A build configurator has n optional features, numbered 0 to n - 1. Some pairs can't be enabled together; they are listed in conflicts as [a, b].
Write countValidSets(n, conflicts) that returns how many sets of features (including the empty set) contain no conflicting pair.
n can be 20, so there are over a million sets: check each one in O(1) on average.
Input: n = 3, conflicts = [[0,1]] Output: 6
Of the 8 sets, only {0, 1} and {0, 1, 2} contain the conflict.
Input: n = 2, conflicts = [] Output: 4
No conflicts, so all 2^2 = 4 sets are valid.
1 <= n <= 200 <= conflicts.length <= 190a !== b, no pair listed twiceRun tries your code on the examples. Submit runs every test, including hidden edge cases, and solves the problem when they all pass.
Ctrl + Enter runs, Ctrl + Shift + Enter submits (⌘ on a Mac).
I'm here to help with #676 Conflict-Free Feature Sets. I give hints, never the solution, so you get to solve it yourself. Ask for a hint, select code to ask about it, or press Ask Tuto why on a failing test.
Your first 3 answers need no account. for free to keep asking.