You have given two strings 'A' and ‘B’ of length ‘N’ each which only consists of lowercase English letters. You have also given an integer ‘K’.Your task is to determine if it is possible to convert string ‘A’ into ‘B’ after performing two types of operations on it:1. Choose an index i (1 <= i <= N - 1) and swap A[i] and A[i+1].2. Choose an index i (1 <= i <= N - K + 1) and if A[i], A[i+1],. . . . , A[i+K-1] all are equal to some character x (x != ‘z’), then you can replace each one with the next character (x + 1) , i.e. ‘a’ is replaced by ‘b’, ‘b’ is replaced by ‘c’ and so on.Note:You are allowed to perform any operation any number of times(possibly zero) only on string 'A'.

Given 2 binary strings A and B of the same length, find the minimum number of operations needed to change string A to B. The operations are defined as follows:AND Operation:Choose a pair of indices i and j such that i != j and perform the following sequence of operations.result = Ai & AjAi = result & AiAj = result & AjOR Operation:Choose a pair of indices i and j such that i != j and perform the following sequence of operations.result = Ai | AjAi = result | AiAj = result | AjXOR Operation:Choose a pair of indices i and j such that i != j and perform the following sequence of operations.result = Ai ^ AjAi = result ^ AiAj = result ^ AjInput FormatThe first line of input contains T - the number of test cases. It's followed by 2T lines, the first line contains string A and the second line contains string B.Output FormatFor each test case, if conversion is possible, print "YES" followed by the minimum number of operations required to convert string A to string B, otherwise print "NO", separated by a new line.Constraints1 <= T <= 10001 <= len(A)==len(B) <= 103ExampleInput210101010101011OutputYES 2YES 1ExplanationTest-Case 1We can choose index-1 and index-2 of string A and perform the XOR operation. The resultant string will look like 110. After that, we can choose index-0 and index-2 of the new string A and perform the AND operation. The resultant string will look like 010.

Compatible Strings Two strings A and B comprising of lower case English letters are compatible if they are equal or can be made equal by following this step any number of times:Select a prefix from the string A, and increase the alphabetical value of all the characters in the prefix by the same valid amount. For example, if the string is ashzb and we select the prefix ash then we can convert it to bti by increasing the alphabetical value by 1. If we select the prefix ashzb, then it will be converted into btiac. Your task is to determine if given strings A and B are compatible. Input formatRead N.Read N pairs of strings, each in a separate line . Output formatFor each pair of input strings,print YES if string A can be converted to string B, otherwise print NO. SAMPLE INPUT3abacacdbdagodzjneaaqlmcmmo SAMPLE OUTPUTYESNONO Explanation for first pair of strings:The string abaca can be converted to bcbda in one move and to cdbda in the next move. No restriction on the number of moves.

You are given two strings, A and B. Answer, what is the smallest number of operations you need totransform A to B?Operations are:Delete one letter from one of stringsInsert one letter into one of stringsReplace one of letters from one of strings with another letterInputT - number of test casesFor each test case:String AString BBoth strings will contain only uppercase characters and they won't be longer than 2000 characters. There will be 10 test cases in data set.OutputFor each test case, one line, minimum number of operations.ExampleInput:1FOODMONEYOutput:4

You are given a collection of N non-empty strings, denoted by S 1 , S 2 , … , S n . Then you are given N - 1 operations which you execute in the order they are given. The i t h operation is has the following format: ‘ a b ’ ( 1 -based indexing, without the quotes), which means that you have to make the following changes: S a = S a + S b , i.e. concatenate a t h string and b t h string and store the result in a t h string, S b = "", i.e. make the b t h string empty, after doing the previous step. You are ensured that after the i t h operation, there will be no future operation that will be accessing S b . Given these operations to join strings, print the last string that will remain at the end of this process. Input The first line contains an integer N ( 1 ≤ N ≤ 10 5 ) denoting the number of strings given. Each of the next N lines contains a string denoting the S i . All the characters in the string S i are lowercase alphabets from ‘a’ to ‘z’. The total number of characters over all the strings is at most 10 6 , i.e ∑ i = 1 N | S i | ≤ 10 6 , where | S i | denotes the length of the i t h string. After these N strings, each of the next N - 1 lines contain two integers a and b , such that a ≠ b and 1 ≤ a , b ≤ N denoting the i t h operation. Output Print the last string which remains at the end of the N - 1 operations. Warning The I/O files are large. Please use fast I/O methods. can you write a java solution, and pick the one that is the most efficient time wise

You are given a string word and an integer k. We consider word to be k-special if |freq(word[i]) - freq(word[j])| <= k for all indices i and j in the string. Here, freq(x) denotes the frequency of the character x in word, and |y| denotes the absolute value of y. Return the minimum number of characters you need to delete to make word k-special. Example 1: Input: word = "aabcaba", k = 0 Output: 3 Explanation: We can make word 0-special by deleting 2 occurrences of "a" and 1 occurrence of "c". Therefore, word becomes equal to "baba" where freq('a') == freq('b') == 2. Example 2: Input: word = "dabdcbdcdcd", k = 2 Output: 2 Explanation: We can make word 2-special by deleting 1 occurrence of "a" and 1 occurrence of "d". Therefore, word becomes equal to "bdcbdcdcd" where freq('b') == 2, freq('c') == 3, and freq('d') == 4. Example 3: Input: word = "aaabaaa", k = 2 Output: 1 Explanation: We can make word 2-special by deleting 1 occurrence of "b". Therefore, word becomes equal to "aaaaaa" where each letter's frequency is now uniformly 6. Constraints: 1 <= word.length <= 105 0 <= k <= 105 word consists only of lowercase English letters. c++