itertools.product is made for this: >>> import itertools >>> [“”.join(seq) for seq in itertools.product(“01”, repeat=2)] [’00’, ’01’, ’10’, ’11’] >>> [“”.join(seq) for seq in itertools.product(“01”, repeat=3)] [‘000’, ‘001’, ‘010’, ‘011’, ‘100’, ‘101’, ‘110’, ‘111’]
Here is a snippet that will do the work. #include <iostream> int main() { const int n = 10; const int k = 5; int combination[k] = {2, 5, 7, 8, 10}; int index = 0; int j = 0; for (int i = 0; i != k; ++i) { for (++j; j != combination[i]; … Read more
I start with a simple but not quite linear-time algorithm. We choose some value between array1[0]+array2[0] and array1[N-1]+array2[N-1]. Then we determine how many pair sums are greater than this value and how many of them are less. This may be done by iterating the arrays with two pointers: pointer to the first array incremented when … Read more
This is a well-studied problem of generating all k-subsets, or k-combinations, which can be easily done without recursion. The idea is to have array of size k keeping sequence of indices of elements from the input array (which are numbers from 0 to n – 1) in increasing order. (Subset then can be created by … Read more
How about using outer? But this particular function concatenates them into one character string. outer( c(“aa”, “ab”, “cc”), c(“aa”, “ab”, “cc”) , “paste” ) # [,1] [,2] [,3] #[1,] “aa aa” “aa ab” “aa cc” #[2,] “ab aa” “ab ab” “ab cc” #[3,] “cc aa” “cc ab” “cc cc” You can also use combn on … Read more
Here’s an ancient algorithm which is exact and doesn’t overflow unless the result is to big for a long long unsigned long long choose(unsigned long long n, unsigned long long k) { if (k > n) { return 0; } unsigned long long r = 1; for (unsigned long long d = 1; d <= … Read more
Very quick and dirty solution in Java: public class JavaApplication1 { public static void main(String[] args) { List<Integer> list = Arrays.asList(1, 3, 7, 6, 8, 3); for (Integer integer : list) { List<Integer> runList = new ArrayList<>(list); runList.remove(integer); Result result = getOperations(runList, integer, 348); if (result.success) { System.out.println(integer + result.output); return; } } } public … Read more
The classic algorithm works like this: Number the teams 1..n. (Here I’ll take n=8.) Write all the teams in two rows. 1 2 3 4 8 7 6 5 The columns show which teams will play in that round (1 vs 8, 2 vs 7, …). Now, keep 1 fixed, but rotate all the other … Read more
First convert the string to a set of unique characters and occurrence numbers e.g. BANANA -> (3, A),(1,B),(2,N). (This could be done by sorting the string and grouping letters). Then, for each letter in the set, prepend that letter to all permutations of the set with one less of that letter (note the recursion). Continuing … Read more
In C++ given the following routine: template <typename Iterator> inline bool next_combination(const Iterator first, Iterator k, const Iterator last) { /* Credits: Thomas Draper */ if ((first == last) || (first == k) || (last == k)) return false; Iterator itr1 = first; Iterator itr2 = last; ++itr1; if (last == itr1) return false; itr1 … Read more