Added solution

Author: avidLearnerInProgress

find_all_anagrams_of_substring.cpp

@@ -0,0 +1,15 @@
+class Solution {
+public:
+    vector<int> findAnagrams(string s, string p) {
+        vector<int> goal(26), cur(26), res;
+        
+        for(char c : p) goal[c - 'a']++;
+        
+        for(int i = 0; i < s.size(); i++) {
+            cur[s[i] - 'a']++;
+            if(i >= p.size()) cur[s[i - p.size()] - 'a']--;
+            if(cur == goal) res.push_back(i - p.size() + 1);
+	    }
+	    return res;
+    }
+};