Construct a string of length n using the first k letters of an alphabet

Find total number of palindromes of length N using K letters such that any prefix of length 2 to N-1 is not a palindrome. tried K*((K-1)^(Math.ceil((N-2)/2))) first place can hold K letters. second can K-1 except one that is at first place. Similarly for third. Since half of the places we need to fill with letters rest will follow the same to make it palindrome. But its not correct solution. Denote M = \ceilf(L) I wrote a Python script to check N and K. Playground: import functools import itertools import string @functools.lru_cache() def f(N, K): if N == 2: return K M = (N + 1)//2 return K**M - sum(K**(M - L) * f(L, K) for L in range(2, M+1)) def is_palindrome(s): return s[:(len(s) + 1)//2] == s[len(s)//2:][::-1] def brutal_f(N, K): alphabet = string.ascii_lowercase[:K] ret = 0 for t in itertools.product(alphabet, repeat=N): s = "".join(t) if is_palindrome(s) and not any(is_palindrome(s[:i]) for i in range(2, N)): ret += 1 return ret print(all( f(N, K) == brutal_f(N, K) for N in range(12) for K in range(4) ))

How would you approach a problem like this? If I were to make words of length 5 from the first 10 letters it would be 10^5 or 10x10x10x10x10, right? But how do I account for the repetition part? repeated does not mean that they have to be next to each other. It just means that the same letter exists more than once. First 10 alphabets of English language are : A,B,C,D,E,F,G,H,I,J. So, if you want to write the 5 letter string without any repetition, then, the ways should be : $$\binom*5!$$ And that's your answer!

Sometimes, rather than initializing the empty string, we need to initialize a string in a different way, vis., we may need to initialize a string with 1st N characters in English alphabets. This can have applications in competitive Programming. Let’s discuss certain ways in which this task can be performed. Method #1: Using join() + list comprehension This task can be performed with the combination of the above functions. The join function can be used to join the string and list comprehension can perform the task of the logic of adding N numbers. Output The string after construction : abcdefghijklmno Time Complexity: O(n) Auxiliary Space: O(n) Method #4:Using the bytearray() function. Step-by-step approach: • Initialize the value of N. • Create a range from 97 to 97+N using the range() function. • Convert the range to a bytearray using the bytearray() function. • Convert the bytearray to a string using the str() function and specifying the encoding as ‘utf-8’. • Print the result. The string after construction : abcdefghijklmno Time Complexity: O(N). Auxiliary Space: O(N). Method #5:Using reduce() and join(): Step-by-step approach: • Set the value of N to the desired length of the string. • Use a list comprehension to create a list of characters from ‘a’ to ‘a’ + N – 1. • Use reduce() to concatenate the characters in the list into a single string. • Print the resulting string. Output The string after construction: abcdefghijklmno Time Complexity: O(N). Auxiliary Space: O(N)...

• I have an alphabet: $ Until I realized that this solution wasn't robust enough-- it doesn't matter WHICH characters are the same, only that k characters are the same. Thanks so much in advance for your help. Let k, l, m be the # of occurrences for different letters in decreasing order. Using the multinomial distribution formula, to illustrate, for string AAABBC, Pr = $\frac$, k+l+m = N $\begingroup$ I'm not sure if this works for a few cases I've been using to test on. For example, if I generate a three-letter string (N=3), the probability that two of them are the same (k = 2) is $\frac)^3 = 1/9$ $\endgroup$

Virtual contest is a way to take part in past contest, as close as possible to participation on time. It is supported only ICPC mode for virtual contests. If you've seen these problems, a virtual contest is not for you - solve these problems in the archive. If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive. Never use someone else's code, read the tutorials or communicate with other person during a virtual contest. Given $$$n$$$ strings, each of length $$$2$$$, consisting of lowercase Latin alphabet letters from ' a' to ' k', output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and the $$$i$$$-th string and the $$$j$$$-th string differ in exactly one position. In other words, count the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$$$$$. The answer may not fit into 32-bit integer type, so you should use 64-bit integers like long long in C++ to avoid integer overflow. Input The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of strings. Then follows $$$n$$$ lines, the $$$i$$$-th of which containing a single string $$$s_i$$$ of length $$$2$$$, consisting of lo...

How would you approach a problem like this? If I were to make words of length 5 from the first 10 letters it would be 10^5 or 10x10x10x10x10, right? But how do I account for the repetition part? repeated does not mean that they have to be next to each other. It just means that the same letter exists more than once. First 10 alphabets of English language are : A,B,C,D,E,F,G,H,I,J. So, if you want to write the 5 letter string without any repetition, then, the ways should be : $$\binom*5!$$ And that's your answer!

Find total number of palindromes of length N using K letters such that any prefix of length 2 to N-1 is not a palindrome. tried K*((K-1)^(Math.ceil((N-2)/2))) first place can hold K letters. second can K-1 except one that is at first place. Similarly for third. Since half of the places we need to fill with letters rest will follow the same to make it palindrome. But its not correct solution. Denote M = \ceilf(L) I wrote a Python script to check N and K. Playground: import functools import itertools import string @functools.lru_cache() def f(N, K): if N == 2: return K M = (N + 1)//2 return K**M - sum(K**(M - L) * f(L, K) for L in range(2, M+1)) def is_palindrome(s): return s[:(len(s) + 1)//2] == s[len(s)//2:][::-1] def brutal_f(N, K): alphabet = string.ascii_lowercase[:K] ret = 0 for t in itertools.product(alphabet, repeat=N): s = "".join(t) if is_palindrome(s) and not any(is_palindrome(s[:i]) for i in range(2, N)): ret += 1 return ret print(all( f(N, K) == brutal_f(N, K) for N in range(12) for K in range(4) ))

Virtual contest is a way to take part in past contest, as close as possible to participation on time. It is supported only ICPC mode for virtual contests. If you've seen these problems, a virtual contest is not for you - solve these problems in the archive. If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive. Never use someone else's code, read the tutorials or communicate with other person during a virtual contest. Given $$$n$$$ strings, each of length $$$2$$$, consisting of lowercase Latin alphabet letters from ' a' to ' k', output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and the $$$i$$$-th string and the $$$j$$$-th string differ in exactly one position. In other words, count the number of pairs $$$(i, j)$$$ ($$$i < j$$$) such that the $$$i$$$-th string and the $$$j$$$-th string have exactly one position $$$p$$$ ($$$1 \leq p \leq 2$$$) such that $$$$$$. The answer may not fit into 32-bit integer type, so you should use 64-bit integers like long long in C++ to avoid integer overflow. Input The first line of the input contains a single integer $$$t$$$ ($$$1 \le t \le 100$$$) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of strings. Then follows $$$n$$$ lines, the $$$i$$$-th of which containing a single string $$$s_i$$$ of length $$$2$$$, consisting of lo...

• I have an alphabet: $ Until I realized that this solution wasn't robust enough-- it doesn't matter WHICH characters are the same, only that k characters are the same. Thanks so much in advance for your help. Let k, l, m be the # of occurrences for different letters in decreasing order. Using the multinomial distribution formula, to illustrate, for string AAABBC, Pr = $\frac$, k+l+m = N $\begingroup$ I'm not sure if this works for a few cases I've been using to test on. For example, if I generate a three-letter string (N=3), the probability that two of them are the same (k = 2) is $\frac)^3 = 1/9$ $\endgroup$

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