Power for liftoff

Problem

The moon base runs on its solar panels: each day they collect some energy, and the lander needs goal units to lift off. The crew wants to know the fewest days in a row that are enough to collect it.

Write a function that takes energy, the list of what was collected each day (integers greater than 0; it may be empty), and goal, an integer greater than 0. Return an integer: the smallest number of consecutive days whose energy adds up to goal or more. The stretch can start on any day, and a single day counts if it is enough. If even all the days together fall short of the goal, or the list is empty, return 0.

With [2, 3, 1, 2, 4, 3] and goal 7, no single day reaches 7, but the last two add up to 4 + 3 = 7: the answer is 2.

Examples

• The example

  [2, 3, 1, 2, 4, 3], 7 → 2

• Three days at the end

  [1, 2, 3, 4, 5], 11 → 3

• One day is enough

  [3, 9, 2], 8 → 1

• Not enough

  [1, 2, 1], 10 → 0

Besides these, the challenge has hidden tests that are revealed when you submit your solution.

You start with this

def shortest_run(energy, goal):
    pass

function shortestRun(energy, goal) {
}