My favorite separation line marker: # ▼▼▼ and # ▲▲▲

TL;DR:

# ▼▼▼ START: Some_documentation ▼▼▼

# ▲▲▲ END: Some_documentation ▲▲▲ 

# Example: 

# solution to https://leetcode.com/problems/rotting-oranges

from collections import deque

from typing import List

class Solution:

    def orangesRotting(self, grid: List[List[int]]) -> int:

        q = deque()

        freshCnt = 0

        m, n = len(grid), len(grid[0])

       

        # Initial pass to find fresh/rotten oranges and check for isolated cells

        for i in range(m):

            for j in range(n):

                if grid[i][j] == 1:

                    freshCnt += 1

                    # ▼▼▼ START: NEW LOGIC FOR EARLY EXIT ▼▼▼

                    has_non_empty_neighbor = False

                    # Check all 4 potential neighbors

                    for di, dj in [(0, 1), (0, -1), (1, 0), (-1, 0)]:

                        ni, nj = i + di, j + dj

                        # First, ensure the neighbor is within the grid bounds

                        if 0 <= ni < m and 0 <= nj < n:

                            # If we find any neighbor that isn't an empty cell,

                            # this orange is NOT isolated.

                            if grid[ni][nj] != 0:

                                has_non_empty_neighbor = True

                                break # No need to check other neighbors

                   

                    # If after checking all valid neighbors, none were found,

                    # this orange is completely surrounded by empty cells.

                    if not has_non_empty_neighbor:

                        return -1 # Early exit

                    # ▲▲▲ END: NEW LOGIC FOR EARLY EXIT ▲▲▲

                elif grid[i][j] == 2:

                    q.append((i, j))

       

        time = 0

        # This BFS part of your logic is correct

        while q and freshCnt > 0:

       

            for _ in range(len(q)):

                i, j = q.popleft()

                for ni, nj in (i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1):

                    if 0 <= ni < m and 0 <= nj < n and grid[ni][nj] == 1:

                        q.append((ni, nj))

                        grid[ni][nj] = 2

                        freshCnt -= 1

            time += 1

       

        return time if freshCnt == 0 else -1