Algorithm
Problem Name: 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 thatdiscuss.leetcode.comwas visited9001times.
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 <= 1001 <= cpdomain[i].length <= 100cpdomain[i]follows either the"repi d1i.d2i.d3i"format or the"repi d1i.d2i"format.repiis an integer in the range[1, 104].d1i,d2i, andd3iconsist of lowercase English letters.
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public List subdomainVisits(String[] cpdomains) {
Map map = new HashMap<>();
for (String cpdomain : cpdomains) {
int spaceIdx = cpdomain.indexOf(' ');
int count = Integer.parseInt(cpdomain.substring(0, spaceIdx));
String[] subdomains = cpdomain.substring(spaceIdx + 1).split("\\.");
StringBuilder sb = new StringBuilder();
for (int i = subdomains.length - 1; i >= 0; i--) {
sb.insert(0, subdomains[i]);
String currDomain = sb.toString();
map.put(currDomain, map.getOrDefault(currDomain, 0L) + count);
sb.insert(0, ".");
}
}
List < String> result = new ArrayList<>();
for (String key : map.keySet()) {
result.add(map.get(key) + " " + key);
}
return result;
}
}
Input
Output
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def subdomainVisits(self, cpdomains):
counter = collections.Counter()
for cpdomain in cpdomains:
count, *domains = cpdomain.replace(" ",".").split(".")
for i in range(len(domains)):
counter[".".join(domains[i:])] += int(count)
return [" ".join((str(v), k)) for k, v in counter.items()]
Input
Output
#3 Code Example with C# Programming
Code -
C# Programming
using System.Collections.Generic;
namespace LeetCode
{
public class _0811_SubdomainVisitCount
{
public IList < string> SubdomainVisits(string[] cpdomains)
{
var counts = new Dictionary();
foreach (var domain in cpdomains)
{
var cpinfo = domain.Split(' ');
int count = int.Parse(cpinfo[0]);
var frags = cpinfo[1].Split('.');
var currentDomain = "";
for (int i = frags.Length - 1; i >= 0; i--)
{
currentDomain = frags[i] + (i < frags.Length - 1 ? "." : "") + currentDomain;
if (counts.TryGetValue(currentDomain, out var existCount))
counts[currentDomain] = existCount + count;
else
counts[currentDomain] = count;
}
}
var answer = new List < string>();
foreach (var domain in counts.Keys)
answer.Add($"{counts[domain]} {domain}");
return answer;
}
}
}
Input
Output