Maximum consecutive occurrences of a string in another given string

Given two strings str1 and str2, the task is to count the maximum consecutive occurrences of the string str2 in the string str1.

Examples:

Input: str1 = “abababcba”, str2 = “ba” 
Output: 2 
Explanation: String str2 occurs consecutively in the substring { str[1], …, str[4] }. Therefore, the maximum count obtained is 2

Input: str1 = “ababc”, str2 = “ac” 
Output: 0 
Explanation:
Since str2 is not present as a substring in str1, the required output is 0.

Approach: Follow the steps below to solve the problem:

• Initialize a variable, say cntOcc, to store the count of occurrences of str2 in the string str1.
• Iterate over the range [CntOcc, 1]. For every i^th value in the iteration, concatenate the string str2 i times and check if the concatenated string is a substring of the string str1 or not. The first i^th value for which it is found to be true, print it as the required answer.

Below is the implementation:

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
#include <string.h>
using namespace std;

int countFreq(string& pat, string& txt)
{
    int M = pat.length();
    int N = txt.length();
    int res = 0;

    // A loop to slide pat[] one by one
    for(int i = 0; i <= N - M; i++)
    {

        // For current index i, check
        // for pattern match
        int j;
        for(j = 0; j < M; j++)
            if (txt[i + j] != pat[j])
                break;

        // If pat[0...M-1] = txt[i, i+1, ...i+M-1]
        if (j == M)
        {
            res++;
            j = 0;
        }
    }
    return res;
}

// Function to count the maximum
// consecutive occurrence of the
// string str2 in the string str1
int maxRepeating(string str1, string str2)
{

    // Stores the count of consecutive
    // occurrences of str2 in str1
    int cntOcc = countFreq(str2, str1);

    // Concatenate str2 cntOcc times
    string Contstr = "";

    for(int i = 0; i < cntOcc; i++)
        Contstr += str2;

    // Iterate over the string str1
    // while Contstr is not present in str1
    size_t found = str1.find(Contstr);

    while (found == string::npos)
    {
        found = str1.find(Contstr);

        // Update cntOcc
        cntOcc -= 1;

        // Update Contstr
        Contstr = "";

        for(int i = 0; i < cntOcc; i++)
            Contstr += str2;
    }
    return cntOcc;
}

// Driver Code
int main()
{
    string str1 = "abababc";
    string str2 = "ba";

    cout << maxRepeating(str1, str2);

    return 0;
}

// This code is contributed by grand_master

Java

// Java program to implement
// the above approach

import java.util.*;
class GFG
{
static int countFreq(String pat, String txt)
{
    int M = pat.length();
    int N = txt.length();
    int res = 0;

    // A loop to slide pat[] one by one
    for(int i = 0; i <= N - M; i++)
    {

        // For current index i, check
        // for pattern match
        int j;
        for(j = 0; j < M; j++)
            if (txt.charAt(i + j) != pat.charAt(j))
                break;

        // If pat[0...M-1] = txt[i, i+1, ...i+M-1]
        if (j == M)
        {
            res++;
            j = 0;
        }
    }
    return res;
}

// Function to count the maximum
// consecutive occurrence of the
// String str2 in the String str1
static int maxRepeating(String str1, String str2)
{

    // Stores the count of consecutive
    // occurrences of str2 in str1
    int cntOcc = countFreq(str2, str1);

    // Concatenate str2 cntOcc times
    String Contstr = "";

    for(int i = 0; i < cntOcc; i++)
        Contstr += str2;

    // Iterate over the String str1
    // while Contstr is not present in str1
    boolean found = str1.contains(Contstr);

    while (!found)
    {
        found = str1.contains(Contstr);

        // Update cntOcc
        cntOcc -= 1;

        // Update Contstr
        Contstr = "";

        for(int i = 0; i < cntOcc; i++)
            Contstr += str2;
    }
    return cntOcc;
}

// Driver Code
public static void main(String[] args)
{
    String str1 = "abababc";
    String str2 = "ba";
    System.out.print(maxRepeating(str1, str2));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to implement
# the above approach

# Function to count the maximum
# consecutive occurrence of the
# string str2 in the string str1
def maxRepeating(str1, str2):

    # Stores the count of consecutive
    # occurrences of str2 in str1
    cntOcc = str1.count(str2)

    # Concatenate str2 cntOcc times
    Contstr = str2 * cntOcc


    # Iterate over the string str1
    # while Contstr is not present in str1
    while(Contstr not in str1):

        # Update cntOcc
        cntOcc -= 1

       # Update Contstr
        Contstr = str2 * cntOcc

    return cntOcc

# Driver Code
if __name__ =="__main__":
  str1 = "abababc"
  str2 = "ba"
  print(maxRepeating(str1, str2))

C#

// C# program to implement
// the above approach
using System;

class GFG
{
static int countFreq(String pat, String txt)
{
    int M = pat.Length;
    int N = txt.Length;
    int res = 0;

    // A loop to slide pat[] one by one
    for(int i = 0; i <= N - M; i++)
    {

        // For current index i, check
        // for pattern match
        int j;
        for(j = 0; j < M; j++)
            if (txt[i + j] != pat[j])
                break;

        // If pat[0...M-1] = txt[i, i+1, ...i+M-1]
        if (j == M)
        {
            res++;
            j = 0;
        }
    }
    return res;
}

// Function to count the maximum
// consecutive occurrence of the
// String str2 in the String str1
static int maxRepeating(String str1, String str2)
{

    // Stores the count of consecutive
    // occurrences of str2 in str1
    int cntOcc = countFreq(str2, str1);

    // Concatenate str2 cntOcc times
    String Contstr = "";

    for(int i = 0; i < cntOcc; i++)
        Contstr += str2;

    // Iterate over the String str1
    // while Contstr is not present in str1
    bool found = str1.Contains(Contstr);
    while (!found)
    {
        found = str1.Contains(Contstr);

        // Update cntOcc
        cntOcc -= 1;

        // Update Contstr
        Contstr = "";

        for(int i = 0; i < cntOcc; i++)
            Contstr += str2;
    }
    return cntOcc;
}

// Driver Code
public static void Main(String[] args)
{
    String str1 = "abababc";
    String str2 = "ba";
    Console.Write(maxRepeating(str1, str2));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>
      // JavaScript program to implement
      // the above approach
      function countFreq(pat, txt) {
        var M = pat.length;
        var N = txt.length;
        var res = 0;

        // A loop to slide pat[] one by one
        for (var i = 0; i <= N - M; i++) {
          // For current index i, check
          // for pattern match
          var j;
          for (j = 0; j < M; j++)
              if (txt[i + j] !== pat[j])
                break;

          // If pat[0...M-1] = txt[i, i+1, ...i+M-1]
          if (j === M) {
            res++;
            j = 0;
          }
        }
        return res;
      }

      // Function to count the maximum
      // consecutive occurrence of the
      // String str2 in the String str1
      function maxRepeating(str1, str2) {
        // Stores the count of consecutive
        // occurrences of str2 in str1
        var cntOcc = countFreq(str2, str1);

        // Concatenate str2 cntOcc times
        var Contstr = "";

        for (var i = 0; i < cntOcc; i++)
            Contstr += str2;

        // Iterate over the String str1
        // while Contstr is not present in str1
        var found = str1.includes(Contstr);
        while (!found) {
          found = str1.includes(Contstr);

          // Update cntOcc
          cntOcc -= 1;

          // Update Contstr
          Contstr = "";

          for (var i = 0; i < cntOcc; i++)
              Contstr += str2;
        }
        return cntOcc;
      }

      // Driver Code
      var str1 = "abababc";
      var str2 = "ba";
      document.write(maxRepeating(str1, str2));
</script>

Output:

2

Time Complexity: O(N²), as we are using nested loops for traversing N*N times.
Auxiliary Space: O(N), as we are using extra space.