Balanced Suffix | codechef| starter 78

Problem

You’re given a string S of length N and an integer K.

Let C denote the set of all characters in S. The string S is called good if, for every suffix of S:

  • The difference between the frequencies of any two characters in C does not exceed K.
    In particular, if the set C has a single element, the string S is good.

Find whether there exists a rearrangement of S which is good.
If multiple such rearrangements exist, print the lexicographically smallest rearrangement.
If no such rearrangement exists, print −1−1 instead.

Note that a suffix of a string is obtained by deleting some (possibly zero) characters from the beginning of the string. For example, the suffixes of S=abca are {a,ca,bca,abca}.

Input Format

  • The first line of input will contain a single integer �T, denoting the number of test cases.
  • Each test case consists of two lines of input.
    • The first line of each test case contains two space-separated integers �N and �K — the length of the string and the positive integer as mentioned in the statement, respectively.
    • The next line consists of a string �S of length �N containing lowercase English alphabets only.

Output Format

For each test case, output on a new line the lexicographically smallest good rearrangement of �S.

If no such rearrangement exists, print −1−1 instead.

Constraints

  • 1≤T≤2000
  • 1≤≤N≤105
  • 1≤�≤�1≤KN
  • S consists of lowercase english alphabets only.
  • The sum of �N over all test cases won’t exceed 2⋅1052⋅105.

Sample 1:

Input

Output

4
3 1
aaa
4 2
baba
4 1
babb
7 2
abcbcac
aaa
aabb
-1
abcabcc

Explanation:

Test case 11: Since �={�}C={a}, the string �S is good.

Test case 22: The set �={�,�}C={a,b}. Consider the rearrangement ����aabb. Let ��fa​ and ��fb​ denote the frequencies of �a and �b respectively:

  • In suffix �b, ��=1fb​=1 and ��=0fa​=0. Thus, ∣��−��∣=1≤�∣fb​−fa​∣=1≤K.
  • In suffix ��bb, ��=2fb​=2 and ��=0fa​=0. Thus, ∣��−��∣=2≤�∣fb​−fa​∣=2≤K.
  • In suffix ���abb, ��=2fb​=2 and ��=1fa​=1. Thus, ∣��−��∣=1≤�∣fb​−fa​∣=1≤K.
  • In suffix ����aabb, ��=2fb​=2 and ��=2fa​=2. Thus, ∣��−��∣=0≤�∣fb​−fa​∣=0≤K.

Thus, the rearrangement ����aabb is good. It is also the lexicographically smallest rearrangement possible of string �S.

Test case 33: It can be proven that there exists no rearrangement of �S which is good.

Test case 44: The set �={�,�,�}C={a,b,c}. Consider the rearrangement �������abcabcc. Let ��,��,fa​,fb​, and ��fc​ denote the frequencies of �,�,a,b, and �c respectively:

  • In suffix �c, ��=0,��=0,fa​=0,fb​=0, and ��=1fc​=1. Thus, ∣��−��∣,∣��−��∣,∣fb​−fa​∣,∣fc​−fb​∣, and ∣��−��∣∣fa​−fc​∣ are all less than or equal to �=2K=2.
  • In suffix ��cc, ��=0,��=0,fa​=0,fb​=0, and ��=2fc​=2. Thus, ∣��−��∣,∣��−��∣,∣fb​−fa​∣,∣fc​−fb​∣, and ∣��−��∣∣fa​−fc​∣ are all less than or equal to �=2K=2.
  • In suffix ���bcc, ��=0,��=1,fa​=0,fb​=1, and ��=2fc​=2. Thus, ∣��−��∣,∣��−��∣,∣fb​−fa​∣,∣fc​−fb​∣, and ∣��−��∣∣fa​−fc​∣ are all less than or equal to �=2K=2.
  • In suffix ����abcc, ��=1,��=1,fa​=1,fb​=1, and ��=2fc​=2. Thus, ∣��−��∣,∣��−��∣,∣fb​−fa​∣,∣fc​−fb​∣, and ∣��−��∣∣fa​−fc​∣ are all less than or equal to �=2K=2.
  • In suffix �����cabcc, ��=1,��=1,fa​=1,fb​=1, and ��=3fc​=3. Thus, ∣��−��∣,∣��−��∣,∣fb​−fa​∣,∣fc​−fb​∣, and ∣��−��∣∣fa​−fc​∣ are all less than or equal to �=2K=2.
  • In suffix ������bcabcc, ��=1,��=2,fa​=1,fb​=2, and ��=3fc​=3. Thus, ∣��−��∣,∣��−��∣,∣fb​−fa​∣,∣fc​−fb​∣, and ∣��−��∣∣fa​−fc​∣ are all less than or equal to �=2K=2.
  • In suffix �������abcabcc, ��=2,��=2,fa​=2,fb​=2, and ��=3fc​=3. Thus, ∣��−��∣,∣��−��∣,∣fb​−fa​∣,∣fc​−fb​∣, and ∣��−��∣∣fa​−fc​∣ are all less than or equal to �=2K=2.

Thus, the rearrangement �������abcabcc is good. It is also the lexicographically smallest good rearrangement of string �S.

Solution:

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solution

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