811. Subdomain Visit Count

A website domain "discuss.leetcode.com" consists of various subdomains. At the top level, we have "com", at the next level, we have "leetcode.com" and at the lowest level, "discuss.leetcode.com". When we visit a domain like "discuss.leetcode.com", we will also visit the parent domains "leetcode.com" and "com" implicitly.

A count-paired domain is a domain that has one of the two formats "rep d1.d2.d3" or "rep d1.d2" where rep is the number of visits to the domain and d1.d2.d3 is the domain itself.

• For example, "9001 discuss.leetcode.com" is a count-paired domain that indicates that discuss.leetcode.com was visited 9001 times.

Given an array of count-paired domains cpdomains, return an array of the count-paired domains of each subdomain in the input. You may return the answer in any order.

 

Example 1:

Input: cpdomains = ["9001 discuss.leetcode.com"]
Output: ["9001 leetcode.com","9001 discuss.leetcode.com","9001 com"]
Explanation: We only have one website domain: "discuss.leetcode.com".
As discussed above, the subdomain "leetcode.com" and "com" will also be visited. So they will all be visited 9001 times.

Example 2:

Input: cpdomains = ["900 google.mail.com", "50 yahoo.com", "1 intel.mail.com", "5 wiki.org"]
Output: ["901 mail.com","50 yahoo.com","900 google.mail.com","5 wiki.org","5 org","1 intel.mail.com","951 com"]
Explanation: We will visit "google.mail.com" 900 times, "yahoo.com" 50 times, "intel.mail.com" once and "wiki.org" 5 times.
For the subdomains, we will visit "mail.com" 900 + 1 = 901 times, "com" 900 + 50 + 1 = 951 times, and "org" 5 times.

 

Constraints:

• 1 <= cpdomain.length <= 100
• 1 <= cpdomain[i].length <= 100
• cpdomain[i] follows either the "repi d1i.d2i.d3i" format or the "repi d1i.d2i" format.
• repi is an integer in the range [1, 104].
• d1i, d2i, and d3i consist of lowercase English letters.

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class Solution:
    def subdomainVisits(self, cpdomains: List[str]) -> List[str]:
        dc={}
        
        while cpdomains:
            s=cpdomains.pop().split(" ")
            count=int(s[0])
            d=s[1].split(".")
            
            for i in range(len(d)):
                sd=".".join(d[i:])
                if sd not in dc:
                    dc[sd]=count
                else:
                    dc[sd]+=count
        
        ans=[]
        for k,v in dc.items():
            ans.append(str(v) + " " + k)
        return ans