dynamic-programming

Given are n boxes in three dimension ( h , w , d ). The goal is to stack them on top of each other to have a maximum height (boxes can be rotated). Each box that you put on top should have a smaller dimension ( w , d ) than the one below. How can we do it with dynamic programming and greedy? This is the box stacking problem - problem 4 there. If you want to think about it yourself, think about how you can adapt the longest increasing subsequence algorithm for solving this. 来源： https://stackoverflow.com/questions/4511086/box-stacking-problem

Here is the problem that tagged as dynamic-programming (Given a number N, find the number of ways to write it as a sum of two or more consecutive integers) and example 15 = 7+8, 1+2+3+4+5, 4+5+6 I solved with math like that : a + (a + 1) + (a + 2) + (a + 3) + ... + (a + k) = N (k + 1)*a + (1 + 2 + 3 + ... + k) = N (k + 1) a + k (k+1)/2 = N (k + 1)*(2*a + k)/2 = N Then check that if N divisible by (k+1) and (2*a+k) then I can find answer in O(sqrt(N)) time Here is my question how can you solve this by dynamic-programming ? and what is the complexity (O) ? P.S : excuse me, if it is a duplicate

Given question: A string of parentheses is said to be balanced if the left- and right-parentheses in the string can be paired off properly. For example, the strings "(())" and "()()" are both balanced, while the string "(()(" is not balanced. Given a string S of length n consisting of parentheses, suppose you want to find the longest subsequence of S that is balanced. Using dynamic programming, design an algorithm that finds the longest balanced subsequence of S in O(n^3) time. My approach: Suppose given string: S[1 2 ... n] A valid sub-sequence can end at S[i] iff S[i] == ')' i.e. S[i] is a

I was practicing for an interview and came across this question on a website: A magical sub-sequence of a string S is a sub-sequence of S that contains all five vowels in order. Find the length of largest magical sub-sequence of a string S . For example, if S = aeeiooua , then aeiou and aeeioou are magical sub-sequences but aeio and aeeioua are not. I am a beginner in dynamic programming and am finding it hard to come up with a recursive formula for this. I did it with an iterative approach rather than recursive one. I started building solution similar to LIS (Longest Increasing Subsequence)

The following is a practice interview question that was given to me by someone, and I'm not sure what the best solution to this is: Given a set of ranges: (e.g. S = {(1, 4), (30, 40), (20, 91) ,(8, 10), (6, 7), (3, 9), (9, 12), (11, 14)} . And given a target range R (e.g. R = (3, 13) - meaning the range going from 3 to 13). Write an algorithm to find the smallest set of ranges that covers your target range. All of the ranges in the set must overlap in order to be considered as spanning the entire target range. (In this example, the answer would be {(3, 9), (9, 12), (11, 14)} . What is the best

The question is to find out sum of distances between every two nodes of BinarySearchTree given that every parent-child pair is separated by unit distance. It is to be calculated after every insertion. ex: ->first node is inserted.. (root) total sum=0; ->left and right node are inserted (root) / \ (left) (right) total sum = distance(root,left)+distance(root,right)+distance(left,right); = 1 + 1 + 2 = 4 and so on..... Solutions I came up with: Brute-force. Steps: perform a DFS and track all the nodes : O(n) . Select every two nodes and calculate : O(nC2)_times_O(log(n))=O(n2log(n)) distance