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Linda Zhou edited this page Oct 29, 2022 · 24 revisions

Problem Highlights

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

  • Established a set (2-3) of test cases to verify their own solution later.
  • Established a set (1-2) of edge cases to verify their solution handles complexities.
  • Have fully understood the problem and have no clarifying questions.
  • Have you verified any Time/Space Constraints for this problem?

  • When do we return true?

    • Return true when the last letter is reached.

  • What if we come across the same letter during the search path?

    • During the search path, set the visited letter as visited to avoid reuse.
word = "ABCCED", -> returns 1,

word = "SEE", -> returns 1,

word = "ABCB", -> returns 1,

word = "ABFSAB" -> returns 1

word = "ABCD" -> returns 0

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

For graph problems, some things we want to consider are:
  • BFS/DFS: Our goal is to find if the word exists in the matrix or not. We only have to look at the adjacent cells (ignore the diagonal ones). Match character-by-character, go ahead and check if the adjacent cells match the next character, and go back if it does not match. How should we traverse the matrix efficiently? We need to think of a traversal approach. BFS? DFS? Both can work. But DFS will be better as it immediately checks the next node of the graph and returns if it is not needed after marking it as visited.
  • Union Find: Are there find and union operations here? Can you perform a find operation where you can determine which subset a particular element is in? This can be used for determining if two elements are in the same subset. Can you perform a union operation where you join two subsets into a single subset? Can you check if the two subsets belong to same set? If no, then we cannot perform union.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

1. Create helper function(node, string):
    1. If the string is empty, return true (success).
    2. If the node's letter is not the first letter of the string, return false (failure).
    3. Otherwise, return true if the helper function returns true for
        1. any neighbor
        2. the remainder of the string
2. Implement main function(board, string):
    1. Return true if any of these helper function calls return true
        1. each node
        2. the full string

⚠️ Common Mistakes

4: I-mplement

Implement the code to solve the algorithm.

def exist(self, board: List[List[str]], word: str) -> bool:        
        if not board or not word:
            return False    
    
        def dfs(row, col, index):
            if index == len(word):
                return True
            
            if row < 0 or row >= len(board) or col < 0 or col >= len(board[row]) or board[row][col] != word[index]:
                return False
            
            temp = board[row][col]
            board[row][col] = ''
            found = dfs(row + 1, col, index + 1) or dfs(row - 1, col, index + 1) or dfs(row, col + 1, index + 1) or dfs(row, col - 1, index + 1)
            board[row][col] = temp
            return found
        
        for row in range(len(board)):
            for col in range(len(board[row])):
                if word[0] == board[row][col] and dfs(row, col, 0):
                    return True

        return False
class Solution {
    public boolean exist(char[][] board, String word) {
        char[] arr = word.toCharArray();
        boolean[][] visited = new boolean[board.length][board[0].length];
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                visited[i][j] = true;
                if (dfs(board,visited, arr,0, i, j)) {
                    return true;
                } else {
                    visited[i][j] = false;
                }
            }
        }
        return false;
        
    }
    
    private boolean dfs(char[][] board, boolean[][] visited, char[] arr, int index, int row, int col) {
        if (arr[index] != board[row][col]) {
            return false;
        } 
        if (index == arr.length - 1) {
            return true;
        }
        int[] rHelp = new int[]{1,-1,0,0};
        int[] cHelp = new int[]{0,0,1, -1};
        for (int i = 0; i < 4; i++) {
            int nextRow = row + rHelp[i];
            int nextCol = col + cHelp[i];
            if (isValid(board, visited, nextRow, nextCol)) {
                visited[nextRow][nextCol] = true;
                if (dfs(board,visited, arr, index + 1, nextRow, nextCol)) {
                    return true;
                } else {
                    visited[nextRow][nextCol] = false;
                }
            }
        }
        return false;
    }
    
    private boolean isValid(char[][] board, boolean[][] visited, int row, int col) {
        int rowMax = board.length - 1;
        int colMax = board[0].length - 1;
        if (row >= 0 && row <= rowMax && col >= 0 && col <= colMax && !visited[row][col]) {
            return true;
        } else {
            return false;
        }
    }
}

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

  • Trace through your code with an input to check for the expected output
  • Catch possible edge cases and off-by-one errorS and verify the code works for the happy and edge cases you created in the “Understand” section

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Time Complexity: O(m * n * 4^w), where Board size = n * m and w = length of the word
Space Complexity: O(1). The space complexity would be the length of the word, since the recursion stack only goes as far as the length. However, if you are using extra space to mark visited cells, then it would be the size of the board.

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