Maximum length substring with highest frequency in a string

Given a string. The task is to find the maximum occurred substring with a maximum length. These occurrences can overlap.

Examples:

Input: str = "abab"
Output: ab
"a", "b", "ab" are occur 2 times. But, "ab" has maximum length

Input: str = "abcd"
Output: a

Approach: The idea is to store the frequency of each substring using a map and print the one with maximum frequency and maximum length.



Below is the implementation of the above approach:

C++

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// CPP program to find maximum
// occurred substring of a string
#include <bits/stdc++.h>
using namespace std;
  
// function to return maximum
// occurred substring of a string
string MaxFreq(string str)
{
    // size of the string
    int n = str.size();
  
    unordered_map<string, int> m;
  
    for (int i = 0; i < n; i++) {
        string s = "";
        for (int j = i; j < n; j++) {
            s += str[j];
            m[s]++;
        }
    }
  
    // to store maximum frequency
    int maxi = 0;
  
    // to store string which has maximum frequency
    string s;
    for (auto i = m.begin(); i != m.end(); i++) {
        if (i->second > maxi) {
            maxi = i->second;
            s = i->first;
        }
        else if (i->second == maxi) {
            string ss = i->first;
            if (ss.size() > s.size())
                s = ss;
        }
    }
  
    // return substring which has maximum frequency
    return s;
}
  
// Driver program
int main()
{
    string str = "ababecdecd";
  
    // function call
    cout << MaxFreq(str);
  
    return 0;
}

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Python3

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# Python3 program to find maximum 
# occured of a string
  
# function to return maximum occurred
# substring of a string
def MaxFreq(s):
      
    # size of string
    n = len(s)
      
    m = dict()
      
    for i in range(n):
        strng = ''
        for j in range(i, n):
            strng += s[j]
            if strng in m.keys():
                m[strng] += 1
            else:
                m[strng] = 1
                  
    # to store maximum freqency
    maxi = 0
      
    # To store string which has 
    # maximum frequency
    maxi_str = ''
      
    for i in m:
        if m[i] > maxi:
            maxi = m[i]
            maxi_str = i
        elif m[i] == maxi:
            ss = i
            if len(ss) > len(maxi_str):
                maxi_str = ss
                  
    # return substring which has maximum freq
    return maxi_str
      
# Driver code
strng = "ababecdecd"
  
print(MaxFreq(strng))
          
# This code is contributed by Mohit kumar 29     

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Output:

ecd


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Improved By : mohit kumar 29