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++
// 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
filter_none
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 |
chevron_right
filter_none
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 |
chevron_right
filter_none
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 |
chevron_right
filter_none
Output:
2
Time Complexity: O(max(M, N)) where M and N are the lengths of the given strings str1 and str2 respectively.
GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details
Recommended Posts:
- Largest substring of str2 which is a prefix of str1
- Count of characters in str1 such that after deleting anyone of them str1 becomes str2
- Check whether str1 can be converted to str2 with the given operations
- Minimum cost to convert str1 to str2 with the given operations
- Generate all possible strings such that char at index i is either str1[i] or str2[i]
- First element that appears even number of times in an array
- Transform string str1 into str2 by taking characters from string str3
- Minimum number of times A has to be repeated such that B is a substring of it
- Maximum difference between two elements such that larger element appears after the smaller number
- Find if a given string can be represented from a substring by iterating the substring “n” times
- Find the only element that appears b times
- Reduce the array such that each element appears at most K times
- Remove characters from a String that appears exactly K times
- Longest subsequence where every character appears at-least k times
- Reduce the array such that each element appears at most 2 times
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.
