CODEWITHSUNDEEP
pythonalgorithmdynamic-programming
RiesenChicken
RiesenChicken

Reputation: 83

Minimum total container size Python

I'm practising some Amazon interview coding exercises and I'm stuck with this exercise.

In a nutshell, given a list l and an integer d, I have to split the list into d parts such that the sum of the max of each part is the smallest, for example, with l = [10,2,20,5,15,10,1], d = 3 should return 31 as the best option is [10,2,20,5,15],[10],[1] -> 20+10+1 = 31.

Link to the problem: Problem

I thought about splitting the list in 2 and calling recursively with the first part and d-1, and taking the max of the second part (this second part will start having one element and we will add an element to this part in each iteration).

I did this and I'm getting the "maximum recursion depth exceeded while calling a Python object" error. I know the solution to this problem can be found easily but I want to understand why is this approach not working.

def getmin(l, d):
    # base case, if the list has d elements each element will be a sublist
    if len(l)==d: 
        return sum(l)
    else:
        minn = float("inf")
        for i in range(1, len(l)-d+1):
            val = getmin(l[0:-i], d-1) + max(l[-i:])
            if minn > val:
                minn = val
    return minn

Upvotes: 3

Views: 1180

Answers (2)

ggorlen
ggorlen

Reputation: 56993

Your list l is not a 2d list as your written example, so your algorithm seems to be having trouble keeping the result list that stores the cuts you made to get to a particular point and the input list straight. It's much easier to keep a separate list and accumulate the result onto it, treating l as read-only. Not to mention, copying it per recursive call is painful from a performance standpoint.

Before going further, Python recommends never using l as a variable name. Prefer L, nums or lst instead--it's really hard to differentiate 1, i and l. I'll use L for the rest of the post.

A brute force solution similar to yours would be to run a counter from 0 to len(L) and use recursion to try every possibility at every node. At each i, we can either append the current number L[i] to the last bucket or start a new bucket with L[i] as the first element. A small optimization is to only keep the max per bucket instead of all the elements.

Now that you have a brute-force solution, you can slap on a cache to memoize repeated work.

from functools import cache
import random

def min_sum_of_max_elems_from_d_cuts(L, d):
    assert len(L) >= d

    @cache
    def find_best_cut(i, cuts):
        if len(cuts) == d and i == len(L):
            return sum(cuts)
        elif len(cuts) > d or i >= len(L):
            return float("inf")

        best = float("inf")

        if cuts: # try extending the current cut by L[i]
            new_cuts = cuts[:-1] + (max(cuts[-1], L[i]),)
            best = min(best, find_best_cut(i + 1, new_cuts))

        #                try making a new cut at L[i]
        return min(best, find_best_cut(i + 1, cuts + (L[i],)))

    return find_best_cut(i=0, cuts=tuple())

if __name__ == "__main__":
    tests = [
        dict(L=[10,2,20,5,15,10,1], d=3, result=31),
        dict(L=[10,2,20,5,15,10,1], d=5, result=43),
        dict(L=[5,4,2,4,3,4,5,4], d=4, result=16),
        dict(L=[22,12,1,14,17], d=2, result=39),
    ]

    for test in tests:
        assert min_sum_of_max_elems_from_d_cuts(test["L"], test["d"]) == test["result"]
    
    L = random.Random(42).choices(range(200), k=25)
    assert min_sum_of_max_elems_from_d_cuts(L, d=15) == 1056

I'll leave optimizations, bottom-up DP and coming up with a recurrence to the reader. It looks like the same problem as LeetCode 1335, so the target constraints are 1 <= len(L) <= 300, 0 <= L[i] <= 1000 and 1 <= d <= 10 so it'll need way more firepower than this to avoid TLE.

Upvotes: 3

Iain Shelvington
Iain Shelvington

Reputation: 32244

For the simple case, where d=1, the solution is just the max of the list

def getmin(l, d):
    if d == 1:
        return max(l)

When d is greater than 1, the solution is to return the smaller of the head or the tail of the list added to the result of a recursive call

def getmin(l, d):
    if d == 1:
        return max(l)
    if l[0] < l[-1]:
        return l[0] + getmin(l[1:], d-1)
    else:
        return getmin(l[:-1], d-1) + l[-1]

Upvotes: 3

Related Questions