Asked in: JPMORGAN
Image of the Question
All Testcases Passed ✔
Solution
def countConnections(matrix):
visited = set()
m,n = len(matrix), len(matrix[0])
for i in range(m):
// ... rest of solution available after purchaseExplanation
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To solve this problem, it is essential to first understand what a "connection" means in the context of a binary matrix. A matrix cell with value 1 represents a valid position where a connection can exist, while a cell with value 0 blocks any connection. The goal is to count how many valid connections can be established between adjacent positions that are both 1s.
What defines adjacency here is key. The problem specifies that we can connect positions horizontally, vertically, and diagonally. That means each cell potentially connects to up to 8 neighbors:
- Up
- Down
- Left
- Right
- Top-left
- Top-right
- Bottom-left
- Bottom-right
For every cell that contains the value 1, we need to check each of these eight directions. If any neighbor in these directions also contains the value 1, we can establish a connection between the two.
However, you must avoid double-counting. Since a connection from cell A to cell B is the same as from B to A, it's important to have a strategy to ensure that each pair of connected cells is only considered once. One straightforward way to handle this is to only check a subset of the directions for each cell. For instance, you can choose to check only the right, bottom-right, bottom, and bottom-left directions. This ensures that every pair of connected 1s is only checked once, since the symmetric directions will be visited from their origin cells in the matrix iteration.
The next part of the strategy is to iterate over every cell in the matrix. For each cell that contains a 1, you will check its possible connections in the chosen subset of directions. For every valid neighboring cell that also contains a 1, you increment a counter that tracks the number of successful connections.
While doing this, ensure you stay within the bounds of the matrix. For example, if you're at the rightmost column, you should not attempt to check the right or diagonal-right directions. Similarly, if you're at the bottom row, do not check any downward directions.
This approach ensures that each connection is identified once, and the process runs in linear time relative to the size of the matrix. Since the maximum number of cells is less than 10,000 (given the constraints of m and n being less than 100), this method is computationally efficient.
To outline your steps clearly:
1. Iterate through each cell in the matrix.
2. For each cell that contains 1:
a. Check the four directions that avoid double-counting (right, bottom-right, bottom, bottom-left).
b. For each of these directions, check if the neighbor is within matrix bounds and also contains 1.
c. If both conditions are true, increment your count of connections.
3. Return the final count after processing all the cells.
This approach is systematic, avoids unnecessary checks, and prevents overcounting by considering only unique directional relationships. It is important not to approach this as a graph traversal or a connected components problem since we are not grouping or exploring full regions. Instead, the task is about counting all valid pairwise connections between directly adjacent 1s.
This type of matrix-based adjacency check is commonly used in image processing, cellular automata, and games like Minesweeper, where each cell’s local neighborhood determines interactions. Understanding how to efficiently iterate and check neighbors in fixed directions is a key pattern that is widely applicable in such problems.
To conclude, the problem is best approached with a nested loop over the matrix and careful directional checks limited to a subset that prevents duplicate evaluations. It's a clean, effective, and highly performant method for counting distinct adjacent pairs based on value conditions.
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