#0286

Word Search

MediumDSA28 min20 XP

Problem

Given a board and word, return True if the word can be built from horizontally or vertically adjacent cells. A cell cannot be reused in the same word path.

Why This Matters

Word Search combines grid DFS with backtracking state cleanup, a frequent interview combination.

Function Signature

def exist(board, word):

Examples

Example 1

Inputboard = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"

OutputTrue

The path A -> B -> C -> C -> E -> D exists through adjacent cells.

Constraints

Return the exact requested value.
Handle edge cases cleanly.
Use the intended DSA pattern when brute force would scale poorly.

CodePython

def exist(board, word):
    """
    Input example: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
    Return: True
    Return the value from the function so the tests can compare it. Do not print unless the prompt asks.
    """
    # write your code here
    return False

def exist(board, word):
    """
    Input example: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
    Return: True
    Return the value from the function so the tests can compare it. Do not print unless the prompt asks.
    """
    # write your code here
    return False

Visible browser tests run here when available.

Testcases2 visible / 3 hidden categories

Word exists

InputABCCED

ExpectedTrue

A valid adjacent path exists.

Cannot reuse cell

InputABCB

ExpectedFalse

The path would need to reuse the B cell.

Hidden Test Categories

single-cell boardword longer than cellssame letter repeated