Stacks and queues · stretch

Next greater number

Write next_greater(nums) returning a list where each position holds the first number to its right that is bigger, or -1 if there is none. [2, 1, 3] gives [3, 3, -1].

Keep a stack of positions still waiting for their answer. Each new number answers every waiting position smaller than it.

right answers
arguments left as they should be
fast enough at scale

def next_greater(nums):
    result = []
    for i in range(len(nums)):
        answer = -1
        for j in range(i + 1, len(nums)):
            if nums[j] > nums[i]:
                answer = nums[j]
                break
        result.append(answer)
    return result

Run adds print(next_greater([2, 1, 3])) after your code, to try it.

Stuck on the idea rather than the code? The Waiting in a stack, waiting in a queue lesson walks through it.