dynamic-programming

In order to find the minimal number of insertions required to convert a given string(s) to palindrome I find the longest common subsequence of the string(lcs_string) and its reverse. Therefore the number of insertions to be made is length(s) - length(lcs_string) What method should be employed to find the equivalent palindrome string on knowing the number of insertions to be made? For example : 1) azbzczdzez Number of insertions required : 5 Palindrome string : azbzcezdzeczbza Although multiple palindrome strings may exist for the same string but I want to find only one palindrome? Let S[i, j]

For instance, if I have a list [1,4,2,3,5,4,5,6,7,8,1,3,4,5,9,10,11] This algorithm should return [1,2,3,4,5,6,7,8,9,10,11]. To clarify, the longest list should run forwards. I was wondering what is an algorithmically efficient way to do this (preferably not O(n^2))? Also, I'm open to a solution not in python since the algorithm is what matters. Thank you. Here is a simple one-pass O(n) solution: s = [1,4,2,3,5,4,5,6,7,8,1,3,4,5,9,10,11,42] maxrun = -1 rl = {} for x in s: run = rl[x] = rl.get(x-1, 0) + 1 print x-run+1, 'to', x if run > maxrun: maxend, maxrun = x, run print range(maxend-maxrun

I'm reading The Algorithm Design Manual by Steven Skiena, and I'm on the dynamic programming chapter. He has some example code for edit distance and uses some functions which are explained neither in the book nor on the internet. So I'm wondering a) how does this algorithm work? b) what do the functions indel and match do? #define MATCH 0 /* enumerated type symbol for match */ #define INSERT 1 /* enumerated type symbol for insert */ #define DELETE 2 /* enumerated type symbol for delete */ int string_compare(char *s, char *t, int i, int j) { int k; /* counter */ int opt[3]; /* cost of the three

Given an array of integers, which can contain both +ve and -ve numbers. I've to maximize the product of any 3 elements of the array. The elements can be non-contiguous. Some examples: int[] arr = {-5, -7, 4, 2, 1, 9}; // Max Product of 3 numbers = -5 * -7 * 9 int[] arr2 = {4, 5, -19, 3}; // Max Product of 3 numbers = 4 * 5 * 3 I've tried solving it using Dynamic Programming , but I'm not getting the expected result. It is returning the result often involving the same number twice in the multiplication. So, for the array - {4, 2, 1, 9} , it is returning - 32 , which is 4 * 4 * 2 . Here's my

From Codechef : A string is considered balanced if and only if all the characters occur in it equal number of times. You are given a string S ; this string may only contain uppercase English letters. You may perform the following operation any number of times (including zero): Choose one letter in S and replace it by another uppercase English letter. Note that even if the replaced letter occurs in S multiple times, only the chosen occurrence of this letter is replaced. Find the minimum number of operations required to convert the given string to a balanced string. Example: For input: ABCB Here

Im studying dynamic programming and am looking to solve the following problem, which can be found here http://www.cs.berkeley.edu/~vazirani/algorithms/chap6.pdf : You are given a rectangular piece of cloth with dimensions X by Y, where X and Y are positive integers, and a list of n products that can be made using the cloth. For each product i in [1,n] you know that a rectangle of cloth of dimensions ai by bi is needed and that the final selling price of the product is ci. Assume that ai, bi, and ci are all positive integers. You have a machine that can cut any rectangular piece of cloth into

I am trying to make a program to find the nth Fibonacci number for 1 < n < 10^19. Here is my code using dynamic programming. memo = {} def fib(n): if n in memo: return memo[n] if n <= 2: f = 1 else: f = fib(n-1) + fib(n-2) memo[n]=f return f print fib(input()) % 1000000007 My code does not seem to work for large numbers. I get invalid response error. Any suggestions? Getting the Nth fibonacci number when N is 10^19 is not goign to work if you do it the naive way (at least i would guess it won't work). There's a much better way to do it. And this technique works with lots of series' like this.