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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T16:44:30+00:00 2026-05-15T16:44:30+00:00

I’m having some trouble developing an algorithm to determine the minimum of a list

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I’m having some trouble developing an algorithm to determine the minimum of a list of n elements. It’s not the case of finding the minimum of an array of length n, that’s simple:

min = A[0]
for i in range(1, len(A)):
    if min > A[i]: min = A[i]
print min

But my list contains objects:

class Object:
    def __init__(self, somelist):
        self.classification = somelist[0] # String
        self.type           = somelist[1] # String
        self.first          = somelist[2] # Integer
        self.last           = somelist[3] # Integer

And for the same ‘classification | type’ objects I have m elements and I want to find the minimum element of the same ‘classification | type’ by comparing the difference between first and last.

Example:

obj1 = Object(['A', 'x', 4, 17])
obj2 = Object(['A', 'y', 5, 20])
obj3 = Object(['B', 'z', 10, 27])
obj4 = Object(['B', 'z', 2, 15])
obj5 = Object(['B', 'z', 20, 40])
obj6 = Object(['A', 'x', 6, 10])
obj7 = Object(['A', 'x', 2, 9])
list = [obj1, obj2, obj3, obj4, obj5, obj6, obj7]

So I need an algorithm to determine the minimums of the list:

A | x –> Object([‘A’, ‘x’, 6, 10])

B | z –> Object([‘B’, ‘z’, 2, 15])

A | y –> Object([‘A’, ‘y’, 5, 20])

Thanks!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T16:44:30+00:00Added an answer on May 15, 2026 at 4:44 pm
    import itertools 
    group_func = lambda o: (o.classification, o.type)
    map(lambda pair: (pair[0], min(pair[1], key=lambda o: o.last - o.first)),
        itertools.groupby(sorted(l, key=group_func), group_func))
    

    group_func returns a tuple key containing the object’s classification, then type (e.g. ('A', 'x')). This is first used to sort the list l (sorted call). We then call groupby on the sorted list, using group_func to group into sublists. Every time the key changes, we have a new sublist. Unlike SQL, groupby requires the list be pre-sorted on the same key. map takes the output of the groupby function. For each group, map returns a tuple. The first element is pair[0], which is the key ('A', 'x'). The second is the minimum of the group (pair[1]), as determined by the last - first key.

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