dynamic-programming

Given an array or object with n keys, I need to find all combinations with length x . Given X is variable. binomial_coefficient(n,x) . Currently I'm using this: function combine(items) { var result = []; var f = function(prefix, items) { for (var i = 0; i < items.length; i++) { result.push(prefix + items[i]); f(prefix + items[i], items.slice(i + 1)); } } f('', items); return result; } var combinations = combine(["a", "b", "c", "d"]); The output is: ["a", "ab", "abc", "abcd", "abd", "ac", "acd", "ad", "b", "bc", "bcd", "bd", "c", "cd", "d"] So if I want the binomial coefficient x=3 from n=4 I

I came across this problem: http://www.geeksforgeeks.org/dynamic-programming-set-7-coin-change/ Given a value N, if we want to make change for N cents, and we have infinite supply of each of S = { S1, S2, .. , Sm} valued coins, how many ways can we make the change? The order of coins doesn’t matter. For example, for N = 4 and S = {1,2,3}, there are four solutions: {1,1,1,1},{1,1,2},{2,2},{1,3}. So output should be 4. For N = 10 and S = {2, 5, 3, 6}, there are five solutions: {2,2,2,2,2}, {2,2,3,3}, {2,2,6}, {2,3,5} and {5,5}. So the output should be 5. I came up with the solution: //

I have looked through various questions on the site and I haven't managed to find anything which implements this by the following reasoning (so I hope this isn't a duplicate). The problem I'm trying to solve via a C program is the following: As the programmer of a vending machine controller your are required to compute the minimum number of coins that make up the required change to give back to customers. An efficient solution to this problem takes a dynamic programming approach, starting off computing the number of coins required for a 1 cent change, then for 2 cents, then for 3 cents, until

I've got several types of coins, each have weight (wi) and cost (ci). So I've got to make a knapsack with weight==W and (!) minimum cost of coins in it. I can`t make recurrence relation to use DP. Just formulate the usual recurrence relation... Designate the minimum cost achievable with total weight k as Min_cost(k). Bootstrap the memoization with: Min_cost(0) = cost of empty set = 0 Then solve for increasing values of k using: Min_cost(i+1) = min [Min_cost(i) + min [ci, for all items with wi = 1], Min_cost(i-1) + min [ci, for all items with wi = 2], Min_cost(i-2) + min [ci, for all items with

I'm trying to understand the solution provided in a book to the following question: "A child is running up a staircase with n steps and can hop either 1 step, 2 steps or 3 steps at a time. Implement a method to count how many possible ways the child can run up the stairs." The book's solution is as follows, stemming from the fact that "the last move may be a single step hop from n - 1, a double step hop from step n - 2 or a triple step hop from step n - 3" public static int countWaysDP(int n, int[] map) { if (n < 0) return 0; else if (n == 0) return 1; else if (map[n] > -1) return map[n]; else