Count of times second string can be formed from the characters of first string
Last Updated :
07 Oct, 2022
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Given two strings str and patt, the task is to find the count of times patt can be formed using the characters of str.
Examples:
Input: str = "geeksforgeeks", patt = "geeks"
Output: 2
"geeks" can be made at most twice from
the characters of "geeksforgeeks".Input: str = "abcbca", patt = "aabc"
Output: 1
Approach: Count the frequency of all the characters of str and patt and store them in arrays strFreq[] and pattFreq[] respectively. Now any character ch which appears in patt can be used in a maximum of strFreq[ch] / pattFreq[ch] words and the minimum of this value among all the characters of patt is the required 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 update the freq[] array
// to store the frequencies of
// all the characters of str
void updateFreq(string str, int freq[])
{
int len = str.length();
// Update the frequency of the characters
for (int i = 0; i < len; i++) {
freq[str[i] - 'a']++;
}
}
// Function to return the maximum count
// of times patt can be formed
// using the characters of str
int maxCount(string str, string patt)
{
// To store the frequencies of
// all the characters of str
int strFreq[MAX] = { 0 };
updateFreq(str, strFreq);
// To store the frequencies of
// all the characters of patt
int pattFreq[MAX] = { 0 };
updateFreq(patt, pattFreq);
// To store the result
int ans = INT_MAX;
// For every character
for (int i = 0; i < MAX; i++) {
// If the current character
// doesn't appear in patt
if (pattFreq[i] == 0)
continue;
// Update the result
ans = min(ans, strFreq[i] / pattFreq[i]);
}
return ans;
}
// Driver code
int main()
{
string str = "geeksforgeeks";
string patt = "geeks";
cout << maxCount(str, patt);
return 0;
}
// Java implementation of the approach
class GFG
{
static int MAX = 26;
// Function to update the freq[] array
// to store the frequencies of
// all the characters of str
static void updateFreq(String str, int freq[])
{
int len = str.length();
// Update the frequency of the characters
for (int i = 0; i < len; i++)
{
freq[str.charAt(i) - 'a']++;
}
}
// Function to return the maximum count
// of times patt can be formed
// using the characters of str
static int maxCount(String str, String patt)
{
// To store the frequencies of
// all the characters of str
int []strFreq = new int[MAX];
updateFreq(str, strFreq);
// To store the frequencies of
// all the characters of patt
int []pattFreq = new int[MAX];
updateFreq(patt, pattFreq);
// To store the result
int ans = Integer.MAX_VALUE;
// For every character
for (int i = 0; i < MAX; i++)
{
// If the current character
// doesn't appear in patt
if (pattFreq[i] == 0)
continue;
// Update the result
ans = Math.min(ans, strFreq[i] / pattFreq[i]);
}
return ans;
}
// Driver code
public static void main(String[] args)
{
String str = "geeksforgeeks";
String patt = "geeks";
System.out.print(maxCount(str, patt));
}
}
// This code is contributed by Rajput-Ji
# Python3 implementation of the approach
MAX = 26
# Function to update the freq[] array
# to store the frequencies of
# all the characters of strr
def updateFreq(strr, freq):
lenn = len(strr)
# Update the frequency of the characters
for i in range(lenn):
freq[ord(strr[i]) - ord('a')] += 1
# Function to return the maximum count
# of times patt can be formed
# using the characters of strr
def maxCount(strr, patt):
# To store the frequencies of
# all the characters of strr
strrFreq = [0 for i in range(MAX)]
updateFreq(strr, strrFreq)
# To store the frequencies of
# all the characters of patt
pattFreq = [0 for i in range(MAX)]
updateFreq(patt, pattFreq)
# To store the result
ans = 10**9
# For every character
for i in range(MAX):
# If the current character
# doesn't appear in patt
if (pattFreq[i] == 0):
continue
# Update the result
ans = min(ans, strrFreq[i] // pattFreq[i])
return ans
# Driver code
strr = "geeksforgeeks"
patt = "geeks"
print(maxCount(strr, patt))
# This code is contributed by Mohit Kumar
// C# implementation of the approach
using System;
class GFG
{
static int MAX = 26;
// Function to update the []freq array
// to store the frequencies of
// all the characters of str
static void updateFreq(String str, int []freq)
{
int len = str.Length;
// Update the frequency of the characters
for (int i = 0; i < len; i++)
{
freq[str[i] - 'a']++;
}
}
// Function to return the maximum count
// of times patt can be formed
// using the characters of str
static int maxCount(String str, String patt)
{
// To store the frequencies of
// all the characters of str
int []strFreq = new int[MAX];
updateFreq(str, strFreq);
// To store the frequencies of
// all the characters of patt
int []pattFreq = new int[MAX];
updateFreq(patt, pattFreq);
// To store the result
int ans = int.MaxValue;
// For every character
for (int i = 0; i < MAX; i++)
{
// If the current character
// doesn't appear in patt
if (pattFreq[i] == 0)
continue;
// Update the result
ans = Math.Min(ans, strFreq[i] / pattFreq[i]);
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
String str = "geeksforgeeks";
String patt = "geeks";
Console.Write(maxCount(str, patt));
}
}
// This code is contributed by 29AjayKumar
<script>
// JavaScript implementation of the approach
const MAX = 26;
// Function to update the freq[] array
// to store the frequencies of
// all the characters of str
function updateFreq(str, freq) {
var len = str.length;
// Update the frequency of the characters
for (var i = 0; i < len; i++) {
freq[str[i].charCodeAt(0) - "a".charCodeAt(0)]++;
}
}
// Function to return the maximum count
// of times patt can be formed
// using the characters of str
function maxCount(str, patt) {
// To store the frequencies of
// all the characters of str
var strFreq = new Array(MAX).fill(0);
updateFreq(str, strFreq);
// To store the frequencies of
// all the characters of patt
var pattFreq = new Array(MAX).fill(0);
updateFreq(patt, pattFreq);
// To store the result
var ans = 21474836473;
// For every character
for (var i = 0; i < MAX; i++) {
// If the current character
// doesn't appear in patt
if (pattFreq[i] == 0) continue;
// Update the result
ans = Math.min(ans, strFreq[i] / pattFreq[i]);
}
return ans;
}
// Driver code
var str = "geeksforgeeks";
var patt = "geeks";
document.write(maxCount(str, patt));
</script>
Output:
2
Time Complexity: O(m+n) where m and n are lengths of the given string str and patt respectively.
Auxiliary Space: O(MAX)
