Skip to main content

Word Search - Leetcode 79 - Python

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.


class Solution:

    def exist(self, b: List[List[str]], w: str) -> bool:

        row, col = len(b), len(b[0])

        p = set()

        

        def dfs(r, c, i):

            if i == len(w):

                return True

            

            if (

                r<0 or 

                c<0 or

                r>=row or

                c>=col or

                w[i]!=b[r][c] or

                (r,c) in p  

               ) : return False

            

            p.add((r,c))

            

            ans = (dfs(r+1,c,i+1) or

                  dfs(r-1,c,i+1) or

                  dfs(r,c+1,i+1) or

                  dfs(r,c-1,i+1))

            

            p.remove((r,c))

            return ans

        

        for r in range(row):

            for c in range(col):

                if dfs(r,c,0): return True

        return False

             

Explaination :



Comments

Popular posts from this blog

Leetcode 371. Sum of Two Integers. C++ / Java

371 .  Sum of Two Integers   Given two integers  a  and  b , return  the sum of the two integers without using the operators   +   and   - .   Example 1: Input: a = 1, b = 2 Output: 3 Example 2: Input: a = 2, b = 3 Output: 5   Constraints: -1000 <= a, b <= 1000 Solution :  C++ : class Solution { public: int getSum(int a, int b) { if (b==0) return a; int sum = a ^ b; int cr = (unsigned int) (a & b) << 1; return getSum(sum, cr); } }; Java :  class Solution { public int getSum(int a, int b) { while(b != 0){ int tmp = (a & b) << 1; a = a ^ b; b = tmp; } return a; } } Explaination :

Number of Connected Components in an Undirected Graph (Python)

66.  Number of Connected Components in an Undirected Graph Question Link :  check here Givennnodes labeled from0ton - 1and a list of undirected edges (each edge is a pair of nodes), write a function to find the number of connected components in an undirected graph. Example 1:      0          3      |          |      1 --- 2    4 Givenn = 5andedges = [[0, 1], [1, 2], [3, 4]], return2. Example 2:      0           4      |           |      1 --- 2 --- 3 Givenn = 5andedges = [[0, 1], [1, 2], [2, 3], [3, 4]], return1. Note: You can assume that no duplicate edges will appear inedges. Since all edges are undirected,[0, 1]is the same as[1, 0]and thus will not appear together inedges. Solution : class Solution: def counComponents(self, n: int, edges : List[List[int]]) -> i...

Leetcode 217. Contains Duplicate. Python (Easiest Approach ✅)

217 .  Contains Duplicate   Given an integer array  nums , return  true  if any value appears  at least twice  in the array, and return  false  if every element is distinct.   Example 1: Input: nums = [1,2,3,1] Output: true Example 2: Input: nums = [1,2,3,4] Output: false Example 3: Input: nums = [1,1,1,3,3,4,3,2,4,2] Output: true   Constraints: 1 <= nums.length <= 10 5 -10 9  <= nums[i] <= 10 9 class Solution: def containsDuplicate(self, nums: List[int]) -> bool: hs = set() for n in nums: if n in hs: return True hs.add(n) return False Explaination :