Added day16 solution

Author: avidLearnerInProgress

30-days-of-code-may/16.findAllAnagramInString.cpp

@@ -0,0 +1,36 @@
+class Solution {
+public:
+    vector<int> findAnagrams(string s, string p) {
+    
+        if(s.size() == 0 || p.size() > s.size()) return {};
+        vector<int> curr_window(26, 0), pattern(26, 0), anagrams_start;
+        
+        for(auto c : p)
+            pattern[c - 'a']++;
+        
+        /*
+        1) eliminate least recently used characters
+        2) length of sliding window should always be <= length of pattern to match. If it exceeds, we have to make the sliding window length = pattern length. For this purpose we use 1)
+        */
+        
+        /*
+        At i = 3, we eliminate char at i = 0 from s to make our sliding window have characters at s[1],s[2],s[3]
+        At i = 4, we eliminate char at i = 1 from s to make our sliding window have characters at s[2],s[3],s[4]
+        and so on..
+        */
+        
+        for(auto i=0; i<s.size(); i++){
+            curr_window[s[i] - 'a']++; //include current unseen character
+            
+            if(i >= p.size()){ //exclude least recently used characters
+                auto exclude = s[i - p.size()];
+                curr_window[exclude - 'a']--; 
+            }
+            
+            if(curr_window == pattern){ //found anagram, store its start index
+                anagrams_start.push_back(i - p.size() + 1); //+1 for 0 based indexing
+            }
+        }
+        return anagrams_start;
+    }
+};