feat : word-search

Author: ekgns33

word-search/ekgns33.java

@@ -0,0 +1,58 @@
+/*
+input : m x n matrix and string word
+output : return true if given word can be constructed
+
+solution 1) brute force
+tc : O(n * 4^k) when n is the number of cells, k is the length of word
+sc : O(k) call stack
+ */
+class Solution {
+  private int[][] directions = new int[][] {{-1,0}, {1,0}, {0,1}, {0,-1}};
+  public boolean exist(char[][] board, String word) {
+    //edge case
+    int m = board.length;
+    int n = board[0].length;
+
+    if(m * n < word.length()) return false;
+
+
+    //look for the starting letter and do dfs
+    for(int i = 0; i < m; i++) {
+
+      for(int j = 0; j < n; j++) {
+
+        if(board[i][j] == word.charAt(0)) {
+          //do dfs and get answer
+          board[i][j] = '0';
+          boolean res = dfsHelper(board, word, i, j, 1);
+          board[i][j] = word.charAt(0);
+          if(res) return true;
+        }
+      }
+    }
+
+    return false;
+  }
+
+  public boolean dfsHelper(char[][] board, String word, int curR, int curC, int curP) {
+
+    //endclause
+    if(curP == word.length()) return true;
+
+    boolean ret = false;
+
+    for(int[] direction : directions) {
+      int nextR = curR + direction[0];
+      int nextC = curC + direction[1];
+
+      if(nextR < 0 || nextR >= board.length || nextC < 0 || nextC >= board[0].length) continue;
+
+      if(board[nextR][nextC] == word.charAt(curP)) {
+        board[nextR][nextC] = '0';
+        ret = ret || dfsHelper(board, word, nextR, nextC, curP + 1);
+        board[nextR][nextC] = word.charAt(curP);
+      }
+    }
+    return ret;
+  }
+}