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# Subdomain Visit Count LeetCode Solution

## Problem – 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.

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.

### Subdomain Visit Count LeetCode Solution in C++

``````    vector<string> subdomainVisits(vector<string>& cpdomains) {
unordered_map<string, int> count;
for (auto cd : cpdomains) {
int i = cd.find(" ");
int n = stoi(cd.substr (0, i));
string s = cd.substr (i + 1);
for (int i = 0; i < s.size(); ++i)
if (s[i] == '.')
count[s.substr(i + 1)] += n;
count[s] += n;
}
vector<string> res;
for (auto k : count)
res.push_back (to_string(k.second) + " " + k.first);
return res;
}``````

### Subdomain Visit Count LeetCode Solution in Java

``````    public List<String> subdomainVisits(String[] cpdomains) {
Map<String, Integer> count = new HashMap();
for (String cd : cpdomains) {
int i = cd.indexOf(' ');
int n = Integer.valueOf(cd.substring(0, i));
String s = cd.substring(i + 1);
for (i = 0; i < s.length(); ++i) {
if (s.charAt(i) == '.') {
String d = s.substring(i + 1);
count.put(d, count.getOrDefault(d, 0) + n);
}
}
count.put(s, count.getOrDefault(s, 0) + n);
}

List<String> res = new ArrayList();
for (String d : count.keySet()) res.add(count.get(d) + " " + d);
return res;
}``````

### Subdomain Visit Count LeetCode Solution in Python

``````    def subdomainVisits(self, cpdomains):
count = collections.Counter()
for cd in cpdomains:
n, s = cd.split()
count[s] += int(n)
for i in range(len(s)):
if s[i] == '.':
count[s[i + 1:]] += int(n)
return ["%d %s" % (count[k], k) for k in count]``````
##### Subdomain Visit Count LeetCode Solution Review:

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