Majority and subarray in Python

Questions 1 to 5 of 6

1. 1. What does this program print? ``python def majority(nums, lo, hi): if lo == hi: return nums[lo] mid = lo + (hi - lo) // 2 left = majority(nums, lo, mid) right = majority(nums, mid + 1, hi) if left is not None and left == right: return left need = (hi - lo + 1) // 2 + 1 if left is not None and count_in(nums, lo, hi, left) >= need: return left if right is not None and count_in(nums, lo, hi, right) >= need: return right return None def count_in(nums, lo, hi, value): return nums[lo:hi + 1].count(value) def majority_of(nums): return majority(nums, 0, len(nums) - 1) a = [7, 7, 3, 7, 3, 3, 3, 7] print('A:', majority_of(a)) b = [4, 4, 1, 4, 1, 2, 4] print('B:', majority_of(b)) c = [9, 9, 5, 9, 5, 5, 5] print('C:', majority_of(c)) ``

   A: 7 B: 4 C: 5A: None B: 4 C: 5A: None B: None C: 5A: 3 B: 4 C: None
2. 2. What does this program print? ``python a = [2, -1, 3, -2] mid = 1 total = 0 best_left = float('-inf') for i in range(mid, -1, -1): total += a[i] best_left = max(best_left, total) total = 0 best_right = float('-inf') for j in range(mid + 1, len(a)): total += a[j] best_right = max(best_right, total) print(f'{best_left} + {best_right} = {best_left + best_right}') ``

   -1 + 3 = 22 + 3 = 51 + 1 = 21 + 3 = 4
3. 3. Why does majority call count_in on left and right before it returns either one?

   Because count_in sorts the range before it countsBecause a value that is the majority of one half can still be too rare across the whole rangeBecause None values must be removed from the halves firstBecause list.count only works on lists that have no repeated values
4. 4. For [5, -1, 5], what does a method return if it takes only the best of the two halves and never checks runs that cross the middle?

   Sum 9, which is correctSum 4, the sum of [5, -1]Sum 5, but the correct sum is 9Sum -1, the sum of [-1]
5. 5. This version skips the count and returns the left candidate straight away. Which input shows the bug? ``python def majority_pick(nums, lo, hi): if lo == hi: return nums[lo] mid = lo + (hi - lo) // 2 left = majority_pick(nums, lo, mid) right = majority_pick(nums, mid + 1, hi) if left is not None: return left return right ``

   [1, 2, 3]: it returns 1, but the correct answer is None[7]: it returns None instead of 7[1, 1, 2]: it returns 2, but the correct answer is 1[4, 4]: it returns None, but the correct answer is 4