Added C++ Anagram Detection (#119)

Author: aniliitb10

problems/anagram-detection/anagram-detection.cpp

@@ -0,0 +1,106 @@
+#include <iostream>
+#include <unordered_map>
+
+// map of all upper case and lower case chars with prime numbers
+const static std::unordered_map<char, int> hash{
+    {'a' , 2},
+    {'b' , 3},
+    {'c' , 5},
+    {'d' , 7},
+    {'e' , 11},
+    {'f' , 13},
+    {'g' , 17},
+    {'h' , 19},
+    {'i' , 23},
+    {'j' , 29},
+    {'k' , 31},
+    {'l' , 37},
+    {'m' , 41},
+    {'n' , 43},
+    {'o' , 47},
+    {'p' , 53},
+    {'q' , 59},
+    {'r' , 61},
+    {'s' , 67},
+    {'t' , 71},
+    {'u' , 73},
+    {'v' , 79},
+    {'w' , 83},
+    {'x' , 89},
+    {'y' , 97},
+    {'z' , 101},
+    {'A' , 103},
+    {'B' , 107},
+    {'C' , 109},
+    {'D' , 113},
+    {'E' , 127},
+    {'F' , 131},
+    {'G' , 137},
+    {'H' , 139},
+    {'I' , 149},
+    {'J' , 151},
+    {'K' , 163},
+    {'L' , 167},
+    {'M' , 173},
+    {'N' , 179},
+    {'O' , 181},
+    {'P' , 191},
+    {'Q' , 193},
+    {'R' , 197},
+    {'S' , 199},
+    {'T' , 211},
+    {'U' , 223},
+    {'V' , 227},
+    {'W' , 229},
+    {'X' , 233},
+    {'Y' , 239},
+    {'Z' , 241 }
+};
+
+// returns the hash of string which is unique for the combination
+// of characters in that string.
+// By unique factorization theorem there won't be any hash collisions
+long long get_hash(const std::string& str) {
+  long long hash_val = 1;
+  for (char c : str) {
+    hash_val *= hash.at(c);
+  }
+
+  return hash_val;
+}
+
+// This funtion calculates the hash of string
+long long anagram_count(const std::string& parent, const std::string& child) {
+  long long hash_child = get_hash(child);
+
+  // to keep count of all anagrams
+  long long count = 0;
+
+  // get the hash of first n characters where n is length of child
+  long long hash_val = get_hash(parent.substr(0, child.size()));
+  if (hash_val == hash_child) {
+    count++;
+  }
+
+  // As we are moving ahead one char at a time, to calculate the hash_val of next string
+  // we can divide the last hash_val by the hash of char which is moving out
+  // and multiply by the hash of char being moving in
+  // e.g. in abcd, if hash_val of "abc" is n, then hash_val of "bcd" is: n / hash.at('a') * hash.at('d');
+  for (size_t index = child.size(); index < parent.size(); ++index) {
+    hash_val /= hash.at(parent[index - child.size()]);
+    hash_val *= hash.at(parent[index]);
+    if (hash_val == hash_child) {
+      count++;
+    }
+  }
+
+  return count;
+}
+
+int main() {
+  std::cout << anagram_count("forbtorfxdofrm", "for") << std::endl; // 3
+  std::cout << anagram_count("aabaabaa", "aaba") << std::endl; // 4
+  std::cout << anagram_count("AdnBndAndBdaBn", "dAn") << std::endl; // 4
+  std::cout << anagram_count("AbrAcadAbRa", "cAda") << std::endl; // 2
+  return 0;
+}