CodeForces （字符串从字母a开始删除k个字母）

 

 

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
u

Output 3

 

思路：

类比桶排序，用桶来存字母的个数

 

代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <string>
#include <math.h>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <math.h>
const int INF=0x3f3f3f3f;
typedef long long LL;
const int mod=1e9+7;
const int maxn=4*1e5+10;
using namespace std;

char str[maxn];
int cnt[27];//原来字母的个数 
int num[27];//处理后字母的个数 

int main()
{
    int n,k;
    scanf("%d %d",&n,&k);
    getchar();
    for(int i=0;i<n;i++)
    {
        scanf("%c",&str[i]);
        cnt[str[i]-'a'+1]++;
        num[str[i]-'a'+1]++;
    }
    str[n]=0;
    for(int i=1;i<=26;i++)//从'a'开始进行k次处理 
    {
        while(num[i]&&k)
        {
            num[i]--;
            k--;
        }
    }
    for(int i=0;str[i];i++)
    {
        if(cnt[str[i]-'a'+1]>num[str[i]-'a'+1])//说明该位置字母已删除 
        {
            cnt[str[i]-'a'+1]--;
            continue;
        }
        else //若该位置字母没被删，输出该字母 
            printf("%c",str[i]);
    }
    return 0;
}