Ali Spittel

Posted on Apr 8, 2019 • Updated on Oct 28, 2019

Coding Puzzles: Week of 4/8

#challenge #codingpuzzle

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!

Top comments (47)

Ali Spittel • Apr 9 '19

Tuesday (7 KYU): Find the stray number

codewars.com/kata/57f609022f4d534f...

Laurie • Apr 9 '19

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

russ • Apr 9 '19

What is this necromancy?

Laurie • Apr 9 '19 • Edited

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”.

russ • Apr 9 '19

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")

Laurie • Apr 9 '19

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

Ali Spittel • Apr 9 '19

They should work in normal image markdown!

Laurie • Apr 9 '19

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

Md Imrul Hassan • Apr 10 '19

Same idea in Haskell:

import Data.Bits

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

Jacob Evans • Apr 24 '19

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

Joshua Gilless • Apr 9 '19

XOR is a great idea, thanks!

Laurie • Apr 9 '19

Math for the win! :D

Ahmed Musallam • Apr 10 '19 • Edited

here my cheap solution:

function stray(array) {
    var sorted = array.sort((a,b) => a - b)
    if (sorted[0] === sorted[1]) 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[0] === sorted[1] ? sorted.pop() : sorted.shift();
}

Jelle Bekker • Apr 10 '19

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;
}

Joshua Gilless • Apr 9 '19

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;
      }
    }
};

Joshua Gilless • Apr 9 '19 • Edited

@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[0];
    for (int i = 1; i < numbers.size(); i++)
    {
        r ^= numbers[i];
    }
    return r;
};

russ • Apr 9 '19

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)

Courtney • Apr 10 '19 • Edited

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[0], 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;
}

Ali Spittel • Apr 9 '19

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

Jacob Evans • Apr 24 '19 • Edited

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[1]).pop()

Ali Spittel • Apr 10 '19

Wednesday (6 KYU): Implement Syntax Highlighting

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

russ • Apr 11 '19 • Edited

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[0], None))
            cmd = cmd[1:]
    return ''.join(['<span style="color: {}">{}</span>'.format(colour, string) if colour else string for string, colour in highlighted])

russ • Apr 13 '19

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)

Laurie • Apr 12 '19

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

Laurie • Apr 10 '19 • Edited

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>';   
  })
}

Laurie • Apr 12 '19 • Edited

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;
      }
  })
}

russ • Apr 12 '19

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

Laurie • Apr 12 '19

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.

russ • Apr 12 '19 • Edited

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+)

Laurie • Apr 12 '19

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 :)

Courtney • Apr 11 '19

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;
  }

Ali Spittel • Apr 8 '19

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

codewars.com/kata/56fc55cd1f5a93d6...

Laurie • Apr 8 '19 • Edited

Javascript (ES6):

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

Jelle Bekker • Apr 8 '19

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;
   }           
}

Jelle Bekker • Apr 9 '19

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;
  }
}

Md Imrul Hassan • Apr 10 '19

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

Ali Spittel • Apr 8 '19 • Edited

My Python solution:

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

Courtney • Apr 10 '19

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);
    }
    return totalStairs * 20;
  }

Ali Spittel • Apr 12 '19

Friday (CodeJam): Foregone Solution

codingcompetitions.withgoogle.com/...

Ali Spittel • Apr 12 '19

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))

Courtney • Apr 14 '19

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++) {
            answerArray.push(noFours(argumentsArray[i]));
        }
        for(let i=0; i<answerArray.length; i++) {
            console.log(`Case #${i+1}: ${answerArray[i][0]} ${answerArray[i][1]}`)
        }
        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 
    }
}

E. Choroba • Apr 12 '19

#! /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";
}

Jacob Evans • Apr 9 '19

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

Ali Spittel • Apr 11 '19

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

codewars.com/kata/559a28007caad2ac...

E. Choroba • Apr 11 '19

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

sub perimeters {
    my ($n) = @_;
    my @f = (1, 1);
    my $s = 0;
    for (0 .. $n) {
        $s += $f[0];
        @f = ($f[1], $f[0] + $f[1]);
    }
    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;

Courtney • Apr 12 '19

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;
    }
}

Jacob Evans • Apr 9 '19

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

View full discussion (47 comments)

DEV Community

Coding Puzzles: Week of 4/8

Top comments (47)

Read next

Unleashing Aurora MySQL: Effortless Clustering with Terraform on AWS

Como iniciar na carreira de SysAdmin

Testing en Go, 2ª parte

Code with Musa - 2