Maximum number of times str1 appears as a nonoverlapping substring in str2 (original) (raw)

Last Updated : 22 Jun, 2022

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;

}

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 substring 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

`

Time Complexity: O(max(M, N)), as we using a loop to traverse N times and M times. Where M and N are the lengths of the given strings str1 and str2 respectively.
Auxiliary Space: O(26), as we are using extra space for the map.