Teleporters Path

Problem Overview

Aspect	Details
Problem	Find path using each teleporter exactly once
Algorithm	Hierholzer’s Algorithm for Eulerian Path
Time Complexity	O(n + m)
Space Complexity	O(n + m)
Key Insight	DFS with path construction on backtracking

Learning Goals

After completing this problem, you should understand:

Eulerian Path vs Eulerian Circuit
- Eulerian Path: visits every edge exactly once
- Eulerian Circuit: Eulerian path that starts and ends at the same vertex
Existence Conditions for Directed Graphs
- When an Eulerian path exists based on vertex degrees
Hierholzer’s Algorithm
- Efficient O(m) algorithm to construct the path

Problem Statement

A game has n levels (numbered 1 to n) and m teleporters. Each teleporter is a one-way connection from level a to level b.

Task: Find a route from level 1 to level n that uses each teleporter exactly once.

Input:

First line: integers n and m
Next m lines: teleporter connections (a, b)

Output:

The path as space-separated level numbers, or “IMPOSSIBLE”

Constraints:

2 <= n <= 10^5
1 <= m <= 2 * 10^5

Eulerian Path Existence Conditions

For a directed graph, an Eulerian path exists if and only if:

Condition 1: At most ONE vertex has out_degree - in_degree = 1
             (This vertex must be the START)

Condition 2: At most ONE vertex has in_degree - out_degree = 1
             (This vertex must be the END)

Condition 3: All OTHER vertices have in_degree = out_degree

Condition 4: The graph is weakly connected
             (All vertices with edges are reachable)

Special Requirements for This Problem

Since we must start at vertex 1 and end at vertex n:

Vertex	Required Condition
Vertex 1	out_degree - in_degree = 1 (start vertex)
Vertex n	in_degree - out_degree = 1 (end vertex)
Others	in_degree = out_degree

If the graph has an Eulerian circuit instead (all vertices balanced), we cannot use it because it would require start = end, but we need start = 1, end = n.

Hierholzer’s Algorithm

The key insight is to build the path during DFS backtracking:

Algorithm Hierholzer(start):
    1. Initialize empty path
    2. DFS from start vertex:
       - While current vertex has unused edges:
           - Pick any unused edge (u, v)
           - Mark edge as used
           - Recursively DFS from v
       - When no more edges: ADD current vertex to path
    3. Reverse the path (built backwards during backtracking)

Why add on backtracking? A vertex is added to the path only when all its outgoing edges have been used, ensuring we don’t get stuck.

Visual Diagram

Example: n=5, m=6
Teleporters: 1->2, 2->3, 3->1, 1->4, 4->5, 5->3

Graph:
          +---+
          | 1 |----+
          +---+    |
         /    \    |
        v      v   |
      +---+  +---+ |
      | 2 |  | 4 | |
      +---+  +---+ |
        |      |   |
        v      v   |
      +---+  +---+ |
      | 3 |<-| 5 | |
      +---+  +---+ |
        |          |
        +----------+

Degree Analysis:
Vertex | in_degree | out_degree | difference
-------|-----------|------------|------------
   1   |     1     |     2      |    +1 (START)
   2   |     1     |     1      |     0
   3   |     2     |     1      |    -1 (END... but we need n=5!)
   4   |     1     |     1      |     0
   5   |     1     |     1      |     0

This example has NO valid path (end should be vertex 5, not 3)

Valid Example: n=4, m=4, edges: 1->2, 2->3, 3->4, 1->3

      +---+
      | 1 |
      +---+
      /   \
     v     v
   +---+  +---+
   | 2 |->| 3 |
   +---+  +---+
            |
            v
          +---+
          | 4 |
          +---+

Vertex | in | out | diff
-------|----|----|-----
   1   |  0 |  2 | +1 (START - correct!)
   2   |  1 |  1 |  0
   3   |  2 |  1 | -1 (... but end should be 4)

Still invalid! Let's use: 1->2, 2->3, 3->4, 2->4

Vertex | in | out | diff
-------|----|----|-----
   1   |  0 |  1 | +1 (START)
   2   |  1 |  2 | +1 (INVALID - two starts!)

Correct Valid Example: n=4, m=3, edges: 1->2, 2->3, 3->4

Path: 1 -> 2 -> 3 -> 4

Dry Run

Input: n=4, m=3
Edges: (1,2), (2,3), (3,4)

Step 1: Build adjacency list (using indices for edges)
adj[1] = [2]
adj[2] = [3]
adj[3] = [4]
adj[4] = []

Step 2: Check degrees
Vertex 1: out=1, in=0  -> diff = +1 (start candidate)
Vertex 2: out=1, in=1  -> diff = 0
Vertex 3: out=1, in=1  -> diff = 0
Vertex 4: out=0, in=1  -> diff = -1 (end candidate)

Start=1, End=4 matches requirements!

Step 3: Hierholzer's Algorithm (iterative with stack)

Stack: [1]     Path: []
  Current=1, has edges -> push 2
Stack: [1,2]   Path: []
  Current=2, has edges -> push 3
Stack: [1,2,3] Path: []
  Current=3, has edges -> push 4
Stack: [1,2,3,4] Path: []
  Current=4, no edges -> pop to path
Stack: [1,2,3]   Path: [4]
  Current=3, no edges -> pop to path
Stack: [1,2]     Path: [4,3]
  Current=2, no edges -> pop to path
Stack: [1]       Path: [4,3,2]
  Current=1, no edges -> pop to path
Stack: []        Path: [4,3,2,1]

Step 4: Reverse path
Final: [1,2,3,4]

Step 5: Validate
- Length = 4 = m + 1? 3 + 1 = 4 YES
- Starts at 1? YES
- Ends at 4 (=n)? YES

Output: 1 2 3 4

Python Solution

import sys
from collections import defaultdict
sys.setrecursionlimit(300000)

def solve():
  input_data = sys.stdin.read().split()
  idx = 0
  n = int(input_data[idx]); idx += 1
  m = int(input_data[idx]); idx += 1

  adj = defaultdict(list)
  in_deg = [0] * (n + 1)
  out_deg = [0] * (n + 1)

  for _ in range(m):
    a = int(input_data[idx]); idx += 1
    b = int(input_data[idx]); idx += 1
    adj[a].append(b)
    out_deg[a] += 1
    in_deg[b] += 1

  # Check Eulerian path conditions for this problem
  # Vertex 1 must be start: out_deg[1] - in_deg[1] = 1
  # Vertex n must be end: in_deg[n] - out_deg[n] = 1
  # All others: in_deg = out_deg

  if out_deg[1] - in_deg[1] != 1:
    print("IMPOSSIBLE")
    return

  if in_deg[n] - out_deg[n] != 1:
    print("IMPOSSIBLE")
    return

  for v in range(2, n):
    if in_deg[v] != out_deg[v]:
      print("IMPOSSIBLE")
      return

  # Hierholzer's algorithm (iterative to avoid stack overflow)
  path = []
  stack = [1]

  while stack:
    u = stack[-1]
    if adj[u]:
      v = adj[u].pop()
      stack.append(v)
    else:
      path.append(stack.pop())

  path.reverse()

  # Validate: path length should be m+1 and end at n
  if len(path) != m + 1 or path[-1] != n:
    print("IMPOSSIBLE")
    return

  print(' '.join(map(str, path)))

solve()

Complexity Analysis

Operation	Time	Space
Build graph	O(m)	O(n + m)
Degree check	O(n)	O(n)
Hierholzer DFS	O(m)	O(m) stack
Total	O(n + m)	O(n + m)

Common Mistakes

Mistake	Why It Fails	Fix
Wrong degree check	General Eulerian path allows any start/end	Must verify vertex 1 is start, vertex n is end
Not reversing path	Hierholzer builds path backwards	Always reverse at the end
Using recursion	Stack overflow for large m	Use iterative version with explicit stack
Missing connectivity check	Disconnected components	Validate path length = m + 1
Checking all vertices equally	Vertex 1 and n have special roles	Separate checks for 1, n, and others

Problem	Type	Key Difference
Mail Delivery	Eulerian Circuit	Undirected graph, start = end
De Bruijn Sequence	Eulerian Path	Implicit graph from string patterns
LeetCode 332: Reconstruct Itinerary	Eulerian Path	Lexicographically smallest path
LeetCode 2097: Valid Arrangement of Pairs	Eulerian Path	Generic start/end