1076A Minimizing the String

迷南。 2022-04-17 04:39 111阅读 0赞

A. Minimizing the String

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a string ss consisting of nn lowercase Latin letters.

You have to remove at most one (i.e. zero or one) character of this string in such a way that the string you obtain will be lexicographically smallest among all strings that can be obtained using this operation.

String s=s1s2…sns=s1s2…sn is lexicographically smaller than string t=t1t2…tmt=t1t2…tm if n<mn<m and s1=t1,s2=t2,…,sn=tns1=t1,s2=t2,…,sn=tn or there exists a number pp such that p≤min(n,m)p≤min(n,m) and s1=t1,s2=t2,…,sp−1=tp−1s1=t1,s2=t2,…,sp−1=tp−1 and sp<tpsp<tp.

For example, "aaa" is smaller than "aaaa", "abb" is smaller than "abc", "pqr" is smaller than "z".

Input

The first line of the input contains one integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the length of ss.

The second line of the input contains exactly nn lowercase Latin letters — the string ss.

Output

Print one string — the smallest possible lexicographically string that can be obtained by removing at most one character from the string ss.

Examples

input

Copy

3
    aaa

output

Copy

input

Copy

5
    abcda

output

Copy

abca

Note

In the first example you can remove any character of ss to obtain the string "aa".

In the second example "abca" < "abcd" < "abcda" < "abda" < "acda" < "bcda".

**题意：给出一个长度为n的字符串，最多删除一个字符，让整体字符串字典序最小。**

**题解：要让字符串字典序最小，一定是前面越小越好，所以从前面一个一个找，如果前面的字符大于后面的就是删除前面的**。

c++：

#include<bits/stdc++.h>
    using namespace std;
    int main()
    {
        int n;
        string s;
        cin>>n>>s;
        for(int i=0; i<n; i++)
            if(s[i]>s[i+1])
            {
                s.erase(i,1);///erase(i,1)删除第i位开始的1个字符
                break;
            }
        cout<<s<<endl;
        return 0;
    }

python:

n=int(input())
    s=input()
    for i in range(n-1):
        if s[i]>s[i+1]:
            print(s[0:i]+s[i+1:n])
            exit()
    print(s[0:n-1])