README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1200-1299/1257.Smallest%20Common%20Region/README_EN.md	1654	Biweekly Contest 13 Q2	Tree Depth-First Search Breadth-First Search Array Hash Table String

1257. Smallest Common Region 🔒

中文文档

Description

You are given some lists of regions where the first region of each list includes all other regions in that list.

Naturally, if a region x contains another region y then x is bigger than y. Also, by definition, a region x contains itself.

Given two regions: region1 and region2, return the smallest region that contains both of them.

If you are given regions r1, r2, and r3 such that r1 includes r3, it is guaranteed there is no r2 such that r2 includes r3.

It is guaranteed the smallest region exists.

 

Example 1:

Input:
regions = [["Earth","North America","South America"],
["North America","United States","Canada"],
["United States","New York","Boston"],
["Canada","Ontario","Quebec"],
["South America","Brazil"]],
region1 = "Quebec",
region2 = "New York"
Output: "North America"

Example 2:

Input: regions = [["Earth", "North America", "South America"],["North America", "United States", "Canada"],["United States", "New York", "Boston"],["Canada", "Ontario", "Quebec"],["South America", "Brazil"]], region1 = "Canada", region2 = "South America"
Output: "Earth"

 

Constraints:

• 2 <= regions.length <= 104
• 2 <= regions[i].length <= 20
• 1 <= regions[i][j].length, region1.length, region2.length <= 20
• region1 != region2
• regions[i][j], region1, and region2 consist of English letters.

Solutions

Solution 1

Python3

class Solution:
    def findSmallestRegion(
        self, regions: List[List[str]], region1: str, region2: str
    ) -> str:
        m = {}
        for region in regions:
            for r in region[1:]:
                m[r] = region[0]
        s = set()
        while m.get(region1):
            s.add(region1)
            region1 = m[region1]
        while m.get(region2):
            if region2 in s:
                return region2
            region2 = m[region2]
        return region1

Java

class Solution {
    public String findSmallestRegion(List<List<String>> regions, String region1, String region2) {
        Map<String, String> m = new HashMap<>();
        for (List<String> region : regions) {
            for (int i = 1; i < region.size(); ++i) {
                m.put(region.get(i), region.get(0));
            }
        }
        Set<String> s = new HashSet<>();
        while (m.containsKey(region1)) {
            s.add(region1);
            region1 = m.get(region1);
        }
        while (m.containsKey(region2)) {
            if (s.contains(region2)) {
                return region2;
            }
            region2 = m.get(region2);
        }
        return region1;
    }
}

C++

class Solution {
public:
    string findSmallestRegion(vector<vector<string>>& regions, string region1, string region2) {
        unordered_map<string, string> m;
        for (auto& region : regions)
            for (int i = 1; i < region.size(); ++i)
                m[region[i]] = region[0];
        unordered_set<string> s;
        while (m.count(region1)) {
            s.insert(region1);
            region1 = m[region1];
        }
        while (m.count(region2)) {
            if (s.count(region2)) return region2;
            region2 = m[region2];
        }
        return region1;
    }
};

Go

func findSmallestRegion(regions [][]string, region1 string, region2 string) string {
	m := make(map[string]string)
	for _, region := range regions {
		for i := 1; i < len(region); i++ {
			m[region[i]] = region[0]
		}
	}
	s := make(map[string]bool)
	for region1 != "" {
		s[region1] = true
		region1 = m[region1]
	}
	for region2 != "" {
		if s[region2] {
			return region2
		}
		region2 = m[region2]
	}
	return region1
}