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Algorithm/LeetCode Daily Challenge

[LeetCode - Daily Challenge] 79. Word Search

배워서 남 주는 Jaime.Lee 2021. 10. 9. 10:30
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소스 코드는 여기 있습니다.
문제는 여기 있습니다.

Problem

Given an m x n grid of characters board and a string word, return true if word exists in the grid.

The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.

Example 1:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
Output: true

Example 2:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
Output: true

Example 3:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
Output: false

Constraints:

  • m == board.length
  • n = board[i].length
  • 1 <= m, n <= 6
  • 1 <= word.length <= 15
  • board and word consists of only lowercase and uppercase English letters.

Follow up: Could you use search pruning to make your solution faster with a larger board?

Solution

m * n 그리드와 단어가 주어질 때 그리드 안에 단어가 존재하는지 확인하는 문제입니다.

단어는 인접한 셀을 이어서 완성해야하고 같은 셀을 두 번 이상 방문할 수 없습니다.

DFS 알고리즘을 사용해 간단히 해결할 수 있습니다.

package io.lcalmsky.leetcode.word_search;

public class Solution {

    public boolean exist(char[][] board, String word) {
        if (word.length() == 0) {
            return true;
        }
        int m = board.length;
        int n = board[0].length;
        boolean[][] visited = new boolean[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (search(board, word, 0, i, j, visited)) {
                    return true;
                }
            }
        }
        return false;
    }

    private boolean search(char[][] board, String word, int n, int i, int j, boolean[][] visited) {
        if (n == word.length()) {
            return true;
        }
        if (i < 0 || i >= board.length || j < 0 || j >= board[0].length) {
            return false;
        }
        if (visited[i][j]) {
            return false;
        }
        if (word.charAt(n) != board[i][j]) {
            return false;
        }
        visited[i][j] = true;
        boolean result = search(board, word, n + 1, i - 1, j, visited)
                || search(board, word, n + 1, i + 1, j, visited)
                || search(board, word, n + 1, i, j - 1, visited)
                || search(board, word, n + 1, i, j + 1, visited);
        visited[i][j] = false;
        return result;
    }
}

Test

package io.lcalmsky.leetcode.word_search;

import org.junit.jupiter.api.Test;

import static org.junit.jupiter.api.Assertions.assertAll;
import static org.junit.jupiter.api.Assertions.assertEquals;

class SolutionTest {
    @Test
    void givenCharacters_whenSearchWord_thenCorrect() {
        char[][] givenArray = {
                {'A', 'B', 'C', 'E'},
                {'S', 'F', 'C', 'S'},
                {'A', 'D', 'E', 'E'}
        };
        assertAll(
                () -> test(givenArray, "ABCCED", true),
                () -> test(givenArray, "SEE", true),
                () -> test(givenArray, "ABCB", false),
                () -> test(new char[][]{
                        {'a'}
                }, "a", true),
                () -> test(new char[][]{
                        {'a', 'b'},
                        {'c', 'd'}
                }, "abcd", false)
        );

    }

    private void test(char[][] givenArray, String givenWord, boolean expected) {
        // when
        Solution solution = new Solution();
        boolean actual = solution.exist(givenArray, givenWord);

        // then
        assertEquals(expected, actual);
    }
}
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