Codeforces Round #490 (Div. 3).C. Alphabetic Removals（思维题+容器排序）

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 $s$ consisting of $n$ lowercase Latin letters. Polycarp wants to remove exactly $k$ characters ($k\le n$) 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 $k$ ($1\le k\le n\le 4\cdot {10}^5$) — 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

Copy

15 3
cccaabababaccbc

output

Copy

cccbbabaccbc

input

Copy

15 9
cccaabababaccbc

output

Copy

cccccc

input

Copy

1 1
u

output

考点：思维题

方法一：

#include <bits/stdc++.h>
using namespace std;
int main()
{
    vector<pair<char,int> > v;
    int n,k;
    string x;
    cin>>n>>k>>x;
    for(int i=0; i<n; i++)
        v.push_back(make_pair(x[i],i));
    sort(v.begin(),v.end());ikui
    for(int i=0; i<k; i++)
        x[v[i].second]='*';
    for(int i=0; i<n; i++)
        if(x[i]!='*') cout<<x[i];
}

方法二：

#include <bits/stdc++.h>
using namespace std;
int n, k, f[400050], id;
string s;
int main()
{
	cin >> n >> k >> s;
	for(int j = 0; j < 26; j++)
		for(int i = 0; i < n; i++)
		    if(s[i] - 'a' == j)
                f[i] = ++id;
	for(int i = 0; i < n; i++)
        if(f[i] > k)
            cout << s[i];
}

方法三：

#include <bits/stdc++.h>
#define ll long long
#define N 1000005
using namespace std;

int n,k;
char s[N];

int main(){
	scanf("%d%d%s",&n,&k,s+1);
	for(char c='a';k>0&&c<='z';++c)
		for(int i=1;k>0&&i<=n;++i)
			if(s[i]==c)
				--k,s[i]=0;
	for(int i=1;i<=n;++i)
		if(s[i])putchar(s[i]);
	return 0;
}