# 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

[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!

### Discussion 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 : evens
}
[
[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) 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.

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 : evens
}
[
[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 😎

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);
}


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


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]


Atleast it works somehow 😅

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

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

if given%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)



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 !== null && lastOddOrEven !== 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];
}


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


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.)

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?

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])


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) + isEven(integers) + isEven(integers) >= 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}
}


Python

def find_outlier(integers):
a= sum(map(int,[integers%2,integers%2,integers%2]))
if a == 0 or a== 1:
for i in integers:
if i % 2 == 1:   return i
else:
for i in integers:
if i % 2 == 0:   return i


Here is the simple solution with PHP:

function find($integers) {$evenArray = [];
$oddArray = []; foreach ($integers as $integer) { if ($integer % 2 === 0) {
$evenArray[] =$integer;
} else {
$oddArray[] =$integer;
}
}

if (count($evenArray) === 1) { return$evenArray;
}

return \$oddArray;
}


With JS

function findOutlier(arr){
let checker = {even:[],odd:[]}
arr.forEach((item) => item%2==0 ? checker['even'].push(item) : checker['odd'].push(item));
return checker['even'].length < checker['odd'].length ? checker['even'] : checker['odd'];
}

findOutlier([2, 4, 0, 100, 4, 11, 2602, 36])



Explanation:

1. Using an object which stores even and odd numbers.
2. just returning the one with less number.

array = [500,502,1002,1234,601]
odd = even = i = 0
oddValue = evenValue = 0
a = len(array)
while (i < a):
if (array[i] % 2):
oddValue = array[i]
odd += 1
else:
evenValue = array[i]
even += 1
i += 1
if (odd > even == 1):
print(evenValue)
elif (odd == 1):
print(oddValue)
else:
print("Wrong list of numbers")

Not elegant, but it'll do the job.

const findOutlier = (arr) => {
const firstPass = arr.filter(el => el % 2 === 0)
const secondPass = arr.filter(el => el % 2 !== 0)

if(firstPass.length === 1) {
return firstPass
}
return secondPass
}


This is the shortest solution I can create.

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

const findOutlier = (arr) => {
const evenArr = arr.filter(isEven)
const oddArr = arr.filter(isOdd)

return evenArr > oddArr ? oddArr : evenArr
}

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



function findOutlier(integers){
let oddCount = 0, evenCount = 0, r1 = 0, r2 = 0
integers.forEach((i)=>{
if(i % 2 == 0) {evenCount++; r1 = i;}
else if(Math.abs(i % 2) == 1) {oddCount++; r2 = i;}
})
return (evenCount>oddCount)?r2:r1
}


let findOutlier = (arr) => {
const g1 = arr.filter(x => x % 2 === 0);
const g2 = arr.filter(x => x % 2 !== 0);
if(g1.length > g2.length )
return g2;
else
return g1;
}


ruby 2.7

def findOutlier(a) a.group_by(&:even?).each_value { return @1 if @2.nil? }; nil end
p [
[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(&self.:findOutlier)

# [11, 160, 15, nil]


This is my proposal:

function findOutlier(arr){
const even=[], odd=[];
arr.find(n => {
(n%2 ? even : odd).push(n);
return even.length ? odd.length > 1 : odd.length ? even.length > 1 : false;
});
return even.length === 1 ? even : odd.length === 1 ? odd : null;
}


In this case, Array.find will stop looping when both the even and odd arrays have at least one item.

There is only one loop on the array values, and this loop stops just when the "strange element" is found.

At the end, the group with just one element has the solution.

In the last test, an array formed entirely by even numbers, null is returned.

Here's my code:
I believe I could have used filters on the arrays method but just decided to follow the long step.



const findOutlier = (arr) => {
const evens = [];
const odds = [];
for(let i in arr) {
const isEven = arr[i] % 2 === 0;
const isOdd = !isEven;
if(isEven){
evens.push(arr[i]);
}
if(isOdd) {
odds.push(arr[i])
}
}

if(odds.length === 0 || evens.length === 0) {
return null;
}

return  odds.length  < evens.length ? odds : evens;
}

console.log([
[160, 3, 1719, 19, 11, 13, -21],
[4, 8, 15, 16, 24, 42],
[16, 6, 40, 66, 68, 28]
].map(findOutlier));



Here's my code:
I believe I could have used filters on the arrays method but just decided to follow the long step.



const findOutlier = (arr) => {
const evens = [];
const odds = [];
for(let i in arr) {
const isEven = arr[i] % 2 === 0;
const isOdd = !isEven;
if(isEven){
evens.push(arr[i]);
}
if(isOdd) {
odds.push(arr[i])
}
}

if(odds.length === 0 || evens.length === 0) {
return null;
}

return  odds.length  < evens.length ? odds : evens;
}

console.log([
[160, 3, 1719, 19, 11, 13, -21],
[4, 8, 15, 16, 24, 42],
[16, 6, 40, 66, 68, 28]
].map(findOutlier));



Here's my code:
I believe I could have used filters on the arrays method but just decided to follow the long step.



const findOutlier = (arr) => {
const evens = [];
const odds = [];
for(let i in arr) {
const isEven = arr[i] % 2 === 0;
const isOdd = !isEven;
if(isEven){
evens.push(arr[i]);
}
if(isOdd) {
odds.push(arr[i])
}
}

if(odds.length === 0 || evens.length === 0) {
return null;
}

return  odds.length  < evens.length ? odds : evens;
}

console.log([
[160, 3, 1719, 19, 11, 13, -21],
[4, 8, 15, 16, 24, 42],
[16, 6, 40, 66, 68, 28]
].map(findOutlier));



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 ] ) )
].


Go:

func FindOutlier(integers []int) int {
var m = map[string][]int{
"odd": []int{},
"even": []int{},
}
for _, i := range integers {
if i % 2 == 0 {
m["even"] = append(m["even"], i)
} else {
m["odd"] = append(m["odd"], i)
}
}
if len(m["even"]) == 1 {
return m["even"]
}
return m["odd"]
}


Dart solution:

findOutlier(List nums) {
var odds = new List();
var evens = new List();

if (odds.length == 0 || evens.length == 0) { return null; }

return (odds.length < evens.length ? odds : evens);
}


## Ruby Language

### With specs!

def outlier values
o = values.partition(&:odd?).sort_by(&:size)
o if o.size == 1
end

require "spec"

describe "#name_shuffler" do
it { expect(outlier [2, 4, 0, 100, 4, 11, 2602, 36]).to eq 11}
it { expect(outlier [160, 3, 1719, 19, 11, 13, -21]).to eq 160}
it { expect(outlier [4, 8, 15, 16, 24, 42]).to eq 15}
it { expect(outlier [16, 6, 40, 66, 68, 28]).to eq nil}
end


## output

>> rspec outlier.rb
....

Finished in 0.0052 seconds (files took 0.15152 seconds to load)
4 examples, 0 failures


Python :

def find_outlier(array):
odd = [i for i in array if i%2 == 0]
even = [i for i in array if i%2]

if not even or not odd:
return "ERROR"
elif len(even) > len(odd):
return odd
else:
return even


array = [500,502,1002,1234,601]
odd = even = i = 0
oddValue = evenValue = 0
a = len(array)
while (i < a):
if (array[i] % 2):
oddValue = array[i]
odd += 1
else:
evenValue = array[i]
even += 1
i += 1
if (odd > even == 1):
print(evenValue)
elif (odd == 1):
print(oddValue)
else:
print("Wrong list of numbers")  