Solution Explanation

This problem asks to find the smallest common region that contains two given regions. The input is a list of lists of strings, where each inner list represents a region and its subregions. The solution uses a graph-like approach to efficiently find the common ancestor.

Approach:

1. Construct a Parent Map: Create a map (dictionary in Python) where keys are subregions and values are their parent regions. This map is built by iterating through the input regions list. For each inner list, the first element is the parent region, and the rest are its children.

2. Trace Ancestors for Region 1: Starting from region1, traverse upwards in the parent map until a region with no parent (the top-level region or root) is encountered. Store all visited regions in a set (s).

3. Trace Ancestors for Region 2: Starting from region2, traverse upwards in the parent map. At each step, check if the current region is present in the set s. If it is, that region is the smallest common region, and the algorithm terminates.

4. Return the Smallest Common Region: If the loop for region2 completes without finding a common region in s, it means that the topmost ancestor of region1 is the smallest common region.

Time Complexity Analysis:

• Constructing the parent map: This takes O(R), where R is the total number of regions across all the input lists. This is because we iterate through each region and its subregions once.
• Tracing ancestors: In the worst case, we might need to traverse all ancestors of both region1 and region2. This depends on the depth of the region hierarchy (tree-like structure). In a balanced tree, this would be O(log R), but in a skewed tree, it could be O(R) in the worst-case scenario. However, considering that the problem statement guarantees the existence of a smallest common region, the overall traversal time is less likely to reach O(R) in practice. Therefore, we can reasonably consider the traversal time as O(D), where D is the maximum depth of the region hierarchy.

Therefore, the overall time complexity is dominated by building the map and traversing the ancestors: O(R + D). In most practical scenarios, D will be significantly smaller than R, making the complexity closer to O(R).

Space Complexity Analysis:

The space complexity is dominated by:

• The parent map: This stores at most R entries (one for each subregion), so it's O(R).
• The set s: This stores at most D entries (the ancestors of region1), so it's O(D).

Therefore, the overall space complexity is O(R + D), which is also approximately O(R) in most cases.

The provided code in Python, Java, C++, and Go implements this approach efficiently. Each language's implementation maintains the same core logic but adjusts syntax and data structures according to the language's specifics.