The Most Difficult Coding Interview at Google: A True Challenge

Google is known for its brutal coding interviews. One of the hardest problems I encountered was a mix of recursion, backtracking, and advanced dynamic programming. Let’s dive deep into one of the most difficult challenges: "The Alien Dictionary".

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The Problem: Alien Dictionary

Problem Statement

You are given a list of words from an alien language, sorted lexicographically according to the rules of this new language. Derive the order of letters in the alien alphabet.

Example:

const words = ["wrt", "wrf", "er", "ett", "rftt"];

Expected Output:

"wertf"

If no valid order exists, return an empty string "".

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Breaking Down the Approach

To solve this efficiently, we need to:

1. Build a Graph – Treat each letter as a node, and edges represent order relationships.
2. Topological Sorting (Kahn's Algorithm) – Use BFS to determine the character order.
3. Handle Edge Cases – If a cycle exists, return an empty string.

Step 1: Building the Graph

We compare adjacent words to establish letter precedence.

function buildGraph(words) {
    let graph = new Map();
    let inDegree = new Map();

    for (let word of words) {
        for (let char of word) {
            graph.set(char, new Set());
            inDegree.set(char, 0);
        }
    }

    for (let i = 0; i < words.length - 1; i++) {
        let [w1, w2] = [words[i], words[i + 1]];
        let minLen = Math.min(w1.length, w2.length);

        for (let j = 0; j < minLen; j++) {
            if (w1[j] !== w2[j]) {
                if (!graph.get(w1[j]).has(w2[j])) {
                    graph.get(w1[j]).add(w2[j]);
                    inDegree.set(w2[j], inDegree.get(w2[j]) + 1);
                }
                break;
            }
        }
    }

    return { graph, inDegree };
}

Step 2: Topological Sorting using BFS

We perform Kahn’s algorithm to derive the order of letters.

function alienOrder(words) {
    let { graph, inDegree } = buildGraph(words);
    let queue = [];
    let order = "";

    for (let [char, degree] of inDegree) {
        if (degree === 0) queue.push(char);
    }

    while (queue.length > 0) {
        let char = queue.shift();
        order += char;
        for (let neighbor of graph.get(char)) {
            inDegree.set(neighbor, inDegree.get(neighbor) - 1);
            if (inDegree.get(neighbor) === 0) queue.push(neighbor);
        }
    }

    return order.length === graph.size ? order : "";
}

Step 3: Running the Solution

console.log(alienOrder(["wrt", "wrf", "er", "ett", "rftt"]));

Expected Output:

"wertf"

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Final Thoughts

This problem is one of those that really make you think. Google interviews are designed to challenge your ability to break down problems into smaller pieces and find efficient solutions. The combination of graph construction, BFS, and cycle detection makes this a tough but rewarding exercise.

If you’re preparing for technical interviews, practicing graph problems, recursion, and dynamic programming will pay off in a big way. The key is to stay calm, break the problem down step by step, and implement a structured approach.

Have you faced any tricky coding challenges? Share your experience in the comments below, and let’s discuss strategies to tackle them! 🚀