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：
1 #include <stdio.h>
2 #include <string.h>
3 #include <iostream>
4 #include <string>
5 #include <math.h>
6 #include <algorithm>
7 #include <vector>
8 #include <stack>
9 #include <queue>
10 #include <set>
11 #include <map>
12 #include <math.h>
13 const int INF=0x3f3f3f3f;
14 typedef long long LL;
15 const int mod=1e9+7;
16 const int maxn=4*1e5+10;
17 using namespace std;
18
19 char str[maxn];
20 int cnt[27];//原来字母的个数
21 int num[27];//处理后字母的个数
22
23 int main()
24 {
25 int n,k;
26 scanf("%d %d",&n,&k);
27 getchar();
28 for(int i=0;i<n;i++)
29 {
30 scanf("%c",&str[i]);
31 cnt[str[i]-‘a‘+1]++;
32 num[str[i]-‘a‘+1]++;
33 }
34 str[n]=0;
35 for(int i=1;i<=26;i++)//从‘a‘开始进行k次处理
36 {
37 while(num[i]&&k)
38 {
39 num[i]--;
40 k--;
41 }
42 }
43 for(int i=0;str[i];i++)
44 {
45 if(cnt[str[i]-‘a‘+1]>num[str[i]-‘a‘+1])//说明该位置字母已删除
46 {
47 cnt[str[i]-‘a‘+1]--;
48 continue;
49 }
50 else //若该位置字母没被删，输出该字母
51 printf("%c",str[i]);
52 }
53 return 0;
54 }
原文地址：https://www.cnblogs.com/jiamian/p/11620309.html