### Daily Coding Puzzles

#### Ali Spittel on July 10, 2018

I think a lot of people know this at this point, but I love writing code. To wake myself up in the morning and get my brain running, I normally s... [Read Full] Day 4

``````sumEvenFib :: Integer -> Integer
sumEvenFib x = sum . takeWhile (<x) . filter even \$ fibs
where fibs = 1 : 1 : zipWith (+) fibs (tail fibs)

main :: IO ()
main = print \$ sumEvenFib 4000000
``````

Okay, just pipe the number of even terms you want into STDIN (an input of 4 will display the sum of the first 4 even terms of fibonacci) and you're good to go.

Befunge 93:

``````188p199p&1                     v
88g99g+88p88g99g+:99p+\1-:!#v_\>
>\.@
``````

Kind of disappointed as this solution uses get and put and therefore is less... "fungical"? but trying to do this all on a single stack got away from me.

A Funge++ solution might look more elegant...

Here we go, MUCH happier with this Funge++ solution:

``````&}1}1      v
{{1-:!#v_}}>:[+:]+
>}.@
``````

Just to show, Day 3 in C, Time: O(n), Space: O(1), the result is processed in place. Of course, this one only works with ASCII…

Quick subquiz for anyone actually reading this: under what circumstances will this program crash or exhibit undefined behaviour?

``````#include <stdio.h>
#include <string.h>
#include <ctype.h>

void esrever(char *s, int from, int to) {
while (from < to) {
char temp = s[from];
s[from] = s[to];
s[to] = temp;
from++;
to--;
}
}

int main(int argc, char *argv[]) {
if (argc < 2) {
return 1;
}

char *s = argv;
int begin = 0, end = 0, len = strlen(s);

do {
while (end < len && ! isspace(s[end])) {
end++;
}
if (end - begin > 4) {
esrever(s, begin, end - 1);
}
begin = end + 1;
end = begin;
} while (end < len);

printf("%s\n", s);
}
``````

Awesome, thanks for sharing. 😁 If you have any interest in learning about bioinformatics (genetics, DNA, inheritance, and the data analysis surrounding it) rosalind.info is a really good site with quite a few puzzles!

These are awesome! Definitely adding them to the rotation

Awesome !
Unfortunately I don't have twitter.. but here are my answers:

Day 1

``````def day_1(sentence: str) -> str:
result = []

for word in sentence.split():
if word.isalpha():
word = word[1:] + word + 'ay'
result.append(word)

return ' '.join(result)
``````

Day 2
Pretty sure it can be waaaay better

``````def day_2(sequences: List[str]) -> List[str]:
opening_seq = ('(', '[', '{')
closing_seq = (')', ']', '}')

matchs = {
closing_seq[i]: opening_seq[i]
for i in range(len(opening_seq))
}

def check(sequence: str) -> bool:
if len(sequence) % 2:
return False

if not all([
sequence.count(i) == sequence.count(matchs[i])
for i in matchs
]):
return False

stack = []
for c in sequence:
if c in opening_seq:
stack.append(c)

elif c in closing_seq:
if not stack:
return False

elif stack.pop() != matchs[c]:
return False

return not stack

return ['YES' if check(s) else 'NO' for s in sequences]
``````

Day 3

``````def day_3(sentence: str) -> str:
return ' '.join([
word if len(word) < 5 else word[::-1]
for word in sentence.split()
])
``````

Day 4

``````def day_4(limit: int) -> int:
seq = [0, 1]
while seq[-1] < limit:
seq.append(sum(seq[-2:]))
return sum([i for i in seq if i % 2 == 0])
``````

Day 4 (Project Euler #2) has been already proposed by @peter : dev.to/peter/project-euler-2---eve...

I gave my answer there 🙂

Others here have proposed pretty good solutions for the other three, but my answer to the fourth problem is distinctively different IMO.

### Day 1

``````const piggy = (str) =>
str.split(" ")
.map(word =>
word.match(/\w/) ?
word.slice(1) + word + 'ay' :
word
)
.join(" ");

console.info(piggy("Pig latin is cool"));
console.info(piggy("Hello world !"));
``````

Time: O(n), Space: O(n)

### Day 2

``````const opposites = {
"}": "{",
")": "(",
"]": "["
};

function isBalanced(str) {
if (str.length & 1) {
return false;
}
const openers = [];
for (const c of str) {
if (c === "{" || c === "(" || c === "[") {
openers.push(c);
}
else {
const opener = openers.pop();
if (opener !== opposites[c]) {
return false;
}
}
}
return true;
}

function balance(input) {
const lines = input.split("\n");
const testCount = parseInt(lines);
let index = 1;
while (index <= testCount) {
const result = isBalanced(lines[index]);
console.info(result ? "YES" : "NO");
index++;
}
}

balance(`7
{[()]}
{[(])}
{{[[(())]]}}
[]{}()
[({})]{}()
({(){}[]})[]
({(){}[]})[](
`);
``````

Time: O(n), Space: O(n)

### Day 3

``````const esrever = (word) =>
word.split("").reverse().join("");

const spinWords = (str) =>
str.split(" ")
.map(word =>
word.length > 4 ?
esrever(word) :
word
)
.join(" ");

console.info(spinWords("Hey fellow warriors"));
console.info(spinWords("This is a test"));
console.info(spinWords("This is another test"));
``````

Time: O(n), Space: O(n)

### Day 4

``````let a = 1, b = 2;
let sum = 0;

while (a <= 4000000) {
if ((a & 1) === 0) {
sum += a;
}

let nb = a + b;
a = b;
b = nb;
}
console.info(sum);
``````

Time: O(n), Space: O(1)

Nice exercise, this also shows how clumsy JS is when dealing with string data. Everything has to be converted to and from arrays. And these exercises do not work with non-BMP Unicode characters, but that was not in the requirement ;)

Thanks for this, Ali. Another source of good programming puzzles is the Daily Programmer subreddit . The tasks are categorized as easy, intermediate and hard and there is a good discussion on how the solutions can be improved.

It's only JS and really hacky, but I've come to like alf.nu.

Oh awesome! Definitely adding those to the rotation -- I like the RegEx series because I don't like writing them so I should get more practice in!

You can try this one:

codingame.com/servlet/urlinvite?u=...

Add me as a friend!!

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