Alphabetical Strings_string s of length n is called

B. Alphabetical Strings

time limit per test2 seconds
memory limit per test512 megabytes
inputstandard input
outputstandard output
A string s of length n (1≤n≤26) is called alphabetical if it can be obtained using the following algorithm:
删除线格式
first, write an empty string to s (i.e. perform the assignment s := “”);
then perform the next step n times;
at the i-th step take i-th lowercase letter of the Latin alphabet and write it either to the left of the string s or to the right of the string s (i.e. perform the assignment s := c+s or s := s+c, where c is the i-th letter of the Latin alphabet).
In other words, iterate over the n first letters of the Latin alphabet starting from ‘a’ and etc. Each time we prepend a letter to the left of the string s or append a letter to the right of the string s. Strings that can be obtained in that way are alphabetical.

For example, the following strings are alphabetical: “a”, “ba”, “ab”, “bac” and “ihfcbadeg”. The following strings are not alphabetical: “z”, “aa”, “ca”, “acb”, “xyz” and “ddcba”.

From the given string, determine if it is alphabetical.

Input
The first line contains one integer t (1≤t≤104) — the number of test cases. Then t test cases follow.

Each test case is written on a separate line that contains one string s. String s consists of lowercase letters of the Latin alphabet and has a length between 1 and 26, inclusive.

Output
Output t lines, each of them must contain the answer to the corresponding test case. Output YES if the given string s is alphabetical and NO otherwise.

You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).

Example
inputCopy
11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba
outputCopy
YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO
Note
The example contains test cases from the main part of the condition.

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<iostream>
#include<queue>
using namespace std;
#define inf 99999999
int dist[1001][1001],b[1001];
int n;
int main() 
{
  int t;
  cin>>t;
  while(t--)
  {
  string s;
  cin>>s;
  int l,r;
  l=0;
  r=s.size()-1;
  while(l<=r)
  {
  	if((s[l]-'a')==r-l)
  	l++;
  	else if((s[r]-'a')==r-l)
  	r--;
  	else
  	break;
  }
  if(l>r)
  cout<<"YES"<<endl;
  else
  cout<<"NO"<<endl;
  }
    return 0;
}
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