• 2018-03-06
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# C. Phone Numbers

And where the are the phone numbers?

You are given a string s consisting of lowercase English letters and an integer k. Find the lexicographically smallest string t of length k, such that its set of letters is a subset of the set of letters of s and s is lexicographically smaller than t.

It’s guaranteed that the answer exists.

Note that the set of letters is a set, not a multiset. For example, the set of letters of abadaba is {a, b, d}.

String p is lexicographically smaller than string q, if p is a prefix of q, is not equal to q or there exists i, such that pi < qi and for all j < i it is satisfied that pj = qj. For example, abc is lexicographically smaller than abcd , abd is lexicographically smaller than abecafa is not lexicographically smaller than ab and a is not lexicographically smaller than a.

## Input

The first line of input contains two space separated integers n and k (1 ≤ n, k ≤ 100 000) — the length of s and the required length of t.

The second line of input contains the string s consisting of n lowercase English letters.

## Output

Output the string t conforming to the requirements above.

It’s guaranteed that the answer exists.

```input
3 3
abc
output
aca```
```input
3 2
abc
output
ac```
```input
3 3
ayy
output
yaa```
```input
2 3
ba
output
baa```

## Note

In the first example the list of strings t of length 3, such that the set of letters of t is a subset of letters of s is as follows: aaaaabaacabaabbabcacaacb. Among them, those are lexicographically greater than abcacaacb. Out of those the lexicographically smallest is aca.

## Code

``````#include <bits/stdc++.h>
using namespace std;
set<char> st;

int main()
{
int n, k;
char s, t;
scanf("%d%d", &n, &k);
getchar();
scanf("%s", s);
int len = strlen(s);
for (int i = 0; i < len; i++)
st.insert(s[i]);
int tot = st.size();
set<char>::iterator it, it2;
if (n < k)
{
printf("%s", s);
char ch = *st.begin();
for (int i = 0; i < k - n; i++)
printf("%c", ch);
printf("\n");
}else
{
for (int i = len - 1 - (n - k); i >= 0; i--)
{
it = st.find(s[i]);
it++;
if (it == st.end())
{
s[i] = *st.begin();
}else
{
s[i] = *it;
break;
}
}
for (int i = 0; i < k; i++)
printf("%c", s[i]);
printf("\n");
}

return 0;
}
``````

Codeforces 940c