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How is this a backtracking solution and not just a restricted DFS?
Arturo Calderón
Nov 13, 2023
Marking the current cell (see line 10 of the solution) does nothing, as it is never checked, nor does it change the result if omitted.
Are we really solving this with backtracking?
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Shubham Voraa year ago
marking the current cell matters, when we check the below condition. It won't use the character at position at board[i][j] as it is already visited.
board[i][j] != word.charAt(k)
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