#0243

Word Ladder Length

HardDSA40 min25 XP

Problem

Each move changes exactly one letter and must create a word in word_list. Return the number of words in the shortest sequence from begin to end, or 0 if impossible.

Why This Matters

Word Ladder is a strong reminder that shortest unweighted path means BFS, even when the graph is hidden inside strings.

Function Signature

def ladder_length(begin, end, word_list):

Examples

Example 1

Inputbegin = "hit", end = "cog", word_list = ["hot", "dot", "dog", "lot", "log", "cog"]

Output5

One shortest path is hit -> hot -> dot -> dog -> cog.

Constraints

Return the exact requested value.
Handle empty or small inputs when the prompt allows them.
Prefer the target DSA pattern over brute force when the input can grow.

CodePython

def ladder_length(begin, end, word_list):
    """
    Input example: begin = "hit", end = "cog", word_list = ["hot", "dot", "dog", "lot", "log", "cog"]
    Return: 5
    Return the value from the function so the tests can compare it. Do not print unless the prompt asks.
    """
    # write your code here
    return 0

def ladder_length(begin, end, word_list):
    """
    Input example: begin = "hit", end = "cog", word_list = ["hot", "dot", "dog", "lot", "log", "cog"]
    Return: 5
    Return the value from the function so the tests can compare it. Do not print unless the prompt asks.
    """
    # write your code here
    return 0

Visible browser tests run here when available.

Testcases2 visible / 3 hidden categories

Reachable path

Inputhit to cog

Expected5

BFS finds the shortest sequence length.

Missing end

Inputend not in list

Expected0

The final word must be available as a move.

Hidden Test Categories

begin equals end style casesmany decoy wordsno connecting bridge