# Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined) D. Dense Subsequence（贪心）

D. Dense Subsequence

You are given a string s, consisting of lowercase English letters, and the integer m.

One should choose some symbols from the given string so that any contiguous subsegment of length m has at least one selected symbol. Note that here we choose positions of symbols, not the symbols themselves.

Then one uses the chosen symbols to form a new string. All symbols from the chosen position should be used, but we are allowed to rearrange them in any order.

Formally, we choose a subsequence of indices 1 ≤ i1 < i2 < ... < it ≤ |s|. The selected sequence must meet the following condition: for every j such that 1 ≤ j ≤ |s| - m + 1, there must be at least one selected index that belongs to the segment [j,  j + m - 1], i.e. there should exist a k from 1 to t, such that j ≤ ik ≤ j + m - 1.

Then we take any permutation p of the selected indices and form a new string sip1sip2... sipt.

Find the lexicographically smallest string, that can be obtained using this procedure.

Input

The first line of the input contains a single integer m (1 ≤ m ≤ 100 000).

The second line contains the string s consisting of lowercase English letters. It is guaranteed that this string is non-empty and its length doesn't exceed 100 000. It is also guaranteed that the number m doesn't exceed the length of the string s.

Output

Print the single line containing the lexicographically smallest string, that can be obtained using the procedure described above.

Examples
input
3cbabc
output
a
input
2abcab
output
aab
input
3bcabcbaccba
output
aaabb
Note

In the first sample, one can choose the subsequence {3} and form a string "a".

In the second sample, one can choose the subsequence {1, 2, 4} (symbols on this positions are 'a', 'b' and 'a') and rearrange the chosen symbols to form a string "aab".

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
char s[maxn];
int vis[maxn];
int n, m;

int work(int c)
{
int pre = -1, now = -1, res = 0;
for(int i = 0; i < n; i++) {
if(s[i] - 'a' < c) pre = i, now = i;
if(s[i] - 'a' == c) now = i;
if(i - pre == m) {
//printf("check %d %d %d\n", i, c, pre);
if(now > pre) {
res++;
pre = now;
}
else return -1;
}
}
return res;
}

int main()
{
scanf("%d %s", &m, s);
memset(vis, 0, sizeof(vis));
n = strlen(s);
for(int i = 0; i < n; i++) {
vis[s[i] - 'a']++;
}
for(int i = 0; i < 26; i++) {
int flag = work(i);
if(flag != -1) {
for(int j = 0; j < i; j++) {
for(int k = 0; k < vis[j]; k++) {
printf("%c", j + 'a');
}
}
for(int j = 0; j < flag; j++) {
printf("%c", i + 'a');
}puts("");
return 0;
}
}
}

posted @ 2016-10-09 13:33  MartinEden  阅读(...)  评论(...编辑  收藏