Word Search LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Word Search in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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LeetCode is for software engineers who are looking to practice technical questions and advance their skills. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.

LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either Easy, Medium, or Hard.

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• Mathematics/Basic Logical Based Questions
• Arrays
• Strings
• Hash Table
• Dynamic Programming
• Stack & Queue
• Trees & Graphs
• Greedy Algorithms
• Breadth-First Search
• Depth-First Search
• Sorting & Searching
• BST (Binary Search Tree)
• Database
• Linked List
• Recursion, etc.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

Link for the Problem – Word Search– LeetCode Problem

Word Search– LeetCode Problem

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.

Word Search– LeetCode Solutions

Word Search in C++:

class Solution {
 public:
  bool exist(vector<vector<char>>& board, string word) {
    for (int i = 0; i < board.size(); ++i)
      for (int j = 0; j < board[0].size(); ++j)
        if (dfs(board, word, i, j, 0))
          return true;
    return false;
  }

 private:
  bool dfs(vector<vector<char>>& board, const string& word, int i, int j,
           int s) {
    if (i < 0 || i == board.size() || j < 0 || j == board[0].size())
      return false;
    if (board[i][j] != word[s] || board[i][j] == '*')
      return false;
    if (s == word.length() - 1)
      return true;

    const char cache = board[i][j];
    board[i][j] = '*';
    const bool isExist = dfs(board, word, i + 1, j, s + 1) ||
                         dfs(board, word, i - 1, j, s + 1) ||
                         dfs(board, word, i, j + 1, s + 1) ||
                         dfs(board, word, i, j - 1, s + 1);
    board[i][j] = cache;

    return isExist;
  }
};

Word Search in Java:

class Solution {
  public boolean exist(char[][] board, String word) {
    for (int i = 0; i < board.length; ++i)
      for (int j = 0; j < board[0].length; ++j)
        if (dfs(board, word, i, j, 0))
          return true;
    return false;
  }

  private boolean dfs(char[][] board, String word, int i, int j, int s) {
    if (i < 0 || i == board.length || j < 0 || j == board[0].length)
      return false;
    if (board[i][j] != word.charAt(s) || board[i][j] == '*')
      return false;
    if (s == word.length() - 1)
      return true;

    final char cache = board[i][j];
    board[i][j] = '*';
    final boolean isExist = dfs(board, word, i + 1, j, s + 1) ||
                            dfs(board, word, i - 1, j, s + 1) ||
                            dfs(board, word, i, j + 1, s + 1) ||
                            dfs(board, word, i, j - 1, s + 1);
    board[i][j] = cache;

    return isExist;
  }
}

Word Search in Python:

class Solution:
  def exist(self, board: List[List[str]], word: str) -> bool:
    m = len(board)
    n = len(board[0])

    def dfs(i: int, j: int, s: int) -> bool:
      if i < 0 or i == m or j < 0 or j == n:
        return False
      if board[i][j] != word[s] or board[i][j] == '*':
        return False
      if s == len(word) - 1:
        return True

      cache = board[i][j]
      board[i][j] = '*'
      isExist = \
          dfs(i + 1, j, s + 1) or \
          dfs(i - 1, j, s + 1) or \
          dfs(i, j + 1, s + 1) or \
          dfs(i, j - 1, s + 1)
      board[i][j] = cache

      return isExist

    return any(dfs(i, j, 0) for i in range(m) for j in range(n))