sorting orders with mergesort incorrect output

Question

I have to design an algorithm to sort a list of orders by selection time (t selection, finding the good in the warehouse and bringing it to the surface) plus shipping time (t shipping, constant). The customer orders can be retrieved (in the same order as placed) from a server database. You should expect between 100-10K elements.

The program takes as input a data-set of orders where the id, t selection, and t shipping are of type unsigned int, n is the number of orders and a space character.

id1, t selection1, t shipping1; ...; idn, t selectionn, t shippingn n

The expected output is a space-separated list of the ids, sorted by t selection + t shipping and terminated by a new line n.

Input: 1, 500, 100; 2, 700, 100; 3, 100, 100n

Output: 3 1 2n

I am trying to do it with merge sort, however my program returns

1 2 3/n instead of 3 1 2/n

I have provided my code below, could anyone help me out?

#!/usr/bin/env python3
import sys


class Order:
    def __init__(self, id: int, selection_time: int, shipping_time: int):
        self.id: int = id
        self.selection_time: int = selection_time
        self.shipping_time: int = shipping_time


def merge(left, right):
    if not len(left) or not len(right):
        return left or right
    result = []
    i, j = 0, 0
    while len(result) < len(left) + len(right):
        if left[i].shipping_time + left[i].selection_time < right[j].shipping_time + right[j].selection_time:
            result.append(left[i])
            i += 1
        else:
            result.append(right[j])
            j += 1
        if i == len(left) or j == len(right):
            result.extend(left[i:] or right[j:])
            break
    return result


def sort(list):
    if len(list) < 2:
        return list
    middle = int(len(list) / 2)
    left = sort(list[:middle])
    right = sort(list[middle:])
    return merge(left, right)


if __name__ == '__main__':
    '''
    Retrieves and splits the input
    '''
    data = input()
    data = data.split('; ')
    order_list = []
    for d in data:
        id, selection_t, shipping_t = d.split(', ', 2)
        order: Order = Order(int(id), int(selection_t), int(shipping_t))
        order_list.append(order)
    sort(order_list)
    for order in order_list:
        sys.stdout.write(str(order.id))
        sys.stdout.write(" ")

Asked By: Ama Sma

||

Source

Answer 1

The simplest (and probably least efficient) sorting algorithm is the Bubble sort. But the question says nothing about performance so it can be simplified like this:

class Order:
    def __init__(self, ident, selection_time, shipping_time):
        self._ident = ident
        self._selection_time = selection_time
        self._shipping_time = shipping_time
    @property
    def selection_time(self):
        return self._selection_time
    @property
    def shipping_time(self):
        return self._shipping_time
    @property
    def ident(self):
        return self._ident


def merge(lst):
    def comboval(order):
        return order.selection_time + order.shipping_time

    if len(lst) > 1:
        mid = len(lst) // 2
        left = lst[:mid]
        right = lst[mid:]
        merge(left)
        merge(right)
  
        i = j = k = 0
  
        while i < len(left) and j < len(right):
            if comboval(left[i]) < comboval(right[j]):
                lst[k] = left[i]
                i += 1
            else:
                lst[k] = right[j]
                j += 1
            k += 1

        for _i in range(i, len(left)):
            lst[k] = left[_i]
            k += 1

        for _j in range(j, len(right)):
            lst[k] = right[_j]
            k += 1

    return lst

inval = '1, 500, 100; 2, 700, 100; 3, 100, 100'

orderlist = []

for order in inval.split(';'):
    orderlist.append(Order(*map(int, order.split(','))))

print(*[order.ident for order in merge(orderlist)])

Output:

3 1 2

Note:

This is an in-place sort

Answered By: Vlad

sorting orders with mergesort incorrect output

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