java,algorithm,divide-and-conquer

The optimum scheme for n>=4 people to share all n pieces of information is 2n-4 as shown in this paper. The divide and conquer approach to this problem is illustrated as follows: For four persons A, B, C and D, say, take conversations AB, and CD, followed by AC and...

c++,arrays,algorithm,math,divide-and-conquer

The approach: Let max_ending_here be an array, whose element max_ending_here[i] denotes the maximum sum of subarrays (could be empty) that ends just before (not included) index i. To calculate it, use the same approach as it in your function maxSubArraySum. The time complexity is O(n), and space complexity is O(n)....

c++,algorithm,divide-and-conquer

It seems to me that the concepts of using recursion to find the solution and storing a value for comparison to the next iteration are opposing concepts in this case. Storing a variable for comparison in a recursive function seems like it would be a bit of a mess (and...

c++,c,arrays,divide-and-conquer

This is my solution for your above problem. But i have modified a few things in your code. #include<iostream> using namespace std; #define NMAX 10 struct something{ char *name; //replaced with char pointer so that i can save values the way i have done }a[NMAX]; int myFunction(char *choice, int l,int...

java,recursion,brute-force,divide-and-conquer,floor

You are trying to solve an integer equation: floor(n/7)*1+floor(n/8)*1+floor(n/10)*2 = 19 You can remove the floor function and solve for n and get a lower bound and upper bound, then search between these two bounds. Solving the following equation: (n/7)*1+(n/8)*1+(n/10)*2 = 19 n=19/(1/7+1/8+2/10) Having found n, which range of value...

c++,algorithm,matrix,divide-and-conquer

Here is the corrected source code. I have added some comments so you can follow it. Furthermore, I switched to a dynamic array of the correct size. #include<fstream> #include<conio.h> #include<iostream> using namespace std; int countOccurences(int** values, int min1, int min2, int max1, int max2, int searchValue) { if (min1 >...

algorithm,recursion,divide-and-conquer

In short: Let A be an array of length n. You want to compute the max subarray of A sou you call Find-Maximum-Subarray(A, 0, n-1). Now try to make the problem easier: Case. high = low: In this case the Array has only 1 Element so the solution is simple...

arrays,algorithm,divide-and-conquer,subset-sum,sub-arrays

here is brute-force java solution you can use, note - complexity of this solution is O(3^N), which is very very slow /** * Returns absolute difference between 3 values in array */ static int getdiff(final int s[]) { return Math.abs(s[0] - s[1]) + Math.abs(s[1] - s[2]) + Math.abs(s[2] - s[0]);...

node.js,algorithm,divide-and-conquer

Under the constraints you specified, you could create a hash table containing the ID's you are looking for as keys, with all values initialized to false. Then, read the table chunk by chunk and for each item in the table, look it up in the hash table. If found, mark...

python,algorithm,divide-and-conquer,clrs

Two really small mistakes: Your list i.e., t is of length 16. This means last index is 15. Thus call maxSubarray(t,0,15) not maxSubarray(t,0,16) while (j <= high). Loop until j<= high not until j<high Also with these two fixes, you don't need to set any default values to max_right or...

c++,dynamic-programming,divide-and-conquer

I answered a similar question 10 minutes ago. The ides here is the same: loop over sorted ranges. Here is the same code as in the other answer adapted to your problem (I just replaced the equality by a smaller-than relation): auto find_pairs(std::vector<int>& arr1, std::vector<int>& arr2, int diff) { std::vector<std::pair<int,int>...

algorithm,data-structures,divide-and-conquer

If a1 > an, then exchange a1 and an. this is a constant operation - so O(1) Run func on {a1, a2, ... ,a2n/3}. You invoke the array recursively on 2n/3 of it, so T(2n/3) Run func on {an/3, a(n/3)+1, ... ,an}. Run func on {a1, a2, ... ,a2n/3}. Similar...

algorithm,optimization,dynamic-programming,divide-and-conquer

Let's make a couple of simple observation: Any balanced subsequence can be represented as A1 X A2 Y, where A1 and A2 are two matched brackets((), [] or {}), X and Y are balanced subsequences(they can be empty). It is true because there is a leftmost bracket in any balanced...

algorithm,graph,divide-and-conquer

If you use an adjacency matrix to represent the graph (let's say M), M^k is the matrix that denotes the number of paths of size k between every pair of nodes. You can calculate M^k using O(log k) matrix multiplications (each taking O(V^3) time). It's a divide and conquer algorithm...

I wouldn't go for third-party library, because the algorithm implementation there would strongly depend on the way the graph is represented (there are plenty of options, like two-dimensional array, list of lists or some other custom class). If I were you, I would implement the algorithm myself. First, you need...

algorithm,time-complexity,binary-search,divide-and-conquer

You are correct. Unless the array is sorted, you still have to examine every element in each half (and each quarter and each eighth as you recur). The only way it can be O(log N) is if you can discard half the search space each recursion level (such as searching...

performance,algorithm,big-o,divide-and-conquer

I think all the answers to your question might come from the Master Theorem It kind of tell you what would be your complexity for almost any divide and conquer solution you have, and yes, it has to do everything with recursion trees, by playing with the parameters you will...

r,algorithm,recursion,divide-and-conquer

It is quite awkward to implement that using D&C. However, let's try. A sequence of length 1 is always sorted. If x is of length >= 2, then divide it into 2 non-empty subsequences x1 and x2. Then x is sorted if x1 is sorted, x2 is sorted, and the...

algorithm,divide-and-conquer,master-theorem

For the recurrence T(n) = 4T(n/2) + n^2 * log n None of the three cases applies because there does not exist an e such that log n = Ω(n^e) or log n = O(n^(-e))...

Don't use postincrement and similar in function calls. Here's the problem condensed down to a really simple piece of code #include <iostream> int test(int n){ if(n == 1){ std::cout << "Function called!"; return test(n++); }else{ return 0; } } int main() { test(1); return 0; } Before you run this,...

algorithm,search,dynamic-programming,binary-search,divide-and-conquer

Drop first egg from floors 1, 3, 6, ... (the partials sums of 1, 2, 3, ...). Then do linear search between last two floors.

python,list,divide-and-conquer

Sort your array and check pairwise for the smallest distance. The sort can be done using a divide and conquer algorithm, like merge_sort in O(nlog(n)) time. For instance you could piggyback on merge sort like this : # For convenience of representation, assume pairs to be defined as # tuples...

python,performance,divide-and-conquer

Without any additional information on the distribution of the values in the list, it is not possible to achieve any algorithmic complexity of less than O(x), where x is the number of elements in the list. Logically, if the dataset is random, such as a brownian noise, a pit can...