Count of times second string can be formed from the characters of first string
Last Updated :
07 Oct, 2022
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++
#include <bits/stdc++.h>
using namespace std;
const int MAX = 26;
void updateFreq(string str, int freq[])
{
int len = str.length();
for ( int i = 0; i < len; i++) {
freq[str[i] - 'a' ]++;
}
}
int maxCount(string str, string patt)
{
int strFreq[MAX] = { 0 };
updateFreq(str, strFreq);
int pattFreq[MAX] = { 0 };
updateFreq(patt, pattFreq);
int ans = INT_MAX;
for ( int i = 0; i < MAX; i++) {
if (pattFreq[i] == 0)
continue ;
ans = min(ans, strFreq[i] / pattFreq[i]);
}
return ans;
}
int main()
{
string str = "geeksforgeeks" ;
string patt = "geeks" ;
cout << maxCount(str, patt);
return 0;
}
|
Java
class GFG
{
static int MAX = 26 ;
static void updateFreq(String str, int freq[])
{
int len = str.length();
for ( int i = 0 ; i < len; i++)
{
freq[str.charAt(i) - 'a' ]++;
}
}
static int maxCount(String str, String patt)
{
int []strFreq = new int [MAX];
updateFreq(str, strFreq);
int []pattFreq = new int [MAX];
updateFreq(patt, pattFreq);
int ans = Integer.MAX_VALUE;
for ( int i = 0 ; i < MAX; i++)
{
if (pattFreq[i] == 0 )
continue ;
ans = Math.min(ans, strFreq[i] / pattFreq[i]);
}
return ans;
}
public static void main(String[] args)
{
String str = "geeksforgeeks" ;
String patt = "geeks" ;
System.out.print(maxCount(str, patt));
}
}
|
Python3
MAX = 26
def updateFreq(strr, freq):
lenn = len (strr)
for i in range (lenn):
freq[ ord (strr[i]) - ord ( 'a' )] + = 1
def maxCount(strr, patt):
strrFreq = [ 0 for i in range ( MAX )]
updateFreq(strr, strrFreq)
pattFreq = [ 0 for i in range ( MAX )]
updateFreq(patt, pattFreq)
ans = 10 * * 9
for i in range ( MAX ):
if (pattFreq[i] = = 0 ):
continue
ans = min (ans, strrFreq[i] / / pattFreq[i])
return ans
strr = "geeksforgeeks"
patt = "geeks"
print (maxCount(strr, patt))
|
C#
using System;
class GFG
{
static int MAX = 26;
static void updateFreq(String str, int []freq)
{
int len = str.Length;
for ( int i = 0; i < len; i++)
{
freq[str[i] - 'a' ]++;
}
}
static int maxCount(String str, String patt)
{
int []strFreq = new int [MAX];
updateFreq(str, strFreq);
int []pattFreq = new int [MAX];
updateFreq(patt, pattFreq);
int ans = int .MaxValue;
for ( int i = 0; i < MAX; i++)
{
if (pattFreq[i] == 0)
continue ;
ans = Math.Min(ans, strFreq[i] / pattFreq[i]);
}
return ans;
}
public static void Main(String[] args)
{
String str = "geeksforgeeks" ;
String patt = "geeks" ;
Console.Write(maxCount(str, patt));
}
}
|
Javascript
<script>
const MAX = 26;
function updateFreq(str, freq) {
var len = str.length;
for ( var i = 0; i < len; i++) {
freq[str[i].charCodeAt(0) - "a" .charCodeAt(0)]++;
}
}
function maxCount(str, patt) {
var strFreq = new Array(MAX).fill(0);
updateFreq(str, strFreq);
var pattFreq = new Array(MAX).fill(0);
updateFreq(patt, pattFreq);
var ans = 21474836473;
for ( var i = 0; i < MAX; i++) {
if (pattFreq[i] == 0) continue ;
ans = Math.min(ans, strFreq[i] / pattFreq[i]);
}
return ans;
}
var str = "geeksforgeeks" ;
var patt = "geeks" ;
document.write(maxCount(str, patt));
</script>
|
Time Complexity: O(m+n) where m and n are lengths of the given string str and patt respectively.
Auxiliary Space: O(MAX)
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...