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Formatted question description: https://leetcode.ca/all/811.html
811. Subdomain Visit Count (Easy)
A website domain like "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.
Now, call a "count-paired domain" to be a count (representing the number of visits this domain received), followed by a space, followed by the address. An example of a count-paired domain might be "9001 discuss.leetcode.com".
We are given a list cpdomains
of count-paired domains. We would like a list of count-paired domains, (in the same format as the input, and in any order), that explicitly counts the number of visits to each subdomain.
Example 1: Input: ["9001 discuss.leetcode.com"] Output: ["9001 discuss.leetcode.com", "9001 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: ["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.
Notes:
- The length of
cpdomains
will not exceed100
. - The length of each domain name will not exceed
100
. - Each address will have either 1 or 2 "." characters.
- The input count in any count-paired domain will not exceed
10000
. - The answer output can be returned in any order.
Solution
Use a map to store each subdomain and the number of times the subdomain is visited. For each count-paired domains in cpdomains
, split the count and the domain, and add the count to each subdomain’s count. Finally, loop over the map to get each subdomain’s number of visits.
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// official solution: https://leetcode.com/problems/subdomain-visit-count/solution/ class Solution { public List<String> subdomainVisits(String[] cpdomains) { Map<String, Integer> counts = new HashMap<>(); for (String domain: cpdomains) { String[] cpinfo = domain.split("\\s+"); String[] frags = cpinfo[1].split("\\."); int count = Integer.valueOf(cpinfo[0]); String cur = ""; for (int i = frags.length - 1; i >= 0; --i) { cur = frags[i] + (i < frags.length - 1 ? "." : "") + cur; counts.put(cur, counts.getOrDefault(cur, 0) + count); } } List<String> ans = new ArrayList<>(); for (String dom: counts.keySet()) ans.add("" + counts.get(dom) + " " + dom); return ans; } } class Solution { public List<String> subdomainVisits(String[] cpdomains) { Map<String, Integer> domainCountMap = new HashMap<String, Integer>(); for (String cpdomain : cpdomains) { String[] cpdomainArray = cpdomain.split(" "); int count = Integer.parseInt(cpdomainArray[0]); String domain = cpdomainArray[1]; while (domain.length() > 0) { int prevCount = domainCountMap.getOrDefault(domain, 0); int newCount = prevCount + count; domainCountMap.put(domain, newCount); if (domain.indexOf('.') < 0) break; domain = domain.substring(domain.indexOf('.') + 1); } } List<String> subdomainVisits = new ArrayList<String>(); Set<String> domainsSet = domainCountMap.keySet(); for (String domain : domainsSet) { int count = domainCountMap.getOrDefault(domain, 0); String subdomainVisit = count + " " + domain; subdomainVisits.add(subdomainVisit); } return subdomainVisits; } } ############ class Solution { public List<String> subdomainVisits(String[] cpdomains) { Map<String, Integer> cnt = new HashMap<>(); for (String s : cpdomains) { int i = s.indexOf(" "); int v = Integer.parseInt(s.substring(0, i)); for (; i < s.length(); ++i) { if (s.charAt(i) == ' ' || s.charAt(i) == '.') { String t = s.substring(i + 1); cnt.put(t, cnt.getOrDefault(t, 0) + v); } } } List<String> ans = new ArrayList<>(); for (var e : cnt.entrySet()) { ans.add(e.getValue() + " " + e.getKey()); } return ans; } }
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// OJ: https://leetcode.com/problems/subdomain-visit-count/ // Time: O(N*S^2) where N is count of cpdomains, S is max length of cpdomain string. // Space: O(N) class Solution { public: vector<string> subdomainVisits(vector<string>& cpdomains) { unordered_map<string, int> m; for (auto cpdomain : cpdomains) { int cnt = stoi(cpdomain), i = cpdomain.find_first_of(" "); string domain = cpdomain.substr(i + 1); while (true) { m[domain] += cnt; i = domain.find_first_of("."); if (i == string::npos) break; domain = domain.substr(i + 1); } } vector<string> v; for (auto p : m) { v.push_back(to_string(p.second) + " " + p.first); } return v; } };
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class Solution: def subdomainVisits(self, cpdomains: List[str]) -> List[str]: cnt = Counter() for s in cpdomains: v = int(s[: s.index(' ')]) for i, c in enumerate(s): if c in ' .': # space ' ' is for the full domain, and '.' for subdomain cnt[s[i + 1 :]] += v return [f'{v} {s}' for s, v in cnt.items()] ############ class Solution(object): def subdomainVisits(self, cpdomains): """ :type cpdomains: List[str] :rtype: List[str] """ domain_counts = collections.defaultdict(int) for cpdomain in cpdomains: times, domains = cpdomain.split() times = int(times) domain_counts[domains] += times while '.' in domains: domains = domains[domains.index('.') + 1:] domain_counts[domains] += times return [str(v) + ' ' + d for d, v in domain_counts.items()]
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func subdomainVisits(cpdomains []string) []string { cnt := map[string]int{} for _, s := range cpdomains { i := strings.IndexByte(s, ' ') v, _ := strconv.Atoi(s[:i]) for ; i < len(s); i++ { if s[i] == ' ' || s[i] == '.' { cnt[s[i+1:]] += v } } } ans := make([]string, 0, len(cnt)) for s, v := range cnt { ans = append(ans, strconv.Itoa(v)+" "+s) } return ans }