感觉这边分享面试题的有点少，把我在国际站做过的题分享过来

Given two arrays representing an elevator problem, one array represents the weight of each person, and the second one represents the floor they are looking to reach.
For example:
[60, 20, 80, 180, 60]
and
[2, 2, 3, 3, 5]
The maximum weight the elevator can carry is 200.
Question: what's the least amount of distance the elevator has to travel? (From floor 1 to floor 2, distance will be 2-1=1)
Assume everyone starts on the first floor, and the elevator has to come back to the first floor to p ick up every group.
Assume people cannot skip the line. ie. 60 and 20 have to go before 80.
This part is relatively simple. Because I can't skip the line, just do a linear scan.
Follow up is much more difficult: what if people can skip the line? I wasn't sure how to solve it. Maybe with backtracking or DP?

这个题没想到什么好办法，回溯暴力解了，抛砖引玉

weight = [60, 20, 80, 180, 60]
level = [2,2,3,3,5]
weight, level = map(list, zip(*sorted(zip(weight, level))))
vis = [False] * len(weight)
target = 200

dis = 0
minDis = float("inf")

# maxH is max height of current group
# curr is sum weight of current group
# dis is total distance of current division
# k is remaning people
def backtrack(start, maxH, curr, dis, k):
    global minDis
    if k == 0:
        minDis = min(minDis, dis + (maxH - 1)*2)
        return

    for i in range(start, len(weight)):
        if vis[i]: continue
        if curr + weight[i] > target:
            backtrack(0, 0, 0, dis + (maxH - 1) * 2, k)
            break
        vis[i] = True
        backtrack(i+1, max(maxH, level[i]), curr + weight[i], dis, k - 1)
        vis[i] = False
        

backtrack(0, 0, 0, 0, 5)
print(minDis)