Merge Two Sorted grids

Question

Given two sorted grids, merge them into a single sorted result.

Codemia · Accepted Answer

The problem involves merging two sorted grids into a single sorted result. We can recognize this as a variant of the merge operation commonly seen in merge sort. The grids are already sorted, which allows us to utilize the two-pointer technique effectively. Each grid can be treated as a flattened list of elements, and we can maintain two pointers to traverse each grid simultaneously, comparing elements and adding the smaller one to the merged result. This approach is optimal because we avoid extra sorting, leveraging the pre-sorted nature of the input.

1. Initialize two pointers, one for each grid (let's call them `i` for grid1 and `j` for grid2).  
2. Create an empty list called `merged` to hold the merged result.  
3. While both pointers are within the bounds of their respective grids, compare the current elements pointed by `i` and `j`.  
   - If grid1[i] is less than or equal to grid2[j], append grid1[i] to `merged` and increment `i`.  
   - Otherwise, append grid2[j] to `merged` and increment `j`.  
4. Once we have traversed one of the grids completely, append any remaining elements from the other grid to `merged`.  
5. Return the `merged` list.  
**Example:** Given grid1 = [[1, 3], [5]], grid2 = [[2, 4], [6]], the merged result would be [1, 2, 3, 4, 5, 6].

```python
def merge_sorted_grids(grid1, grid2):
    i, j = 0, 0
    merged = []
    # Flatten the grids
    flat_grid1 = [item for sublist in grid1 for item in sublist]
    flat_grid2 = [item for sublist in grid2 for item in sublist]
    # Merge process
    while i < len(flat_grid1) and j < len(flat_grid2):
        if flat_grid1[i] <= flat_grid2[j]:
            merged.append(flat_grid1[i])
            i += 1
        else:
            merged.append(flat_grid2[j])
            j += 1
    # Add remaining elements
    while i < len(flat_grid1):
        merged.append(flat_grid1[i])
        i += 1
    while j < len(flat_grid2):
        merged.append(flat_grid2[j])
        j += 1
    return merged
```

The time complexity of this solution is O(n + m), where n is the total number of elements in grid1 and m is the total number of elements in grid2. This is because we are essentially iterating through both grids once during the merge process. The space complexity is O(n + m) as well, since we are storing the flattened grids and the merged result. Given that the grids can vary in size, this approach efficiently handles merging without requiring additional sorting.

1. **One or both grids are empty:** If grid1 is empty, the merged result should simply be the flattened version of grid2, and vice versa. If both are empty, the result should be an empty list.  
2. **All elements are the same:** If both grids have the same repeated elements, the merged list should maintain the same element order.  
3. **Different dimensions of grids:** The grids can have varying row sizes; the algorithm should handle this by flattening them completely before merging.  
4. **Large grids:** The solution should efficiently handle large input sizes without performance degradation, given the linear time complexity.

Merge Two Sorted grids

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