Codeforces (letter string that begins with the letter k deleting a)

 

 

You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k≤n) from the string s. Polycarp uses the following algorithm k times:

• if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
• if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
• ...
• remove the leftmost occurrence of the letter 'z' and stop the algorithm.

This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.

Help Polycarp find the resulting string.

Input

The first line of input contains two integers n and k (1≤k≤n≤4⋅10⁵) — the length of the string and the number of letters Polycarp will remove.

The second line contains the string s consisting of n lowercase Latin letters.

Output

Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.

If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).

Examples

Input 1

15 3
cccaabababaccbc

Output 1

cccbbabaccbc

Input 2

15 9
cccaabababaccbc

Output 2

cccccc

Input 3

1  1 
in

Output 3

 

Ideas:

Analogy bucket sort with the number of buckets stored letter

 

Code:

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <iostream>
 4 #include <string>
 5 #include <math.h>
 6 #include <algorithm>
 7 #include <vector>
 8 #include <stack>
 9 #include <queue>
10 #include <set>
11 #include <map>
12 #include <math.h>
13 const int INF=0x3f3f3f3f;
14 typedef long long LL;
15 const  int MOD = 1E9 + . 7 ;
 16  const  int MAXN = . 4 * 1E5 + 10 ;
 . 17  the using  namespace STD;
 18 is  
. 19  char STR [MAXN];
 20 is  int CNT [ 27 ]; // number of original letters 
21 is  int NUM [ 27 ] ; // number of letters treatment 
22 is  
23 is  int main ()
 24  {
 25      int n-, K;
 26 is      Scanf ( " % D% D " , & n-, & K);
27     getchar();
28     for(int i=0;i<n;i++)
29     {
30         scanf("%c",&str[i]);
31         cnt[str[i]-'a'+1]++;
32         num[str[i]-'a'+1]++;
33     }
34     str[n]=0;
35     for(int i=1;i<=26; I ++) // 'a' and the k-th process starts 
36      {
 37 [          the while (NUM [I] && k)
 38 is          {
 39              NUM [I] - ;
 40              K-- ;
 41 is          }
 42 is      }
 43 is      for ( int I = 0 ; STR [I]; I ++ )
 44 is      {
 45          IF (CNT [STR [I] - ' A ' + . 1 ]> NUM [STR [I] - ' A ' + . 1 ]) // description of the position of the letters is delete 
46         {
 47              CNT [STR [I] - ' A ' + . 1 ] - ;
 48              Continue ;
 49          }
 50          the else  // If this is not the position of the letter is deleted, the output of the letter 
51 is              the printf ( " % C " , STR [I] );
 52 is      }
 53 is      return  0 ;
 54 is }