Find All Anagrams in a String in C++

C++Server Side ProgrammingProgramming

Suppose we have a string s and a non-empty string p, we have to find all the start indices of p's anagrams in s. The strings consist of lowercase letters only and the length of both strings s and p will not be larger than 20 and 100. So for example, if s: "cbaebabacd" p: "abc", then the output will be [0, 6], at index 0, it is “cba”, and another is “bac”, these are the anagrams of “abc”.

To solve this, we will follow these steps −

  • Define a map m, n := size of s, set left := 0, right := 0, counter := size of p

  • define an array ans

  • store the frequency of characters in p into the map m

  • for right := 0 to n – 1

    • if m has s[right] and m[s[right]] is non zero, then decrease m[s[right]] by 1, decrease counter by 1 and if counter = 0, then insert left into ans

    • otherwise

      • while left < right,

        • if s[left] is not present in m, then increase counter by 1, and increase m[s[left]] by 1

        • increase left by 1

        • if m has s[right] and m[s[right]] is non zero, then decrease right by 1, and come out from the loop

      • if m has no s[left], then set left := right + 1

  • return ans

Example

Let us see the following implementation to get better understanding −

class Solution {
   public:
   vector<int> findAnagrams(string s, string p) {
      map <char, int> m;
      int n = s.size();
      int left = 0, right = 0;
      int counter = p.size();
      vector <int> ans;
      for(int i = 0; i < p.size(); i++) m[p[i]]++;
      for(int right = 0; right < n; right++){
         //cout << left << " " << right << endl;
         if(m.find(s[right]) != m.end() && m[s[right]]){
            m[s[right]]--;
            counter--;
            //cout << "Counter : " << counter << endl;
            if(counter == 0)ans.push_back(left);
         } else {
            while(left<right){
               if(m.find(s[left]) != m.end()) {
                  //if(m[s[left]] == 0)
                     counter++;
                  m[s[left]]++;
               }
               left++;
               if(m.find(s[right]) != m.end() && m[s[right]]){
                  right--;
                  break;
               }
            }
            if(m.find(s[left])==m.end())left = right + 1;
         }
         //printMap(m);
      }
      //if(counter == 0)ans.push_back(left);
      return ans;
   }
};

Input

"cbaebabacd"
"abc"

Output

[null,true,false,true,1,true,false,2]
raja
Published on 17-Mar-2020 06:17:12
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