Maximum number of times str1 appears as a non-overlapping substring in str2

Given two strings str1 and str2, the task is to find the maximum number of times str1 occurs in str2 as a non-overlapping substring after rearranging the characters of str2

Examples:

Input: str1 = “geeks”, str2 = “gskefrgoekees”
Output: 2
str = “geeksforgeeks



Input: str1 = “aa”, str2 = “aaaa”
Output: 2

Approach: The idea is to store the frequency of characters of both the strings and comparing them.

  • If there is a character whose frequency in the first string is greater than its frequency in the second string then the answer is always 0 because string str1 can never occur in str2.
  • After storing the frequency of the characters of both the strings, perform integer division between the non-zero frequency of characters of str1 and str2. The minimum value would be the answer.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
const int MAX = 26;
  
// Function to return the maximum number
// of times str1 can appear as a
// non-overlapping substring in str2
int maxSubStr(string str1, int len1, string str2, int len2)
{
  
    // str1 cannot never be substring of str2
    if (len1 > len2)
        return 0;
  
    // Store the frequency of the characters of str1
    int freq1[MAX] = { 0 };
    for (int i = 0; i < len1; i++)
        freq1[str1[i] - 'a']++;
  
    // Store the frequency of the characters of str2
    int freq2[MAX] = { 0 };
    for (int i = 0; i < len2; i++)
        freq2[str2[i] - 'a']++;
  
    // To store the required count of substrings
    int minPoss = INT_MAX;
  
    for (int i = 0; i < MAX; i++) {
  
        // Current character doesn't appear in str1
        if (freq1[i] == 0)
            continue;
  
        // Frequency of the current character in str1
        // is greater than its frequency in str2
        if (freq1[i] > freq2[i])
            return 0;
  
        // Update the count of possible substrings
        minPoss = min(minPoss, freq2[i] / freq1[i]);
    }
    return minPoss;
}
  
// Driver code
int main()
{
    string str1 = "geeks", str2 = "gskefrgoekees";
    int len1 = str1.length();
    int len2 = str2.length();
  
    cout << maxSubStr(str1, len1, str2, len2);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach 
class GFG 
{
    final static int MAX = 26
      
    // Function to return the maximum number 
    // of times str1 can appear as a 
    // non-overlapping substring in str2 
    static int maxSubStr(char []str1, int len1,
                         char []str2, int len2) 
    
      
        // str1 cannot never be substring of str2 
        if (len1 > len2) 
            return 0
      
        // Store the frequency of the characters of str1 
        int freq1[] = new int[MAX]; 
          
        for (int i = 0; i < len1; i++) 
            freq1[i] = 0
              
        for (int i = 0; i < len1; i++) 
            freq1[str1[i] - 'a']++; 
      
        // Store the frequency of the characters of str2 
        int freq2[] = new int[MAX]; 
          
        for (int i = 0; i < len2; i++) 
            freq2[i] = 0
              
        for (int i = 0; i < len2; i++) 
            freq2[str2[i] - 'a']++; 
      
        // To store the required count of substrings 
        int minPoss = Integer.MAX_VALUE;
      
        for (int i = 0; i < MAX; i++)
        
      
            // Current character doesn't appear in str1 
            if (freq1[i] == 0
                continue
      
            // Frequency of the current character in str1 
            // is greater than its frequency in str2 
            if (freq1[i] > freq2[i]) 
                return 0
      
            // Update the count of possible substrings 
            minPoss = Math.min(minPoss, freq2[i] / freq1[i]); 
        
        return minPoss; 
    
      
    // Driver code 
    public static void main (String[] args)
    
        String str1 = "geeks", str2 = "gskefrgoekees"
        int len1 = str1.length(); 
        int len2 = str2.length(); 
      
        System.out.println(maxSubStr(str1.toCharArray(), len1, 
                                     str2.toCharArray(), len2)); 
    
}
  
// This code is contributed by AnkitRai01

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
import sys
MAX = 26;
  
# Function to return the maximum number
# of times str1 can appear as a
# non-overlapping sustring bin str2
def maxSubStr(str1, len1, str2, len2):
  
    # str1 cannot never be 
    # substring of str2
    if (len1 > len2):
        return 0;
  
    # Store the frequency of
    # the characters of str1
    freq1 = [0] * MAX;
    for i in range(len1):
        freq1[ord(str1[i]) - 
              ord('a')] += 1;
  
    # Store the frequency of 
    # the characters of str2
    freq2 = [0] * MAX;
    for i in range(len2):
        freq2[ord(str2[i]) - 
              ord('a')] += 1;
  
    # To store the required count 
    # of substrings
    minPoss = sys.maxsize;
  
    for i in range(MAX):
  
        # Current character doesn't appear
        # in str1
        if (freq1[i] == 0):
            continue;
  
        # Frequency of the current character 
        # in str1 is greater than its 
        # frequency in str2
        if (freq1[i] > freq2[i]):
            return 0;
  
        # Update the count of possible substrings
        minPoss = min(minPoss, freq2[i] / 
                               freq1[i]);
    return int(minPoss);
  
# Driver code
str1 = "geeks"; str2 = "gskefrgoekees";
len1 = len(str1);
len2 = len(str2);
  
print(maxSubStr(str1, len1, str2, len2));
  
# This code is contributed by 29AjayKumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the above approach 
using System;
      
class GFG 
{
    readonly static int MAX = 26; 
      
    // Function to return the maximum number 
    // of times str1 can appear as a 
    // non-overlapping substring in str2 
    static int maxSubStr(char []str1, int len1,
                         char []str2, int len2) 
    
      
        // str1 cannot never be substring of str2 
        if (len1 > len2) 
            return 0; 
      
        // Store the frequency of the characters of str1 
        int []freq1 = new int[MAX]; 
          
        for (int i = 0; i < len1; i++) 
            freq1[i] = 0; 
              
        for (int i = 0; i < len1; i++) 
            freq1[str1[i] - 'a']++; 
      
        // Store the frequency of the characters of str2 
        int []freq2 = new int[MAX]; 
          
        for (int i = 0; i < len2; i++) 
            freq2[i] = 0; 
              
        for (int i = 0; i < len2; i++) 
            freq2[str2[i] - 'a']++; 
      
        // To store the required count of substrings 
        int minPoss = int.MaxValue;
      
        for (int i = 0; i < MAX; i++)
        
      
            // Current character doesn't appear in str1 
            if (freq1[i] == 0) 
                continue
      
            // Frequency of the current character in str1 
            // is greater than its frequency in str2 
            if (freq1[i] > freq2[i]) 
                return 0; 
      
            // Update the count of possible substrings 
            minPoss = Math.Min(minPoss, freq2[i] / freq1[i]); 
        
        return minPoss; 
    
      
    // Driver code 
    public static void Main (String[] args)
    
        String str1 = "geeks", str2 = "gskefrgoekees"
        int len1 = str1.Length; 
        int len2 = str2.Length; 
      
        Console.WriteLine(maxSubStr(str1.ToCharArray(), len1, 
                                    str2.ToCharArray(), len2)); 
    
}
  
// This code is contributed by 29AjayKumar

chevron_right


Output:

2

Time Complexity: O(max(M, N)) where M and N are the lengths of the given strings str1 and str2 respectively.



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : AnkitRai01, 29AjayKumar