Amazon Interview Question for SDE-2s


Country: United States
Interview Type: In-Person




Comment hidden because of low score. Click to expand.
2
of 2 vote

import java.util.Arrays;
import java.util.LinkedHashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Objects;
import java.util.Set;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

public class TotalNHighestValueProductCombo {

	public static void main(String[] args) {
		System.out.println(getHighestSumCombos(3, new int[] { 1, 5, 3, 9, 7 }, new int[] { 3, 7, 9, 4, 1 }));
	}

	static List<List<Integer>> getHighestSumCombos(int n, int[] aisle1_prod_rpices, int[] aisle2_prod_rpices) {
		if (n < 1) {
			return Arrays.asList();
		}
		List<List<Integer>> ls = new LinkedList<List<Integer>>();
		IntStream.of(aisle1_prod_rpices).forEach(val1 -> {
			IntStream.of(aisle2_prod_rpices).forEach(val2 -> {
				ls.add(Arrays.asList(val1, val2));
			});
		});

		Set<String> h = new LinkedHashSet<String>();
		return ls.stream()
				// implement custom sorting in reverse order
				.sorted((l1, l2) -> Integer.compare((l2.get(0) + l2.get(1)), (l1.get(0) + l1.get(1))))
				// remove reverse combinations, comment if you want reverse combinations as well
				// e.g 6,4 and 4,6
				.map(list -> {
					String v = String.valueOf(list.get(1)) + String.valueOf(list.get(0));
					String _v = String.valueOf(list.get(0)) + String.valueOf(list.get(1));
					if (h.contains(v)) {
						return null;
					} else {
						h.add(_v);
						return list;
					}
				})
				// remove nulls
				.filter(Objects::nonNull)
				// get only first n from list
				.limit(n).collect(Collectors.toList());
	}

}

- prithvish.mukherjee@techolution.com January 04, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Kind of challenging to do this from scratch by writing algorithms, but using in-built libraries, we can come up with elegant code that gets the job done! :) For this problem, we need to use a mix of dictionaries/hashmaps + combinations generator + max-heaps. If we were doing this from scratch, obviously, this would amount to a lot of code! Therefore, I have outlined an elegant solution below.

Elegant solution in Python below:

import itertools
from collections import defaultdict
import heapq

class Product:
  def __init__(self, aisleId, productId, price):
    self.aisleId = aisleId
    self.productId = productId
    self.price = price

def getKHighestPrices(products, k):
  # Structure of the data structure will be as follows:
  # Aisle IDs: [Prices]
  aisleMap = defaultdict(list)
  for product in products:
    aisleMap[product.aisleId].append(product.price)

  heap = []
  for combination in itertools.product(*aisleMap.values()):
    # The negation is for max-heaps
    heapq.heappush(heap, (-sum(combination), combination))
  result = []
  while k > 0:
    priceList = heapq.heappop(heap)[1]
    formattedStringPriceList = ','.join([('$%s' % price) for price in sorted(priceList,reverse=True)])
    result.append(formattedStringPriceList)
    k = k - 1

  return result

Test code:

# Aisle 1 Products
p11 = Product(1,100,4)
p12 = Product(1,101,2)
p13 = Product(1,102,5)

# Aisle 2 Products
p21 = Product(2, 200, 1)
p22 = Product(2, 201, 6)

totalProducts = [p11, p12, p13, p21, p22]
print(getKHighestPrices(totalProducts, 2)) #['$6,$5', '$6,$4']
print(getKHighestPrices(totalProducts, 5)) #['$6,$5', '$6,$4', '$6,$2', '$5,$1', '$4,$1']

- prudent_programmer March 09, 2018 | Flag Reply


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