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++ 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 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 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# 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
<script>
// Javascript implementation of the approach
const MAX = 26;
// Function to return the maximum number
// of times str1 can appear as a
// non-overlapping substring in str2
function 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
let freq1 = new Array(MAX).fill(0);
for (let i = 0; i < len1; i++)
freq1[str1.charCodeAt(i) - 'a'.charCodeAt(0)]++;
// Store the frequency of the characters of str2
let freq2 = new Array(MAX).fill(0);
for (let i = 0; i < len2; i++)
freq2[str2.charCodeAt(i) - 'a'.charCodeAt(0)]++;
// To store the required count of substrings
let minPoss = Number.MAX_SAFE_INTEGER;
for (let 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, Math.floor(freq2[i] / freq1[i]));
}
return minPoss;
}
// Driver code
let str1 = "geeks", str2 = "gskefrgoekees";
let len1 = str1.length;
let len2 = str2.length;
document.write(maxSubStr(str1, len1, str2, len2));
// This code is contributed by _saurabh_jaiswal.
</script>
Output:
2
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.
