Word Ladder

📋

Problem

Imagine you want to transform one word into another by changing one letter at a time, like a puzzle where each step must be a valid English word. How quickly can you find the shortest transformation?

Given two words, beginWord and endWord, and a dictionary wordList, find the length of the shortest transformation sequence from beginWord to endWord such that: 1. Only one letter can be changed at a time. 2. Each transformed word must exist in the wordList. Return 0 if no such sequence exists. Input: beginWord (string), endWord (string), wordList (list of strings) Output: integer representing the length of the shortest transformation sequence.

1 ≤ length of beginWord, endWord ≤ 101 ≤ number of words in wordList ≤ 10^5All words have the same lengthAll words contain only lowercase English letters

Edge cases: endWord not in wordList → 0beginWord equals endWord → 1wordList empty → 0

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IDE

def word_ladder_bruteforce(beginWord: str, endWord: str, wordList: list) -> int:public int wordLadderBruteforce(String beginWord, String endWord, List<String> wordList)int wordLadderBruteforce(string beginWord, string endWord, vector<string> wordList)function wordLadderBruteforce(beginWord, endWord, wordList)

def word_ladder_bruteforce(beginWord, endWord, wordList):
    # Write your solution here
    pass

class Solution {
    public int wordLadderBruteforce(String beginWord, String endWord, List<String> wordList) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int wordLadderBruteforce(string beginWord, string endWord, vector<string> wordList) {
    // Write your solution here
    return 0;
}

function wordLadderBruteforce(beginWord, endWord, wordList) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: 0Not checking if beginWord equals endWord, returning 0 instead of 1.✅ Add condition: if beginWord == endWord: return 1

Wrong: non-zero (e.g. 1 or 2) when no path existsNot verifying endWord presence in wordList before BFS or incorrect BFS termination.✅ Return 0 if endWord not in wordList before BFS starts.

Wrong: Longer path length or incorrect pathUsing greedy or DFS instead of BFS, missing shortest path guarantee.✅ Implement BFS with queue and visited set to find shortest path.

Wrong: Infinite loop or repeated nodes causing wrong outputMissing visited set causing revisiting nodes multiple times.✅ Maintain visited set to mark words already processed in BFS.

Wrong: TLE or timeoutUsing brute force DFS or inefficient neighbor generation on large inputs.✅ Use BFS with efficient neighbor generation and pruning; avoid DFS on large inputs.

✓

Test Cases

Handle base cases explicitly before BFS to avoid unnecessary computation.

Test your BFS against tricky cases that break greedy or DFS solutions.

Optimize your BFS with efficient neighbor generation and pruning.

t1_01basic

Input{"beginWord":"hit","endWord":"cog","wordList":["hot","dot","dog","lot","log","cog"]}

Expected5

⏱ Performance - must finish in 2000ms

One shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog", length 5.

💡 Think of the problem as finding the shortest path in a graph where nodes are words.

💡 Use BFS to explore neighbors differing by one letter to find the shortest transformation.

💡 Track levels in BFS; the first time you reach endWord, return the current level count.

Why it failed: Returned incorrect length; likely BFS or neighbor generation logic is wrong. Fix by ensuring only one letter changes and BFS level counting is correct.

✓ Correctly found shortest transformation length using BFS.

t1_02basic

Input{"beginWord":"game","endWord":"thee","wordList":["fame","fate","gate","gape","tape","tame","thee"]}

Expected0

⏱ Performance - must finish in 2000ms

No valid transformation sequence from "game" to "thee" exists with one letter changes only.

💡 Check if endWord is in wordList; if not, return 0 immediately.

💡 Ensure BFS terminates when no path is found and returns 0.

💡 Verify neighbor generation does not produce invalid words outside wordList.

Why it failed: Returned non-zero when no path exists; fix by verifying endWord presence and BFS termination conditions.

✓ Correctly returned 0 when no transformation path exists.

t2_01edge

Input{"beginWord":"same","endWord":"same","wordList":["same"]}

Expected1

⏱ Performance - must finish in 2000ms

BeginWord equals endWord, so shortest transformation length is 1 (no changes needed).

💡 Check if beginWord equals endWord and return 1 immediately.

💡 This is a base case that avoids unnecessary BFS.

💡 Add a condition at the start of your function to handle this case.

Why it failed: Returned 0 or incorrect value when beginWord equals endWord; fix by adding explicit equality check returning 1.

✓ Correctly handled beginWord equals endWord case.

t2_02edge

Input{"beginWord":"hit","endWord":"cog","wordList":[]}

Expected0

⏱ Performance - must finish in 2000ms

Empty wordList means no transformations possible; return 0.

💡 If wordList is empty, no transformations can occur.

💡 Check for empty wordList and return 0 immediately.

💡 Avoid running BFS if wordList is empty.

Why it failed: Returned non-zero with empty wordList; fix by adding empty list check returning 0.

✓ Correctly returned 0 for empty wordList.

t2_03edge

Input{"beginWord":"hit","endWord":"hot","wordList":["hot"]}

Expected2

⏱ Performance - must finish in 2000ms

One letter difference between beginWord and endWord, and endWord in wordList; shortest path length is 2.

💡 Check neighbors differing by one letter from beginWord.

💡 If endWord is a direct neighbor, shortest path length is 2.

💡 Ensure BFS counts levels correctly starting from 1.

Why it failed: Returned incorrect length for direct neighbor case; fix BFS level counting and neighbor checks.

✓ Correctly handled direct neighbor transformation.

t3_01corner

Input{"beginWord":"aaa","endWord":"bbb","wordList":["aab","abb","bbb"]}

Expected4

⏱ Performance - must finish in 2000ms

Shortest path is "aaa" -> "aab" -> "abb" -> "bbb", length 4. Tests BFS correctness over greedy approach.

💡 Beware of greedy approaches that pick first neighbor without exploring all paths.

💡 Use BFS to guarantee shortest path, not DFS or greedy.

💡 Track visited words properly to avoid cycles and incorrect paths.

Why it failed: Greedy or DFS approach returns longer or incorrect path length; fix by implementing BFS with proper visited tracking.

✓ Correctly found shortest path length using BFS, avoiding greedy traps.

t3_02corner

Input{"beginWord":"abc","endWord":"xyz","wordList":["xbc","xyc","xyz"]}

Expected4

⏱ Performance - must finish in 2000ms

Tests confusion between 0/1 knapsack style unbounded vs 0/1 BFS: each word can be used once only, path length 4.

💡 Ensure each word is used at most once by tracking visited set.

💡 Do not revisit words even if they appear as neighbors multiple times.

💡 Implement BFS with visited set to avoid cycles and infinite loops.

Why it failed: Incorrectly revisiting words or infinite loops due to missing visited set; fix by adding visited set to BFS.

✓ Correctly handled 0/1 usage of words with visited set.

t3_03corner

Input{"beginWord":"hit","endWord":"cog","wordList":["hot","dot","dog","lot","log"]}

Expected0

⏱ Performance - must finish in 2000ms

endWord "cog" not in wordList, so no transformation possible; return 0.

💡 Check if endWord is in wordList before BFS.

💡 Return 0 immediately if endWord missing.

💡 Avoid unnecessary BFS if endWord not present.

Why it failed: Returned non-zero when endWord missing; fix by checking endWord presence before BFS.

✓ Correctly returned 0 when endWord not in wordList.

t4_01performance

Input{"beginWord":"aaaaa","endWord":"zzzzz","wordList":["aaaaa","aaaab","aaabb","aabbb","abbbb","bbbbb","bbbbc","bbbcc","bbccc","bcccc","ccccc","ccccd","ccddd","cdddd","ddddd","ddddz","dddzz","ddzzz","dzzzz","zzzzz","aaaac","aaacc","aaccc","acccc","ccccc","ccccd","ccddd","cdddd","ddddd","ddddd","zzzzz","zzzza","zzzaz","zzazz","zazzz","azzzz"]}

Expectednull

⏱ Performance - must finish in 2000ms

Large input with n=40 words, word length=5, BFS O(n*L*26) must complete within 2s.

💡 Use BFS with efficient neighbor generation.

💡 Avoid recomputing neighbors by preprocessing or pruning.

💡 Use sets for O(1) lookups and visited tracking.

Why it failed: TLE due to exponential or DFS approach; fix by implementing BFS with pruning and visited set.

✓ Efficient BFS implementation confirmed by passing performance test.

Practice

1. You are given a graph with n nodes and n-1 edges, where exactly one node is connected to all other nodes, forming a star shape. Which approach guarantees finding the center node in O(n) time without unnecessary graph traversal?

easy

A. Count the frequency of each node in the edges and return the node appearing n-1 times.

B. Apply dynamic programming to store intermediate results of node degrees and find the maximum.

C. Use a greedy approach by picking the first node in the first edge as the center without verification.

D. Perform a depth-first search (DFS) from each node to count its neighbors and identify the center.

Solution

1. Step 1: Understand the star graph property

   In a star graph with n nodes, the center node appears in every edge, so it has degree n-1.

2. Step 2: Use frequency counting

   Counting how many times each node appears in the edges lets us find the node with degree n-1 efficiently in O(n) time.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Frequency counting directly identifies the center without traversal [OK]

Hint: Center node appears in all edges [OK]

Common Mistakes:

• Assuming the first node in edges is always center
• Using DFS from every node causing O(n²) time
• Applying DP unnecessarily for a simple counting problem

2. You are given an undirected graph and need to determine if its vertices can be divided into two groups such that no edge connects vertices within the same group. Which algorithmic approach guarantees an efficient and correct solution for this problem?

easy

A. Greedy algorithm that assigns groups based on vertex degree

B. Dynamic Programming on subsets of vertices to find valid partitions

C. Depth-First Search (DFS) to find cycles and count their lengths

D. Breadth-First Search (BFS) coloring to assign two colors and check for conflicts

Solution

1. Step 1: Understand the problem as a 2-coloring challenge

   The problem requires checking if the graph is bipartite, which means it can be colored using two colors such that no adjacent vertices share the same color.

2. Step 2: Identify BFS coloring as the optimal approach

   BFS coloring systematically assigns colors level by level and detects conflicts efficiently, guaranteeing correctness and linear time complexity.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   BFS coloring is the standard method for bipartite checking [OK]

Hint: BFS coloring detects bipartite conflicts efficiently [OK]

Common Mistakes:

• Assuming greedy degree-based grouping works
• Confusing cycle detection with bipartite check

3. The following code attempts to check if a graph is a valid tree. Which line contains a subtle bug that can cause incorrect results on disconnected graphs?

medium

A. Line checking if edge count equals n - 1

B. Line returning true without verifying all nodes visited

C. Line skipping neighbor if it equals parent

D. Line adding neighbor to queue

Solution

1. Step 1: Identify final return condition

   The code returns true without checking if all nodes were visited, missing disconnected components.

2. Step 2: Understand impact

   Disconnected graphs with correct edge count and no cycles will incorrectly return true.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Must verify visited nodes count equals n [OK]

Hint: Always check all nodes visited for connectivity [OK]

Common Mistakes:

• Forgetting connectivity check after BFS/DFS
• Misinterpreting cycle detection as full validation

4. The following BFS code attempts to find the max area of an island in a grid. Identify the line that contains a subtle bug that can cause incorrect area calculation or runtime errors.

from collections import deque

def maxAreaOfIsland(grid):
    rows, cols = len(grid), len(grid[0])
    visited = set()
    max_area = 0

    for r in range(rows):
        for c in range(cols):
            if grid[r][c] == 1 and (r, c) not in visited:
                queue = deque([(r, c)])
                area = 0
                while queue:
                    row, col = queue.popleft()
                    # Bug: Missing visited check here
                    visited.add((row, col))
                    area += 1
                    for nr, nc in [(row+1,col),(row-1,col),(row,col+1),(row,col-1)]:
                        if grid[nr][nc] == 1 and (nr, nc) not in visited:
                            queue.append((nr, nc))
                max_area = max(max_area, area)
    return max_area

medium

A. Line that adds neighbors to the queue without boundary checks

B. Line that adds (row, col) to visited without checking if already visited

C. Line that initializes the queue with the starting cell

D. Line that updates max_area after BFS completes

Solution

1. Step 1: Check neighbor addition line

   The code adds neighbors without verifying if nr, nc are within grid boundaries, risking index out of range errors.

2. Step 2: Confirm other lines are correct

   Visited check before adding to queue is present, queue initialization and max_area update are correct.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Boundary checks missing cause runtime errors [OK]

Hint: Always check grid boundaries before accessing neighbors [OK]

Common Mistakes:

• Forgetting boundary checks
• Not checking visited before adding to queue
• Adding visited check after popping queue

5. What is the worst-case space complexity of the in-place DFS solution for the Number of Islands problem on an m x n grid, considering the recursion stack usage?

medium

A. O(1) -- since the grid is modified in-place, no extra space is needed.

B. O(min(m, n)) -- recursion depth limited by smaller grid dimension.

C. O(m*n) -- recursion stack can grow up to the size of the entire grid in worst case.

D. O(m + n) -- recursion stack limited by sum of rows and columns.

Solution

1. Step 1: Identify auxiliary space usage

   The grid is modified in-place, so no extra data structures are used except recursion stack.

2. Step 2: Analyze recursion stack depth

   In worst case (e.g., one large island), DFS can recurse over all cells, leading to recursion stack depth O(m*n).

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Recursion stack can be as large as all grid cells in worst case [OK]

Hint: Recursion stack can be as large as island size [OK]

Common Mistakes:

• Assuming constant space due to in-place modification
• Confusing recursion depth with grid dimensions