Count the number of strings of length 5 over A = {a, b, c, d} that begins with a, ends with d and have exactly two b’s.
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Count the number of strings of length 5 over A = {a, b, c, d} that begins with a, ends with d and have exactly two b’s.
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- LAB RESTRICTIONS, PLEASE READ: Do not add any imports, the ones that you need will be given to you. You may not use any lists or list methods, or any while loops. Within your loops, you MUST NOT use any break or continue statements. Please also do not use try-except statements, you should be able to anticipate or prevent any errors from happening at all!Maximum Length Common Suffix Problem is defined as follows: Input: a list of L of n strings over the lowercase English alphabet of various lengths, maximum 10 characters Output: the longest common suffix among all the strings, or if they do not have a common suffix. A suffix is a list of characters attached at the end of a word; for example, the word "letter" has as suffixes the empty string, the word itself, and "r", "er", "ter", "tter", and "etter". Design a divide-and-conquer algorithm for it similar to merge sort algorithm and write a pseudocode for it. If you write multiple drafts, mark clearly your final draft. Hint: I expect your algorithm's time complexity ends up being O(n log n). For your convenience, the seudocode of decrease-by-half is given below. Examples: Input: n=4, L ={ "salty", "beauty", "meaty", "beatty") Output:"ty" Input: n=8, L-{"lot", "plot", "bylot", "dulot", "camelot", "effilot", "hairlot", "lelot"} Output="lot" Input: n=9, L={"lol", "tutu", "aba", "ddr",…please code in python You are given an array of strings arr. Your task is to construct a string from the words in arr, starting with the 0th character from each word (in the order they appear in arr), followed by the 1st character, then the 2nd character, etc. If one of the words doesn't have an ith character, skip that word.
- 2- Let a = [123; 2 3 4; 12 5], b = [ 2133 points ; 50 -2; 2 3 -1] prod = a* b. What is the value of prod?def longest_unique_substring(s: str) -> str:"""Given a string <s>, return the longest unique substring that occurs within<s>.A unique substring is a substring within <s> which DOES NOT have anyrepeating characters. As an example, "xd" is unique but "xxd" is not.If there are two equal length unique substrings within <s>, return the onethat starts first (i.e., begins at a smaller index).tips:In order to get your function to run fast, consider using a dictionary tostore the indexes of previously seen characters, from there, you canfollow a set of rules based on each new character you see to determinethe length of the longest unique substring seen so far.>>> longest_unique_substring('aab')'ab'""" RESTRICTIONS: - Do not add any imports and do it on python .Do not use recursion. Do not use break/continue.Do not use try-except statements.1. Let L = {w {a, b}* : w contains bba as a substring that starts in an odd numbered position in the string (where numbering starts at 1)}. For example, bbbbaaaab L, while bbbaaaab L. Show an NDFSM that accepts L
- def longest_unique_substring(s: str) -> str:"""Given a string <s>, return the longest unique substring that occurs within<s>.A unique substring is a substring within <s> which DOES NOT have anyrepeating characters. As an example, "xd" is unique but "xxd" is not.If there are two equal length unique substrings within <s>, return the onethat starts first (i.e., begins at a smaller index).tips:In order to get your function to run fast, consider using a dictionary tostore the indexes of previously seen characters, from there, you canfollow a set of rules based on each new character you see to determinethe length of the longest unique substring seen so far.>>> longest_unique_substring('aab')'ab' >>> longest_unique_substring('abcabcbb')'abc'""" RESTRICTIONS: - Do not add any imports and do it on python .Do not use recursion. Do not use break/continue.Do not use try-except statements.Convert NFA L2 = {set of all strings over (a, b) that starts with ‘a’ and ends with ‘b’ to its equivalent DFA.def longest_unique_substring(s: str) -> str:"""Given a string <s>, return the longest unique substring that occurs within<s>.A unique substring is a substring within <s> which DOES NOT have anyrepeating characters. As an example, "xd" is unique but "xxd" is not.If there are two equal length unique substrings within <s>, return the onethat starts first (i.e., begins at a smaller index).tips:In order to get your function to run fast, consider using a dictionary tostore the indexes of previously seen characters, from there, you canfollow a set of rules based on each new character you see to determinethe length of the longest unique substring seen so far.>>> longest_unique_substring('aab')'ab' >>> longest_unique_substring('abcabcbb')'abc'"""
- def substring_with_largest_sum(string: str) -> str:"""Given a string <string>, return the substring with the largest sum (ascalculated by sum_string()) that exists within <string>. If there aremultiple substrings with equally large sums, return the first one thatoccurs. The substring could just be the whole string itself, or it couldbe empty. To calculate the sum of the substring, please use the function sum_string()which you have implemented above as a helper. For example, given the string "321", there are 7 possible substrings:"", "3", "2", "1", "32", "21", "321". For each one of them, their sum canbe calculated as per the sum_string() function which you have alreadyimplemented. By using your understanding of loops, implement this functionto return the substring with the largest sum by checking all of thesubstrings of the input string. So, "32" evaluates to 1, "21" alsoevaluates to 1, "321" evaluates to 2, resulting in "3" being the largestpossible substring as it…The set of all strings beginning with b and no of 'a's is odd on {a,b} ba(ab'a+b)" bb'a(aba+b) bab (ab'a+b)" bb'a(aba+aab+baa+b) NoneThe program written for binary search, calculates the midpoint of the span as mid: (Low + High)/2. The program works well if the number of elements in the list is small (about 32,000) but it behaves abnormally when the number of elements is large. This can be avoided by performing the calculation as: =