# Coding Puzzles: Week of 4/8

### Ali Spittel twitter logo github logo Apr 8・1 min read

I'm bringing back the coding puzzles after a few month hiatus!

Every day, I post coding puzzles. These are quick coding challenges that increase in difficulty across the span of the week -- with Monday being the most beginner friendly and Friday being super tough. I love seeing other people's solutions as well, and so people post their solution to the problem in any programming language.

Here's more about them.

I wanted to try posting these here. I'm going to post each question from this week as a comment below, and then we will thread answers under those questions.

I'll be adding in the questions each day, so stay tuned and come back for more 😊

Excited to see your solutions!

DISCUSS (47) Tuesday (7 KYU): Find the stray number

codewars.com/kata/57f609022f4d534f...

``````function stray(arr) {
return arr.reduce((a, b) => a ^ b)
}
``````

What is this necromancy?

Haha, bitwise xor. Since it's immutable and commutative it'll reduce down to the stray!

I should add that this only works because it’s an odd number of elements in the array. An even number of matching elements cancel each other out to result in the “stray”.

Oh my, it even works in python!

``````def stray(arr):
count = {}
for index, i in enumerate(arr):
count[i] = count.setdefault(i, 0) + 1
if index >= 2 and len(count.keys()) > 1:
break
return next(k for k, v in count.items() if v == 1)

from functools import reduce

def intstray(arr: [int]):
return reduce(lambda x,y: x ^ y, arr)

assert(intstray([1, 1, 2]) == 2)
#nope..!
#assert(intstray([1, 1, 2, 1]) == 2)
assert(intstray([17, 17, 3, 17, 17, 17, 17]) == 3)
assert(intstray([1, 2, 2]) == 1)
assert(stray(["bob", "bob", "bob", "steve", "bob"]) == "steve")
``````

This is one of those moments when I wish gifs worked better on Dev. But yay!

Whaaa?! How did I not know this! ...game changer

Same idea in Haskell:

``````import Data.Bits

stray :: [Int] -> Int
stray = foldl1 xor
``````

XOR is a great idea, thanks!

Math for the win! :D

Well I can't think of a better answer for this particular problem domain.

Thanks for this! I don't really know C++, but I figured I'd give it a shot:

``````int stray(std::vector<int> numbers) {
int n1, n2, n3;
for (int i = 2; i < numbers.size(); i++) {
n1 = numbers[i];
n2 = numbers[i - 1];
n3 = numbers[i - 2];
if (n1 == n2 && n1 != n3) {
return n3;
} else if (n1 == n3 && n1 != n2) {
return n2;
} else if (n1 != n2 && n2 == n3) {
return n1;
}
}
};
``````

@laurieontech did something really cool in JS with a bitwise XOR (her answer is above), so I figured I'd update this C++ answer with a bitwise XOR since I love it!

``````int stray(std::vector<int> numbers) {
int r = numbers;
for (int i = 1; i < numbers.size(); i++)
{
r ^= numbers[i];
}
return r;
};
``````

here my cheap solution:

``````function stray(array) {
var sorted = array.sort((a,b) => a - b)
if (sorted === sorted) return sorted.pop();
else return sorted.shift()
}
``````

or shorter and more unreadable:

``````function stray(array) {
var sorted = array.sort((a,b) => a - b)
return sorted === sorted ? sorted.pop() : sorted.shift();
}

``````

in c# using linq:

``````public static int Stray(int[] numbers)
{
return numbers.Aggregate((x, y) => x ^ y);
}
``````

or:

``````public static int Stray(int[] numbers)
{
return numbers.GroupBy(x => x).Single(num => num.Count() == 1).Key;
}
``````

Looks a bit messy, but I was going for something that might not be too inefficient with a large input array

``````def stray(arr):
count = {}
for index, i in enumerate(arr):
count[i] = count.setdefault(i, 0) + 1
if index >= 2 and len(count.keys()) > 1:
break
return next(k for k, v in count.items() if v == 1)
``````

This is the first I'm finding this, excited to play along! Here is yesterday's (edited to make the colors show up):

``````function stray(numbers) {
let num1 = numbers, num2;
let dupOf1 = false;
for(let i=1; i<numbers.length; i++) {
if(numbers[i] === num1) {
if(num2) return num2;
else dupOf1 = true;
}
else if(numbers[i] === num2) return num1;
else num2 = numbers[i];
}
return dupOf1 ? num2 : num1;
}
``````

``````def stray(arr):
for item in arr:
if arr.count(item) == 1:
return item
``````

I don't think this is better than the bitwise solution but it's a different one lol

``````const stray =(n)=> n.filter((ele,_,ar) => ele !== ar).pop()
``````

Wednesday (6 KYU): Implement Syntax Highlighting

codewars.com/kata/roboscript-numbe...

``````import re

def highlight(cmd):
colour_map = [
(r'F+', 'pink'),
(r'L+', 'red'),
(r'R+', 'green'),
(r'\d+', 'orange'),
(r'()+', None),
]
highlighted = []
while(len(cmd)):
for regex, colour in colour_map:
match = re.match(regex, cmd)
if match:
idx_start, idx_end = match.span()
highlighted.append((cmd[idx_start:idx_end], colour))
cmd = cmd[idx_end:]
break
else:
highlighted.append((cmd, None))
cmd = cmd[1:]
return ''.join(['<span style="color: {}">{}</span>'.format(colour, string) if colour else string for string, colour in highlighted])
``````

And now with added re.sub with a callable, which I had no idea was a thing! These coding things are pretty nifty for leaning new tricks I must say!

``````import re

def highlight(cmd):
colour_map = [
(r'F+', 'pink'),
(r'L+', 'red'),
(r'R+', 'green'),
(r'\d+', 'orange'),
(r'()+', None),
]

def replacer(match : re.Match):
substr = match.group()
colour = next((colour for regex, colour in colour_map if re.match(regex, substr)), None)
return '<span style="color: {}">{}</span>'.format(colour, substr) if colour else substr

return re.sub(r'(\D)\1*|(\d+)', replacer, cmd)
``````

Oooh, I like this. I was thinking about a dictionary but didn't think about a dictionary with the regex as a key!

Booo regex

``````function highlight(code) {
return code.replace(/(\D)\1+|(\d+|\D)/g, (substr) => {
var color;
let testChar = substr.charAt(0);

if (testChar === 'F') {
color = 'pink';
} else if (testChar === 'L') {
color = 'red';
} else if (testChar === 'R') {
color = 'green';
} else if (!isNaN(testChar)) {
color = 'orange';
} else {
return substr;
}
return '<span style="color: ' + color + '">' + substr + '</span>';
})
}
``````

Quick edit with a dictionary.

``````function highlight(code) {
const colorDict = {'F': 'pink', 'L': 'red', 'R':'green'};

return code.replace(/(\D)\1*|\d+/g, (substr) => {
let testChar = substr.charAt(0);

if (testChar in colorDict) {
return '<span style="color: ' + colorDict[testChar] + '">' + substr + '</span>';
} else if (!isNaN(testChar)) {
return '<span style="color: orange">' + substr + '</span>';
} else {
return substr;
}
})
}
``````

I had no idea you could use \1 in the same expression as the tagged group, never seen that before!

Stackoverflow helped me with that one. I knew it was possible, but wasn't sure how. I actually started with the (.)\1* and iterated to the right regex from there.

You should be able to simplify it a little by changing the + to a * so you don't need the second \D I think?
(\D)\1*|(\d+)

Ooh, you're right. I had it that way to start and then ran into problems with a test case that had 663. It was splitting that grouping because it hit the matching case first. But when I removed the capture group generic and made it non-digit that solved that. So can revert back. Thanks :)

The answers to this kata on codewars are blowing my mind.

``````function highlight(code) {
let codeString = '';
let leftPointer, currentItem;
let colorObj = {'F': '<span style=\"color: pink\">', 'L': '<span style=\"color: red\">', 'R': '<span style=\"color: green\">'};
for(let i=0; i<code.length; i++){
currentItem = code[i];
if(!isNaN(code[i])) {
codeString += '<span style=\"color: orange\">'
leftPointer = i;
while(i+1 < code.length && !isNaN(code[i+1])) {
i++;
}
codeString += code.slice(leftPointer, i+1) + '</span>';
}
else if(colorObj[currentItem]) {
codeString += colorObj[currentItem];
while(i+1 < code.length && currentItem === code[i+1]) {
i++;
codeString += currentItem;
}
codeString += currentItem  + '</span>';
}
if(currentItem === '(' || currentItem === ')') codeString += currentItem;
}
return codeString;
}
``````

Monday (8KYU): How many stairs will Suzuki climb in 20 years?

codewars.com/kata/56fc55cd1f5a93d6...

Not a one liner like you guys but C#:

``````using System;
public class Kata
{
public static long StairsIn20(int[][] stairs)
{
long sum = 0;
foreach (var weekday in stairs)
foreach (var steps in weekday)
sum += steps;

return sum * 20;
}
}
``````

Okay, I couldn't help myself:

``````using System;
using System.Linq;

public class Kata
{
public static long StairsIn20(int[][] stairs)
{
return stairs.SelectMany(x => x).Sum() * 20;
}
}
``````

Javascript (ES6):

``````function stairs_in_20(stairs) {
return stairs.reduce((steps, day) => steps.concat(day)).reduce((sum, count) => sum + count) * 20;
}
``````

``````def stairs_in_20(stairs):
return 20*sum(sum(stairs, []))
``````

Practicing TypeScript for my upcoming internship. And I figured out how to add code highlighting!

``````function stairsIn20(stairs : number[][]) : number {
let totalStairs : number = 0;
for(let day of stairs) {
totalStairs += day.reduce((total, num) => total + num);
}
}
``````

My Python solution:

``````def stairs_in_20(stairs):
return sum(i for sublist in stairs for i in sublist)  * 20
``````

Friday (CodeJam): Foregone Solution

``````n_cases = int(input())
for i in range(1, n_cases + 1):
check_amount = int(input())
unfound = True
fours = [int(i) for i in str(check_amount)]
others = ["1" if n == 4 else "0" for n in fours]
n = int(''.join(others))
print("Case #{}: {} {}".format(i, n, check_amount - n))
``````

brute force for me so far.

``````const readline = require('readline');

function foregone() {
let argumentsArray = [], answerArray = [];
let linecounter = 0, tests;

const rl = readline.createInterface({
input: process.stdin,
output: process.stdout
});
rl.on('line', (line) => {
linecounter++;
if(linecounter === 1) tests = line;
else {
argumentsArray.push(line);
if(linecounter > tests) rl.close();
}
}).on('close', () => {
for(let i=0; i< argumentsArray.length; i++) {
}
for(let i=0; i<answerArray.length; i++) {
}
process.exit(0);
});

function noFours(num) {
let a = num - 1, b = num - a;
while(a.toString().indexOf('4') !== -1 || b.toString().indexOf('4') !== -1 && a >= b) {
a--, b++;
}
if(a.toString().indexOf('4') === -1 && b.toString().indexOf('4') === -1) return [a, b];
else return undefined
}
}
``````

``````#! /usr/bin/perl
use warnings;
use strict;
use feature qw{ say };

use Math::BigInt;

chomp( my \$cases = <> );
for my \$case (1 .. \$cases) {
chomp( my \$n = <> );
(my \$i = \$n) =~ tr/4/3/;
my \$j = 'Math::BigInt'->new(\$n) - 'Math::BigInt'->new(\$i);
say "Case #\$case: \$i \$j";
}
``````

``````const stairsIn20 =(s) => [...s].map((ele)=> ele.reduce((ag, nx)=> ag+nx))
.reduce((ag, nx)=> ag+nx) * 20
``````

Thursday (5 KYU): Perimeter of squares in a rectangle

TypeScript. Great practice!

``````export class G964 {
public static perimeter = (n) => {
let fibArray : number[] = [0, 1, 1];
for(let i=3; i<=n+1; i++) {
fibArray[i] = fibArray[i-1] + fibArray[i-2];
}
return fibArray.reduce((a, b) => a + b) * 4;
}
}
``````

``````#!/usr/bin/perl
use warnings;
use strict;

sub perimeters {
my (\$n) = @_;
my @f = (1, 1);
my \$s = 0;
for (0 .. \$n) {
\$s += \$f;
@f = (\$f, \$f + \$f);
}
return 4 * \$s
}

# In a good TDD tradition, I started with these lines:
use Test::More tests => 2;
is perimeters(5), 80;
is perimeters(7), 216;
``````

``````const stairsIn20 =(s) => [...s].map((ele)=> ele.reduce((ag, nx)=> ag+nx)).reduce((ag, nx)=> ag+nx) * 20
``````
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