CodeForces - 999C Alphabetic Removals

C - Alphabetic Removals

You are given a string

s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k≤n≤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 kkletters 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

15 3
cccaabababaccbc

Output

cccbbabaccbc

Input

15 9
cccaabababaccbc

Output

cccccc

Input

1 1
u

Output

这个题就是给你一个已知长度的字符串，给你一个数k，要求要从这里面删去k个字符，删去的这k个字符要求是从a开始删除，如果a没有删完，那么就不能删除b，依次进行，直到删完k个字符就可以了。

#include <bits/stdc++.h>

using namespace std;

char s[400005];
int a[1000];  // a用来标记每个字母出现的次数
int main()
{
    int n, k;
    while(~scanf("%d %d",&n, &k))
    {
        getchar();
        memset(a, 0,sizeof(a));
        for(int i = 0; i < n; i ++)
        {
            scanf("%c",&s[i]);
            a[s[i]] ++;
        }
        if(k >= n) printf("\n");  //如果需要删除的k大于n，就全部删除没了
        else
        {
            int num = 0;
            for(int i = 97; ; i ++) // 从a开始删除，如果满足k>=a[i],把所有的字符都删去即可
            {
                if(k >= a[i])
                {
                    num ++;
                    k = k - a[i];
                }
                else break;
            }
            for(int i = 0; i < n; i ++)
            {

                if(s[i] < 97 + num);  //删除掉的不再输出
                else
                {
                    if(s[i] == 97 + num)  // 没有全部删除完成，继续删除这个字符
                    {
                        if(k > 0)k--;
                        else printf("%c",s[i]);
                    }
                    else printf("%c",s[i]); // 输出不需要删除的
                }

            }
            printf("\n");
        }
    }
    return 0;
}

发现可以省去好多代码：

#include <bits/stdc++.h>
using namespace std;
int n,m;
char str[400005];

int main()
{
  cin>>n>>m;
  cin>>str;
  for(int i=0;i<=25;i++){
    for(int j=0;j<n&&m;j++){
      if(str[j] == 'a' + i){
        str[j] = '0';
        m--;
      }
    }
  }
  for(int i=0;i<n;i++){
    if(str[i] != '0')cout<<str[i];
  }
  return 0;
}