Merge pull request #1854 from aditya-bhaumik/811

added solution for leetcode problem 811

Author: ajay-dhangar

dsa-solutions/lc-solutions/0800- 0899/0811 - Subdomain-Visit-Count.md

@@ -0,0 +1,93 @@
+---
+id: subdomain-visit-count
+title: Subdomain Visit Count
+sidebar_label: 811- Subdomain Visit Count
+tags:
+  - Array
+  - String
+  - Hash Table
+description: Given an array of count-paired domains, return an array showing the total visit counts for each subdomain and its parent domains.
+sidebar_position: 0811
+---
+
+## Problem Description
+
+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.`
+
+
+### Constraints
+
+- `1 <= cpdomain.length <= 100`
+- `1 <= cpdomain[i].length <= 100`
+
+## Approach
+
+Just store each & every valid substring in a HashMap with its frequency count.
+Update the frequency count whenever the same string is encounterd.
+Finally, create a new string using each key-value pair of the HashMap & add that new string to the final list.
+
+### Solution Code
+
+#### Java
+```Java
+class Solution {
+    HashMap<String, Integer> hm= new HashMap<>();
+    
+    public List<String> subdomainVisits(String[] cpdomains) {
+        List<String> result= new ArrayList<>();
+		
+        for(String s: cpdomains){
+            addToList(s);
+        }
+        
+        for(String s: hm.keySet()){
+            StringBuilder sb= new StringBuilder( Integer.toString(hm.get(s)) );
+            sb.append(" ");
+            sb.append( s );
+            result.add( sb.toString() );
+        }
+        return result;
+    }
+    
+    public void addToList(String s){
+        String[] split1= s.split(" ");
+        int n= Integer.parseInt(split1[0]);
+        String[] split2= split1[1].split("\\.");
+        int l= split2.length;
+        
+        for(int i=0; i<l; i++){
+            StringBuilder sb= new StringBuilder();
+            int index= i;
+            
+            while(index < l){
+                sb.append( split2[index] );
+				
+                if(index != l-1)
+                    sb.append( "." );
+                index++;
+            }
+            String str= sb.toString();
+            hm.put( str, hm.getOrDefault( str, 0 )+n);
+        }
+    }
+}
+        
+```
+
+- Time Complexity
+The time complexity is O(N⋅L2).
+
+- Space Complexity
+The space complexity is O(1).