Maximum Number of Occurrences of a Substring in C++

Suppose we have a string s, we have to find the maximum number of occurrences of any substring that satisfies the following rules −

The number of distinct characters in the substring must be less than or equal to maxLetters.
The substring size must be in range minSize and maxSize inclusive.

So if the input is like − “aababcaab”, maxLetters = 2, minSize = 3 and maxSize = 4, then the output will be 2. The substring "aab" has 2 occurrences in the original string. This satisfies the conditions, 2 unique letters and size 3 (between minSize and maxSize).

To solve this, we will follow these steps −

Define a map m
for sz in range minSize to maxSize
- make a map x, create a temp, initially empty
- for i in range 0 to sz – 1
  - increase x[s[i]] by 1
  - temp := temp + s[i]
- for j is 0, i in range sz to size of s – 1, increase i and j by 1
  - if size of x <= maxLetters, then increase m[temp] by 1
  - decrease x[temp[0]] by 1
  - if x[temp[0]] is 0, then delete temp[0] from x
  - delete first and second element of temp from the temp itself
  - increase x[s[i]] by 1
  - temp := temp + s[i]
- if size of x <= maxLetters, then increase m[temp] by 1
ans := 0
while map m has some elements, then
- ans := max of ans and value of ith key-value pair
return ans

Example(C++)

Let us see the following implementation to get a better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   int maxFreq(string s, int maxLetters, int minSize, int maxSize) {
      unordered_map <string ,int > m;
      for(int sz = minSize; sz <= minSize; sz++){
         unordered_map <char, int> x;
         string temp ="";
         for(int i = 0; i < sz; i++){
            x[s[i]]++;
            temp += s[i];
         }
         for(int j = 0, i = sz; i < s.size(); i++, j++){
            if(x.size() <= maxLetters){
               m[temp]++;
            }
            x[temp[0]]--;
            if(x[temp[0]] == 0)x.erase(temp[0]);
            temp.erase(temp.begin(),temp.begin() + 1);
            x[s[i]]++;
            temp += s[i];
         }
         if(x.size() <= maxLetters){
            m[temp]++;
         }
      }
      int ans = 0;
      unordered_map <string ,int > :: iterator i = m.begin();
      while(i != m.end()){
         ans = max (ans, i->second);
         i++;
      }
      return ans;
   }
};
main(){
   Solution ob;
   cout << (ob.maxFreq("aababcaab",2,3,4));
}