Codeforces Round #490 (Div. 3).C. Alphabetic Removals（思维题+容器排序）_codeforces round #490 (div. 3)c. alphabetic remova-优快云博客

本文介绍了一道关于从字符串中按字母顺序移除特定数量字符的算法题目，并提供了三种不同的实现方法，包括使用向量排序、数组标记及直接修改等技术。

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You are given a string $s$ consisting of $n$ lowercase Latin letters. Polycarp wants to remove exactly $k$ characters ( $k \leq n$ ) 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 $k$ ( $1 \leq k \leq n \leq 4 \cdot 10^{5}$ ) — 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
inputCopy
15 3
cccaabababaccbc
outputCopy
cccbbabaccbc
inputCopy
15 9
cccaabababaccbc
outputCopy
cccccc
inputCopy
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;
}