Leetcode 200. Number of Islands. Python

Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands.

An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

Input: grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
Output: 1

Example 2:

Input: grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
Output: 3

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] is '0' or '1'.

class Solution:
    def numIslands(self, g: List[List[str]]) -> int:
        if not g:
            return 0
        
        rows, cols = len(g), len(g[0])
        vs = set()
        ans = 0
        
        def bfs(r, c):
            q = collections.deque()
            vs.add((r,c))
            q.append((r,c))
            
            while q:
                row, col = q.popleft()
                d = [[1,0],[-1,0],[0,1],[0,-1]]
                
                for dr, dc in d:
                    r, c = row + dr, col + dc
                    if ( r in range(rows) and 
                         c in range(cols) and
                         g[r][c] == "1" and
                         (r,c) not in vs):
                        q.append((r,c))
                        vs.add((r,c))
                        
        for r in range(rows):
            for c in range(cols):
                if g[r][c] == "1" and (r,c) not in vs:
                    bfs(r,c)
                    ans += 1
        
        return ans

Explaination :

Comments

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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 :

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322 . Coin Change You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money. Return the fewest number of coins that you need to make up that amount . If that amount of money cannot be made up by any combination of the coins, return -1 . You may assume that you have an infinite number of each kind of coin. Example 1: Input: coins = [1,2,5], amount = 11 Output: 3 Explanation: 11 = 5 + 5 + 1 Example 2: Input: coins = [2], amount = 3 Output: -1 Example 3: Input: coins = [1], amount = 0 Output: 0 Constraints: 1 <= coins.length <= 12 1 <= coins[i] <= 2 31 - 1 0 <= amount <= 10 4 Solution : class Solution: def coinChange(self, coins: List[int], amount: int) -> int: dp = [amount + 1] * (amount + 1) #[0...7] dp[0] = 0 ...

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