    }

    #[test]
    #[ignore]
    fn test_2097_example_2() {
        let pairs = vec![vec![1, 3], vec![3, 2], vec![2, 1]];

        let result = vec![vec![1, 3], vec![3, 2], vec![2, 1]];

        assert_eq!(Solution::valid_arrangement(pairs), result);
    }

    #[test]
    #[ignore]
    fn test_2097_example_3() {
        let pairs = vec![vec![1, 2], vec![1, 3], vec![2, 1]];

        let result = vec![vec![1, 2], vec![2, 1], vec![1, 3]];

        assert_eq!(Solution::valid_arrangement(pairs), result);
    }
}

// Accepted solution for LeetCode #2097: Valid Arrangement of Pairs
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2097: Valid Arrangement of Pairs
// class Solution {
//   public int[][] validArrangement(int[][] pairs) {
//     List<int[]> ans = new ArrayList<>();
//     Map<Integer, Deque<Integer>> graph = new HashMap<>();
//     Map<Integer, Integer> outDegree = new HashMap<>();
//     Map<Integer, Integer> inDegrees = new HashMap<>();
// 
//     for (int[] pair : pairs) {
//       final int start = pair[0];
//       final int end = pair[1];
//       graph.putIfAbsent(start, new ArrayDeque<>());
//       graph.get(start).push(end);
//       outDegree.merge(start, 1, Integer::sum);
//       inDegrees.merge(end, 1, Integer::sum);
//     }
// 
//     final int startNode = getStartNode(graph, outDegree, inDegrees, pairs);
//     euler(graph, startNode, ans);
//     Collections.reverse(ans);
//     return ans.stream().toArray(int[][] ::new);
//   }
// 
//   private int getStartNode(Map<Integer, Deque<Integer>> graph, Map<Integer, Integer> outDegree,
//                            Map<Integer, Integer> inDegrees, int[][] pairs) {
//     for (final int u : graph.keySet())
//       if (outDegree.getOrDefault(u, 0) - inDegrees.getOrDefault(u, 0) == 1)
//         return u;
//     return pairs[0][0]; // Arbitrarily choose a node.
//   }
// 
//   private void euler(Map<Integer, Deque<Integer>> graph, int u, List<int[]> ans) {
//     Deque<Integer> stack = graph.get(u);
//     while (stack != null && !stack.isEmpty()) {
//       final int v = stack.pop();
//       euler(graph, v, ans);
//       ans.add(new int[] {u, v});
//     }
//   }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.