Daily Challenge #35 - Find the Outlier

In this challenge, you'll be given an array with a length of at least three, containing (possibly quite large) integers. The array is either comprised of entirely odd integers or even integers with one exception. Write a method that returns the integer that is not like the others.

Example:
findOutlier([2, 4, 0, 100, 4, 11, 2602, 36]) => 11

Additional practice arrays:
[160, 3, 1719, 19, 11, 13, -21]
[4, 8, 15, 16, 24, 42]
[16, 6, 40, 66, 68, 28]

Good luck, happy coding~!

---

This challenge comes from user obnounce. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Comments

[Sung M. Kim]

JavaScript 😀

Not efficient, as it's iterating the array twice (I could've used reduce to iterate once but well... this looks more readable)

BTW, the last one looks weird, as it contains only even numbers.

const isEven = n => n % 2 === 0
const isOdd = n => n % 2 === 1
const findOutlier = arr => {
    const odds = arr.filter(isOdd)
    const evens = arr.filter(isEven)
    return odds.length < evens.length ? odds[0] : evens[0]
}
[
  [2, 4, 0, 100, 4, 11, 2602, 36],
  [160, 3, 1719, 19, 11, 13, -21],
  [4, 8, 15, 16, 24, 42],
  [16, 6, 40, 66, 68, 28]
].map(findOutlier)

[Hao]

There's a potential bug.

-21 will fail both isOdd and isEven function.

The function isOdd should be

const isOdd = n => n % 2 !== 0

Since negative odd numbers mod 2 equals to -1 instead of 1.

[Sung M. Kim]

Whoa.... you've just caught years worth of bugs I have possibly created... 😅

Thank you @Hao for finding the issue. 😀
I really didn't think that thru. 👍

Here is the updated code

let isEven = n => n % 2 === 0
let isOdd = n => n % 2 !== 0
const findOutlier = arr => {
    const odds = arr.filter(isOdd)
    const evens = arr.filter(isEven)
    return odds.length < evens.length ? odds[0] : evens[0]
}
[
  [2, 4, 0, 100, 4, 11, 2602, 36],
  [160, 3, 1719, 19, 11, 13, -21],
  [4, 8, 15, 16, 24, 42],
  [16, 6, 40, 66, 68, 28],
  [-21, 1, 2, 3]
].map(findOutlier)

I think I would probably use !isEven next time 😎

[Vincent Thibault]

You only need to fetch the firsts 3 items to know what you are searching for.

function findOutlier (list: number[]): number | void {
  const search = list.slice(0,3).filter(v => v % 2).length > 1 ? 0 : 1;
  return list.find(v => v % 2 === search);
}

[Dane Hillard]

This goes through the list of numbers only a single time, stopping early if possible. Could probably be code golfed further 😛

def find_outlier(nums):
    odds = evens = 0
    for num in nums:
        is_even = num % 2 == 0
        is_odd = not is_even

        odds += 1 if is_odd else 0
        evens += 1 if is_even else 0

        # Two of one kind prove the other kind is the outlier
        if (odds > 1 and is_even) or (evens > 1 and is_odd):
            return num

[MetaDave 🇪🇺]

I have done a Ruby one, based on counting the number of even values in the first three elements of the array to switch the logic between finding the first even and the first odd element.

[
  [2, 4, 0, 100, 4, 11, 2602, 36],
  [160, 3, 1719, 19, 11, 13, -21],
  [4, 8, 15, 16, 24, 42],
  [16, 6, 40, 66, 68, 28]
].map do |array|
  case array.first(3).select(&:even?).size
    when 1 
      array.find(&:even?)
    else 
      array.find(&:odd?)
  end
end

=> [11, 160, 15, nil] 

[Palash Bauri 👻]

Atleast it works somehow 😅

given = [2, 4, 0, 100, 4, 11, 2602, 36] #input array
odds = evens = 0

if given[0]%2 == 0:
    evens  += 1
else:
    odds += 1

if given[1]%2 == 0:
    evens  += 1
else:
    odds += 1

if given[-1]%2 == 0:
    evens  += 1
else:
    odds += 1

if odds > evens:
    for i in given:
        if i%2==0:
            print(i)

else:
    for i in given:
        if not i%2==0:
            print(i)

[Jerome Cukier]

in JavaScript. o(n), uses constant space.

in the worst case (outlier is at the end of the loop) it will go through all items. but if the outlier is found before, the loop stops.

function findOutlier(nums) {
  let lastOddOrEven = [null, null];
  for (let i = 0; i < nums.length; i++) {
    const parity = nums[i] % 2;
    if (lastOddOrEven[0] !== null && lastOddOrEven[1] !== null) {
      // we've already seen one odd and one even number,
      // so the outlier is the one number we've seen with a
      // different parity
      return lastOddOrEven[1 - parity];
    }
    lastOddOrEven[parity] = nums[i];
  }
  // if the loop can't return the outlier, this is because it's the last item
  return nums[nums.length - 1];
}

[willsmart]

A javascript one that has pretty reasonable complexity

function findOutlier(integers){
  // The first three ints can be used to determine whether we are 
  // looking for an odd or an even
  // If at least two are even, then we're looking for the odd integer
  const typicalIntegerIsEven = isEven(integers[0]) + isEven(integers[1]) + isEven(integers[2]) >= 2
  // Return the first integer that is non-typical
  return integers.find(v => isEven(v) != typicalIntegerIsEven)

  // Helper function to determine if a number is even or odd
  // Note that (-1)%2 === -1 and some integers might be negative.
  // (-2)%2 === -0 and -0 === 0, so this check will work for any integer.
  // v&1 is better in many ways, but only goes up to 32 bit ints
  function isEven(v) {return v%2 === 0}
}

[Oleksii Filonenko]

Elixir:

defmodule Outlier do
  require Integer

  def find(list) do
    if list
       |> Enum.take(3)
       |> Enum.filter(&Integer.is_even/1)
       |> Enum.count() == 1,
       do: Enum.find(list, &Integer.is_even/1),
       else: Enum.find(list, &Integer.is_odd/1)
  end
end

[Tanguy Andreani]

Bad solution! See comments below!

Ruby solution, I tried to do something different.

def findOutlier numbers
  sum = numbers.sum
  if sum.odd?
    numbers.find(&:odd?)
  else
    numbers.find(&:even?)
  end
end

puts findOutlier([2, 4, 0, 100, 4, 11, 2602, 36])
puts findOutlier([160, 3, 1719, 19, 11, 13, -21])
puts findOutlier([4, 8, 15, 16, 24, 42])
puts findOutlier([16, 6, 40, 66, 68, 28])
puts findOutlier([5, 9, 4, 155, 7])

When the list has no exception, the first element is returned; except when numbers looks like [3,3] (bug that you can fix by adding .uniq when computing the sum.)

[Dane Hillard]

I believe this would fail if the outlier is even and the rest are an odd number of odd numbers. The examples all have an even number of odd numbers. Can you check that?

[Tanguy Andreani]

It seems that I can’t make it right and as simple as it was. So I’ll stick with this solution:

def findOutlier numbers
  if numbers.first(3).count(&:even?) > 1
    numbers.find(&:odd?)
  else
    numbers.find(&:even?)
  end
end

puts findOutlier([2, 4, 0, 100, 4, 11, 2602, 36])
puts findOutlier([160, 3, 1719, 19, 11, 13, -21, 33])
puts findOutlier([160, 3, 1719, 19, 11, 13, -21])
puts findOutlier([4, 8, 15, 16, 24, 42])
puts findOutlier([16, 6, 40, 66, 68, 28])
puts findOutlier([3,3])

[Olivier “Ölbaum” Scherler]

Erlang.

• If the first two numbers have the same parity, I search the rest of the array for the other parity;
• If they have different parities, I rotate them with the third number and check the resulting three-element array.

-module( outlier ).

-include_lib("eunit/include/eunit.hrl").

outlier( [ A, B, C | Rest ] ) ->
    case { abs( A rem 2 ), abs( B rem 2 ) } of
        { S, S } -> outlier( [ C | Rest ], 1 - S );
        { _, _ } -> outlier( [ B, C, A ] )
    end.
outlier( [ A | _ ], S ) when abs( A rem 2 ) == S ->
    A;
outlier( [ _ | Rest ], S ) ->
    outlier( Rest, S ).    

outlier_test_() -> [
    ?_assertEqual( 11, outlier( [ 2, 4, 0, 100, 4, 11, 2602, 36 ] ) ),
    ?_assertEqual( 160, outlier( [ 160, 3, 1719, 19, 11, 13, -21 ] ) ),
    ?_assertEqual( 15, outlier( [ 4, 8, 15, 16, 24, 42 ] ) ),

    ?_assertEqual( 2, outlier( [ 1, 2, 3, 5 ] ) ),
    ?_assertEqual( 1, outlier( [ 1, 2, 4, 6 ] ) ),
    ?_assertEqual( 2, outlier( [ 2, 1, 3, 5 ] ) ),
    ?_assertEqual( 1, outlier( [ 2, 1, 4, 6 ] ) ),

    ?_assertError( function_clause, outlier( [ 16, 6, 40, 66, 68, 28 ] ) ),
    ?_assertError( function_clause, outlier( [ 16, 6 ] ) )
].