You can build a backtracking based solution to formulate all the possible sub-sequence for the given array. Sharing an ideone link for the same: https://ideone.com/V0gCDX all = [] def gen(A, idx = 0, cur = []): if idx >= len(A): if len(cur): all.append(cur) return gen(A, idx + 1, list(cur)) incl = list(cur) incl.append(A[idx]) gen(A, idx … Read more
For what it’s worth, the function is looping through the given array [9,8,6,2] and adding the numbers together. It’s essentially a sum of the values of the given array. The var s is initially zero (0) (int s = 0). It loops through each value in the array and adds it to itself (s = … Read more
To access the array, you need to use array index. Instead of using @messages_test = options[‘Messages’][‘IBMessageID’].to_s you need to use, to access to first element the arrayoptions[‘Messages’] array, below code @messages_test = options[‘Messages’][0][‘IBMessageID’].to_s You can iterate the array, if you wish, by using options[‘Messages’] each do |item| puts item[“IBMessageID”] # for illustration end
Three problems with your request: my declares a lexical variable at compile-time, so asking to pass a name to my at run-time makes no sense. Symbolic references can’t be used to access lexical variables. It’s a stupid thing to do.
You can first slice your array with a range. Then filter the sliced array by checking if the elements can be type cast to string. If the count of filtered array is equal to your range then you can say first n elements are of type String. This is test code I have tried: let … Read more
use File::Find::Rule; my @array = File::Find::Rule->directory->in(‘/home’); File::Find::Rule
Pop k-1 elements in k-1 operations. Now for the last operation, either you can pop one more element, or can push one element from the popped elements. So compare both the cases, if you pop element in kth operation, the element on top should be more than all first k-1 popped elements, otherwise for kth … Read more
var dictionaryArray: [[String : Any?]] = [[:]] After that you can append as many dictionaries as you want