Add One Source to Minimize Grid Inconvenience

Question

The question asks the candidate to determine the optimal position for an additional dye source in a 2D grid to minimize the maximum inconvenience across all grid cells.

Codemia · Accepted Answer

To solve this problem, we need to evaluate the inconvenience of each cell in the grid based on the distance to the nearest dye source. The main goal is to find a cell where placing an additional dye source minimizes the maximum inconvenience across the grid. We will perform a breadth-first search (BFS) from all existing dye sources to calculate the initial inconvenience for each cell.

1. **Initialize the grid**: Create a 2D array to store the inconvenience values for each cell. 
2. **Perform BFS**: Start from all existing dye sources and propagate the distances to all reachable cells. 
3. **Evaluate each cell**: For each cell in the grid, calculate the maximum inconvenience if a dye source is placed there. 
4. **Find the minimum maximum inconvenience**: Keep track of the minimum value found during the evaluation.

The time complexity of the BFS is O(n * m) for traversing the grid, where n is the number of rows and m is the number of columns. Evaluating each cell to find the maximum inconvenience also takes O(n * m). Therefore, the overall complexity is O(n * m) for the BFS and O(n * m) for the evaluation, leading to a total complexity of O(n * m).

```python
from collections import deque

def min_inconvenience(grid):
    n, m = len(grid), len(grid[0])
    distances = [[float('inf')] * m for _ in range(n)]
    queue = deque()

# Initialize the queue with existing dye sources
    for i in range(n):
        for j in range(m):
            if grid[i][j] == 1:  # Assuming 1 indicates a dye source
                queue.append((i, j))
                distances[i][j] = 0

directions = [(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 1), (1, -1), (1, 0), (1, 1)]

# BFS to calculate distances
    while queue:
        x, y = queue.popleft()
        for dx, dy in directions:
            nx, ny = x + dx, y + dy
            if 0 <= nx < n and 0 <= ny < m:
                if distances[nx][ny] > distances[x][y] + 1:
                    distances[nx][ny] = distances[x][y] + 1
                    queue.append((nx, ny))

# Evaluate the maximum inconvenience for each cell
    min_max_inconvenience = float('inf')
    for i in range(n):
        for j in range(m):
            if grid[i][j] == 0:  # Only consider empty cells
                max_inconvenience = 0
                for x in range(n):
                    for y in range(m):
                        max_inconvenience = max(max_inconvenience, distances[x][y])
                min_max_inconvenience = min(min_max_inconvenience, max_inconvenience)

return min_max_inconvenience
```

Consider scenarios where all cells are already dyed or where the grid is very small. The algorithm should handle these cases gracefully, returning appropriate values without errors.

Add One Source to Minimize Grid Inconvenience

Amazon

What the Interviewer Expects

Key Topics to Cover

How to Approach This

Possible Follow-up Questions

Sharpen Your Skills on Codemia

Sample Answer

Approach

Algorithm

Submit Your Answer

Amazon Software Engineer Interview Guide

Related Questions

Binary Search on graph

Count kth largest element in array

Find connected components in string

Merge Three Sorted strings

Binary Search on string