# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ 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 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 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# 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 `

Output:
```2
```

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

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Improved By : AnkitRai01, 29AjayKumar

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