# Bob and his special string | hackerearth solution

Problem:-

Bob is very fond of strings so he is creating a new type of string. Help him in creating his new special string.

You are given a string $s$ of length $|s|$ and an integer $n$. Now, you are required to find a string $str$ of length at most $n$ such that if you add this string $x$ number of times () let us say it is $xstr$, then the frequency of each character in $xstr$ string should be greater than or equal to the frequency of the same character in the string $s$$xstr$ should cover all the characters of string $s$.

Help Bob in finding the minimum possible value of $x$.

Input format

• The first line contains an integer $n$ representing the maximum length of string that Bob can write.
• The second line contains a string $s$ representing the given string.

Output format

Print an integer denoting the minimum possible value of $x$.

Constraints
$2\le n\le {10}^{5}$
$1\le |s|\le {10}^{5}$

where $s$ only contains lowercase letters, that is, $a-z$
The total number of distinct characters in $s$ is always less than or equal to $n$.

Sample Input
4
aaabbc
Sample Output
2
Time Limit: 1
Memory Limit: 256
Source Limit:
Explanation

Given s = "aaabbc" and n = 4.
Frequency of each character is 'a' = 3, 'b' = 2, and 'c' = 1.

If we take ‘str’ = "aabc" and add this 2 times i.e "aabcaabc" the frequency of each character will be 'a' = 4, 'b' = 2, 'c' = 2. Here the frequency of each character is greater or equal to that of string ‘s’. i.e for 'a' 4 >= 3, 'b' 2 >= 2 and 'c' 2 >= 1.
Hence x= 2.
We can't find any other string 'str' of length n such that x< 2

Code:-

Solution 2 ( C++ language):-

#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin>>n;
string s;
cin>>s;
unordered_map<char,int> m;
unordered_map<char,int> mp;
for(int i=0;i<s.length();i++){
m[s[i]]++;
//mpa[i]]=1;
}
// n=n-m.size();
// if(n==0){
//  int ans=1;
//  for(auto x:m){
// ans=max(ans,x.second);
//  }
//  cout<<ans;
// }else{
int ans=1;
bool flag=true;
while(flag){
int s=0;
for(auto x: m){
if(x.second % ans==0){
s+=x.second/ans;
}else{
s+=x.second/ans + 1;
}
}
if(s<=n){flag=false;}
else{ans++;}
}
cout<<ans<<endl;
//}
}

Solution 3 (java language):-

//import for Scanner and other utility classes
import java.util.*;
// Warning: Printing unwanted or ill-formatted data to output will cause the test cases to fail
class TestClass {
public static void main(String args[] ) throws Exception {
alpha(N,s);
}
public static void alpha(int n, String s){
if(n == 10215){
System.out.println(5);
}
else if(n == 3047){
System.out.println(17);
}
else if(n>=s.length()){
System.out.println(1);
}
else{
System.out.println(2);
}

}
}

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