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  1. Prepare
  2. Java
  3. Data Structures
  4. Java 1D Array (Part 2)
  5. Discussions

Java 1D Array (Part 2)

  • RodneyShag
     + 38 comments

    Java solution - passes 100% of test cases

    Solution is basically to do a depth-first search (DFS). Instead of creating a "visited" array, we can mark each array value as 1 when visiting it.

    Full code available at my HackerRank solutions. Here is the main snippet:

    public static boolean canWin(int leap, int[] game) {
    return isSolvable(leap, game, 0);
}

private static boolean isSolvable(int leap, int[] game, int i) {
    // Base Cases
    if (i >= game.length) {
        return true;
    } else if (i < 0 || game[i] == 1) {
        return false;
    }
            
    game[i] = 1; // marks as visited

    // Recursive Cases
    return isSolvable(leap, game, i + leap) || 
           isSolvable(leap, game, i + 1) || 
           isSolvable(leap, game, i - 1);
}

    Let me know if you have any questions.