Skip to content
Related Articles
Maximum number of times str1 appears as a non-overlapping substring in str2
• Last Updated : 04 Jun, 2021

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:
str = “geeksforgeeks
Input: str1 = “aa”, str2 = “aaaa”
Output:

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++

 `// C++ implementation of the approach``#include ``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;``}`

## Java

 `// 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`

## Python3

 `# 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`

## C#

 `// 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`

## Javascript

 ``
Output:
`2`

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live

My Personal Notes arrow_drop_up