The Most Creative Problem I’ve Ever Solved

A Deep Dive into an Unconventional Coding Challenge

Introduction

Every programmer encounters problems that seem impossible at first—until a sudden flash of insight changes everything. This is the story of one such problem: a puzzle that forced me to think outside conventional algorithms and discover a solution so elegant that it felt like cracking a secret code.

The Problem: "Grid Escape with Teleporters"

(Inspired by a contest problem, simplified for clarity)

Scenario

You’re trapped in an N x N grid where:

• Some cells contain teleporters (marked T).


• Others are blocked (#) or empty (.).


• You start at (0, 0) and must reach (N-1, N-1).

Catch:

• Stepping on a teleporter instantly transports you to every other teleporter in the grid.


• You can choose which teleporter to exit from (like a network of portals).

Goal: Find the shortest path to escape the grid.

First Attempt: Standard BFS (And Why It Failed)

My initial thought: "This is just a BFS problem with a twist."

Why BFS Broke Down

1. Exponential States: If there are k teleporters, stepping on one creates k possible positions instantly.


2. Infinite Loops: You could teleport between the same two points forever.


3. Visited Tracking: Traditional visited checks fail because teleporting resets your possible locations.

Grid:
.T.
T.T
...

Stepping on (0,1) teleports you to all Ts: (0,1), (1,0), (1,2). From (1,0), you could teleport back to (0,1), creating a loop.

The Insight: Teleporters as a "Super Node"

After hours of frustration, I realized:

All teleporters are effectively one node in the traversal graph.

Key Observations

• Teleporters are Linked: Entering any T means you can exit from any other T in zero steps.


• BFS Layers Adjust: Moving to a teleporter doesn’t increment the path length until you exit one.

Solution Sketch

1. Preprocess Teleporters: Collect all (i,j) positions of T into a list.


2. Augmented BFS: Treat all teleporters as a single state (e.g., "I’m in the teleporter network").

The "Aha!" Moment: Pseudocode

from collections import deque

def shortest_path(grid):
    N = len(grid)
    teleporters = [(i, j) for i in range(N) for j in range(N) if grid[i][j] == 'T']
    
    q = deque()
    q.append((0, 0, False))  # (i, j, used_teleporter)
    visited = set()
    visited.add((0, 0, False))
    steps = 0
    
    while q:
        for _ in range(len(q)):
            i, j, used = q.popleft()
            
            if (i, j) == (N-1, N-1):
                return steps
            
            if grid[i][j] == 'T' and not used:
                for (ti, tj) in teleporters:
                    if (ti, tj) not in visited:
                        visited.add((ti, tj))
                        q.append((ti, tj, True))  # Teleport for free
            
            for di, dj in [(0,1), (1,0), (-1,0), (0,-1)]:
                ni, nj = i + di, j + dj
                if 0 <= ni < N and 0 <= nj < N and grid[ni][nj] != '#':
                    if (ni, nj, used) not in visited:
                        visited.add((ni, nj, used))
                        q.append((ni, nj, used))
        steps += 1
    return -1

Why This Works

• State = (Position, Teleporter Usage)


• (i, j, False): Haven’t used a teleporter yet.


• (i, j, True): Already used one (prevents loops).


• Teleporting is a Zero-Cost Move


• Efficiency: Visited checks prevent redundant work. Worst case: O(N²) like standard BFS.

Edge Cases That Tripped Me Up

1. No Teleporters: Falls back to classic BFS.


2. Start/End on Teleporters: Must handle (0,0) or (N-1,N-1) being T.


3. All Cells Blocked Except Teleporters: Path exists only via teleporting.

Final Thoughts

This problem taught me:

• Sometimes, the hardest part is redefining the problem.


• Graph theory hides in unexpected places (even teleporters!).


• Creative state representation unlocks solutions.

Try It Yourself

Input:

[
  "...T",
  ".T#.",
  "T##T",
  "...."
]

Output: 5 (Path: Start → (0,3) → (3,3) → End).

Can you find the path? 🚀