Bandit Best Arm Gap

Problem

Return the suboptimality gap of every arm: max(means) - means[a]. An empty list returns [].

Hint

Small gaps make arms harder to distinguish, so regret bounds depend on them.

case 1

in: [[0.2,0.5,0.7]]

expected: [0.49999999999999994,0.19999999999999996,0]

case 2

in: [[1]]

expected: [0]

case 3

in: [[0.3,0.3]]

expected: [0,0]

case 4

in: [[]]

expected: []

Starter codePython

def bandit_best_arm_gap(means):
    # Your code here
    pass

4 test cases · run your code against them in the editor