CF1673B.A Perfectly Balanced String?
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题目描述
Let's call a string s perfectly balanced if for all possible triplets (t,u,v) such that t is a non-empty substring of s and u and v are characters present in s , the difference between the frequencies of u and v in t is not more than 1 .
For example, the strings "aba" and "abc" are perfectly balanced but "abb" is not because for the triplet ("bb",'a','b'), the condition is not satisfied.
You are given a string s consisting of lowercase English letters only. Your task is to determine whether s is perfectly balanced or not.
A string b is called a substring of another string a if b can be obtained by deleting some characters (possibly 0 ) from the start and some characters (possibly 0 ) from the end of a .
输入格式
The first line of input contains a single integer t ( 1≤t≤2⋅104 ) denoting the number of testcases.
Each of the next t lines contain a single string s ( 1≤∣s∣≤2⋅105 ), consisting of lowercase English letters.
It is guaranteed that the sum of ∣s∣ over all testcases does not exceed 2⋅105 .
输出格式
For each test case, print "YES" if s is a perfectly balanced string, and "NO" otherwise.
You may print each letter in any case (for example, "YES", "Yes", "yes", "yEs" will all be recognized as positive answer).
输入输出样例
输入#1
5 aba abb abc aaaaa abcba
输出#1
YES NO YES YES NO
说明/提示
Let ft(c) represent the frequency of character c in string t .
For the first testcase we have
t ft(a) ft(b) a 1 0 ab 1 1 aba 2 1 b 0 1 ba 1 1 It can be seen that for any substring t of s , the difference between ft(a) and ft(b) is not more than 1 . Hence the string s is perfectly balanced.For the second testcase we have
t ft(a) ft(b) a 1 0 ab 1 1 abb 1 2 b 0 1 bb 0 2 It can be seen that for the substring t=bb , the difference between ft(a) and ft(b) is 2 which is greater than 1 . Hence the string s is not perfectly balanced.For the third testcase we have
t ft(a) ft(b) ft(c) a 1 0 0 ab 1 1 0 abc 1 1 1 b 0 1 0 bc 0 1 1 c 0 0 1 It can be seen that for any substring t of s and any two characters u,v∈{a,b,c} , the difference between ft(u) and ft(v) is not more than 1 . Hence the string s is perfectly balanced.