Number of Enclaves Python Solution with Tests

Tested Python solution for LeetCode 1020 with 12 pytest cases. Generate a practice environment with lcpy.

LeetCode 1020, Medium. Topics: Array, Depth-First Search, Breadth-First Search, Union Find, Matrix. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 12 parametrized pytest cases, and a playground notebook:

lcpy gen -n 1020   # by problem number
lcpy gen -s number_of_enclaves   # by problem name

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Problem

You are given an m x n binary matrix grid, where 0 represents a sea cell and 1 represents a land cell.

A move consists of walking from one land cell to another adjacent (4-directionally) land cell or walking off the boundary of the grid.

Return the number of land cells in grid for which we cannot walk off the boundary of the grid in any number of moves.

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Examples

Input: grid = [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
Output: 3
Explanation: There are three 1s that are enclosed by 0s, and one 1 that is not enclosed because its on the boundary.

Input: grid = [[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
Output: 0
Explanation: All 1s are either on the boundary or can reach the boundary.

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Constraints

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 500
• grid[i][j] is either 0 or 1.

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(m * n)
    # Space: O(m * n)
    def num_enclaves(self, grid: list[list[int]]) -> int:
        m, n = len(grid), len(grid[0])

        def drain(r: int, c: int) -> None:
            stack = [(r, c)]
            grid[r][c] = 0
            while stack:
                r, c = stack.pop()
                for dr, dc in ((1, 0), (-1, 0), (0, 1), (0, -1)):
                    nr, nc = r + dr, c + dc
                    if 0 <= nr < m and 0 <= nc < n and grid[nr][nc] == 1:
                        grid[nr][nc] = 0
                        stack.append((nr, nc))

        for r in range(m):
            for c in (0, n - 1):
                if grid[r][c] == 1:
                    drain(r, c)
        for c in range(n):
            for r in (0, m - 1):
                if grid[r][c] == 1:
                    drain(r, c)
        return sum(row.count(1) for row in grid)

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Complexity

Time	Space
O(m * n)	O(m * n)

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Tags

NeetCode All.