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 $\mathrm{ss consisting\; of nn lowercase\; Latin\; letters.\; Polycarp\; wants\; to\; remove\; exactly kk characters\; (k\le nk\le n)\; from\; the\; string ss.\; Polycarp\; uses\; the\; following\; algorithm kk times:}$

• 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 $\mathrm{kk times,\; thus\; removing\; exactly kk characters.}$

Help Polycarp find the resulting string.

Input

The first line of input contains two integers $\mathrm{nn and kk (1\le k\le n\le 4\cdot 1051\le k\le n\le 4\cdot 105)\; —\; the\; length\; of\; the\; string\; and\; the\; number\; of\; letters\; Polycarp\; will\; remove.}$

The second line contains the string $\mathrm{ss consisting\; of nn lowercase\; Latin\; letters.}$

Output

Print the string that will be obtained from $\mathrm{ss after\; Polycarp\; removes\; exactly kk letters\; using\; the\; above\; algorithm kk times.}$

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(26);
int main() {


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

    for(int i=0; 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=26;
    for(int i=0; i<26; i++) {
        if(k>=cnt[i]) {
            k-=cnt[i];
        } else {
            pos=i;
            break;
        }
    }

    string ans;
    int rem=k;
    for(int i=0; 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==0)) {
            ans+=str[i];
        } else if(cur==pos) {
            rem--;
        }
    }
    cout<<ans<<endl;

    return 0;
}