code forces 999C Alphabetic Removals

C. Alphabetic Removals

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a string ss consisting of nn lowercase Latin letters. Polycarp wants to remove exactly kk characters (k≤nk≤n) from the string ss. Polycarp uses the following algorithm kk 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 kk times, thus removing exactly kk characters.

Help Polycarp find the resulting string.

Input

The first line of input contains two integers nn and kk (1≤k≤n≤4⋅1051≤k≤n≤4⋅105) — the length of the string and the number of letters Polycarp will remove.

The second line contains the string ss consisting of nn lowercase Latin letters.

Output

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

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

Examples

input

Copy

15 3
cccaabababaccbc

output

Copy

cccbbabaccbc

input

Copy

15 9
cccaabababaccbc

output

 

cccccc

input

 

1 1
u

output

u

 

题意：给你一个字符串，你可以对这个字符串做k次操作，每次操作遵循这个规则：从a开始删除、删完a后删b...一直删到z

题解：用一个数组来存字符串中各个字母的个数，然后从a开始删，记录到不能删除的那个位置然后用另一个字符串来保存输出

代码如下

#include<bits/stdc++.h>
using namespace std;
int n;
string str;
int k;
vector<int> cnt();
int main() {

    cin>>n>>k;
    cin>>str;

    for(int i=; i<n; i++) {
        cnt[str[i]-'a']++;
    }
    //记录下每个字母的数量
    //从a开始减去存在的字母
    //如果没有了  就记录下还可以输出的字母
    int pos=;
    for(int i=; i<; i++) {
        if(k>=cnt[i]) {
            k-=cnt[i];
        } else {
            pos=i;
            break;
        }
    }

    string ans;
    int rem=k;
    for(int i=; i<n; i++) {
        int cur=str[i]-'a';
        //在pos后面的字母可以用
        //用完了k的字母可以用

        if(cur>pos||(cur==pos&&rem==)) {
            ans+=str[i];
        } else if(cur==pos) {
            rem--;
        }
    }
    cout<<ans<<endl;

    return ;
}