1257. Smallest Common Region

Problem Description

You are provided with a list that represents a hierarchy of regions where each list's first element is a region that includes all the following elements in the same list. Every region is considered to contain itself and any region it's listed with, where the first region is the largest and it is followed by regions it includes in descending order of size.

Now, you are given two specific regions, region1 and region2, and your task is to find the smallest common region that contains both of these regions.

Imagine a scenario where you have regions defined in a hierarchical manner, like in a geographical context—continents include countries, countries include states, states include cities, and so on. If given two cities, you must find the smallest state or country that includes both cities.

Note that for any region r3 included in r1, there will be no other r2 that includes r3. This simplifies the problem, ensuring that the tree of regions is well-defined and does not have any overlapping or conflicting ownership. The problem guarantees that there is always a smallest common region that exists for any two given regions.

Intuition

To arrive at the solution, we need to identify the hierarchy or the path of parents from each region to the top-most region in the hierarchy. Once we have the path for both regions, we can traverse these paths to find the smallest common region, which will be the first common ancestor.

We start by creating a map m that holds the parent-child relationship, where every region except the very first has a corresponding parent region that includes it—this constructs a map that resembles a tree.

Consider this tree as a family tree, where every person apart from the family's progenitor has a parent. region1 and region2 then become two family members, and we are looking for their closest common ancestor.

We populate a set s with ancestors of region1 by following the parent links up the hierarchy. Then, we start at region2 and move up its ancestral line until we encounter the first region already present in the set s, which will be the least common ancestor. If region2 or any of its ancestors do not appear in the set s, then region1 itself is the smallest region (this can only happen if region1 is the top-most region).

The use of the set ensures that we can efficiently check whether the region is a common ancestor, and iterating up the ancestral lines ensures we find the smallest such region containing both given regions.

Learn more about Tree, Depth-First Search and Breadth-First Search patterns.

Solution Approach

The solution follows a direct approach using a map and a set, which are simple yet powerful data structures in Python. Here's a step-by-step breakdown of the implementation based on the provided code:

1. Create a map m to store the parent-child relationships. After processing all the regions' lists, m will contain a mapping where the key is a 'child' region, and the value is its immediate 'parent' region. This is achieved with a nested loop: the outer loop iterates over each list of regions, and the inner loop (starting from the second element) maps each region to its 'parent', which is the first element in the current list.

   1m = {}
   2for region in regions:
   3    for r in region[1:]:
   4        m[r] = region[0]

2. Instantiate a set s, which will hold all the ancestors of region1. We then traverse up from region1 to the top-most ancestor (which will be a region with no parent in map m). As we traverse, we populate the set with each ancestor, creating a pathway from region1 to its top-most ancestor.

   1s = set()
   2while m.get(region1):
   3    s.add(region1)
   4    region1 = m[region1]

3. Starting with region2, the code checks if region2 or any of its ancestors are in the set s, which would indicate a common ancestor has been found.

   1while m.get(region2):
   2    if region2 in s:
   3        return region2
   4    region2 = m[region2]

4. If region2 itself or one of its ancestors was not in set s, then the loop stops once region2 becomes a region without a parent, which, as mentioned before, can only be the top-most ancestor (the very first region). In such a case, region1 had to be the top-most ancestor, and it's returned as the smallest region that contains both region1 and region2.

The implementation uses hashmap and hashset data structures, as map m and set s respectively, which facilitate the efficient lookup in constant time, O(1), for checking the presence of ancestors and mapping children to parents. This is how the Python dictionary and set are generally implemented.

This approach is efficient in terms of time complexity because it has to process each region only once when creating the map, which is O(N) where N is the total number of regions, and then the ancestral path for both regions, each being at most O(H) where H is the height of the hierarchy. Therefore, the overall complexity is O(N + H). The space complexity is O(N) for storing the map and the set of ancestors for region1.

Example Walkthrough

Let's illustrate the solution approach with an example. Suppose we have five regions arranged in the following hierarchy:

1[
2 ["Earth", "North America", "United States"],
3 ["United States", "California"],
4 ["United States", "New York"],
5 ["California", "Los Angeles"],
6 ["New York", "New York City"]
7]

And we are given two regions to find the smallest common region for: region1 = "Los Angeles" and region2 = "New York City".

1. First, we create the map m listing each child's parent. It looks like this after processing the lists:

   1m = {
   2    "North America": "Earth",
   3    "United States": "North America",
   4    "California": "United States",
   5    "New York": "United States",
   6    "Los Angeles": "California",
   7    "New York City": "New York"
   8}

2. Next, we add all ancestors of region1 ("Los Angeles") into the set s. We follow up the hierarchy from "Los Angeles" to "California", and then to "United States", which is an ancestor to both "California" and "New York":

   1s = set(["Los Angeles", "California", "United States"])

3. After this, we check if region2 ("New York City") or any of its ancestors are in set s. We start with "New York City", whose parent is "New York", and then the parent of "New York" is "United States" which is in s. Thus, we find that "United States" is the smallest common region for "Los Angeles" and "New York City":

   1# "New York City" is not in s
   2# "New York"'s parent is "United States" which is in s
   3smallest_common_region = "United States"

4. Now we have the result: the smallest common region containing both "Los Angeles" and "New York City" is "United States".

The example confirms our approach, showing how it finds the least common ancestor efficiently by constructing a map to represent the hierarchy and then using a set to track the ancestors of one region to compare with the ancestral line of the other region.

Solution Implementation

Time and Space Complexity

The given Python code defines a function to find the smallest common region between region1 and region2.

• Time complexity:

The time complexity of the first loop is O(N) where N is the total number of regions inside all the lists combined. This is because each region list is traversed exactly once, and each traversal takes O(K), where K is the number of sub-regions within the current region. Hence, the overall time for this loop is proportional to the total number of regions in all lists.

For the second and third while loops, the time complexity is O(H1 + H2), where H1 is the height of the region hierarchy (region tree) from region1 to the top, and H2 is the height from region2 to the top. In the worst-case scenario, this would be O(H) where H is the height of the tree.

Combining the two parts, the overall time complexity of the function can be expressed as O(N + H), assuming H can be considered smaller than N in a large dataset.

• Space complexity:

The space complexity consists of the additional space used by the map m and set s. The map m has O(N) complexity as it contains all regions except the global root. The set s has O(H1) complexity in the worst case as it stores the path from region1 to the root.

Hence, the total space complexity of the function is O(N + H1). Since H1 is generally smaller compared to N, we can approximate the space complexity as O(N).

Learn more about how to find time and space complexity quickly using problem constraints.