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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 sol...

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Sam Ferree • Jul 25 '18

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...

Sam Ferree • Jul 25 '18 • Edited

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

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

zenmumbler • Jul 10 '18 • Edited

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[1];
    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);
}

Ryan Palo • Jul 10 '18

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!

Ali Spittel • Jul 11 '18

These are awesome! Definitely adding them to the rotation

zenmumbler • Jul 10 '18 • Edited

Day 1

const piggy = (str) =>
    str.split(" ")
    .map(word =>
        word[0].match(/\w/) ?
            word.slice(1) + word[0] + '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[0]);
    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 ;)

Pierre Bouillon • Jul 10 '18

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[0] + '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])

Massimo Artizzu • Jul 10 '18

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.

Raunak Ramakrishnan • Jul 11 '18

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.