DEV Community: Ha Anna

[#3] KoreanCoding 101: for loops and arrays

Ha Anna — Fri, 12 May 2023 14:01:30 +0000

Hey there! This week we will be learning vocabulary related to arrays and for loops! These are the bread and butter for us developers, something we deal with almost every single day. Arrays are like handy containers that let us store and organize data, while for loops are these amazing tools that help us work with arrays in super flexible and speedy ways. We can loop through arrays, find specific elements, or filter out data that meets certain criteria using for loops. (By the way, some cool JavaScript methods work similarly but we will talk about those next week!) So, let's get into it!

Arrays | 배열

Arrays are like boxes that can hold multiple things at once - we can store strings, numbers, booleans, objects, and even other arrays inside of them. Arrays help us keep things organized and easily accessible. Just like how you can pick out a specific item from your box, we can retrieve individual elements (원소) from an array (배열) using their position or index (인덱스). Index counting starts from 0 (zero-based indexing), so the first element's position in an array called array will be array[0], and the last item can be accessed by using length property:

const array = ['a', 'b', 'c', 'd']
console.log(array.length) // Logs out 4
console.log(array[0]) // Logs out 'a'
console.log(array[array.length - 1])  // Logs out 'd'

The length property returns the total number of elements present in the array. To access the last element, we subtract 1 from the length (길이). Why? Well, since index (인덱스) counting starts from 0, the last element's position is actually one less than the length. So, to access the last element (원소), we use array[array.length - 1].

`for` loop | `for` 문

Now it's time to understand how for loops work. A for loop allows us to repeat (반복) a block of code multiple times. It consists of three essential components: initialization (초기화), condition (조건), and increment/decrement (증감식).

Initialization (초기화): Set an initial value before the loop starts, like initializing a variable or setting a counter.
Condition (조건): Define a condition that is checked before each loop iteration. The loop continues as long as the condition remains true and stops when it becomes false.
Increment/Decrement (증감식): Specify how the loop variable should change after each iteration, like increasing or decreasing its value (값).

for (let i = 0; i < 5; i++) {
  console.log("Iteration: " + i);
}
// Logs out 'Iteration: 0'
//          'Iteration: 1'
//          'Iteration: 2'
//          'Iteration: 3'
//          'Iteration: 4'

We start with i initialized to 0. The loop will continue as long as i is less than 5. After each iteration, i is incremented (증감) by 1. The loop will run five times, with i taking the values 0, 1, 2, 3, and 4.

It's worth noting that we can set i to be different values (값) depending on our needs, or we can even make it go in reverse order, like this:

for (let i = 5; i > 0; i--) {
  console.log("Iteration: " + i);
}
// Logs out 'Iteration: 5'
//          'Iteration: 4'
//          'Iteration: 3'
//          'Iteration: 2'
//          'Iteration: 1'

Other than for loop, there are many other kinds of loops, such as while, do... while, for... in, for... of, etc. You can learn more about them here: JavaScript Loops.

Vocabulary

Nouns:

배열 - array

문자열 배열 - an array of strings

원소, 요소 - element

인덱스 - index

길이 - length

값 - value

정수 - integer

짝수 - even number

홀수 - odd number

유사도 - similarity

소문자 - lowercase letters

중복된 원소 - duplicate elements

반복문 - iteration 

for문 - for loop

중첩 반복문 - nested loop 

초기화 - initialization

조건 - condition

증감 - increment

반복 횟수 - iteration count

Verbs:

추가하다 - to append, to add

삭제하다 - to delete

정렬하다 - to sort

합치다 - to merge

반복하다 - to repeat

Phrases:

1씩 감소하는 수들을 차례로 담은 리스트

a list of numbers decreasing by one

정수가 담긴 리스트

a list of integers

(something)의 개수

the number of (something)

원소의 개수

the number of elements

Challenges

카운트 다운

Description

정수 start와 end가 주어질 때, start에서 end까지 1씩 감소하는 수들을 차례로 담은 리스트를 return하도록 solution 함수를 완성해주세요.

제한사항

0 ≤ end ≤ start ≤ 50

입출력 예

start end result

10 3 [10, 9, 8, 7, 6, 5, 4, 3]

입출력 예 설명

입출력 예 #1

10부터 3까지 1씩 감소하는 수를 담은 리스트는 [10, 9, 8, 7, 6, 5, 4, 3]입니다.

start	end	result
10	3	[10, 9, 8, 7, 6, 5, 4, 3]

Description:

카운트 다운

Countdown

정수 start와 end가 주어질 때,

Integers start and end are given as parameters,

start에서 end까지 1씩 감소하는 수들을 차례로 담은 리스트를 return하도록 solution 함수를 완성해주세요.

complete the solution function by returning a list of numbers that decrease by one from start to end.

제한사항 (constraints):

The constraints section of the challenge informs us that both end and start can have values ranging from 0 to 50. The end's value will always be less or even start.

입출력 예 (input-output examples):

#1: When start = 10 and end = 3, the result will be [10, 9, 8, 7, 6, 5, 4, 3].

입출력 예 설명 (explanation of input-output examples):

Based on the example #1:

The list containing the numbers from 10 to 3 in increments of 1 is [10, 9, 8, 7, 6, 5, 4, 3].

JavaScript Solution

First, we create an empty array called result.

Then, we use a for loop to iterate over numbers from start to end. The loop starts with the value of start and continues as long as i is greater than or equal to end. The i's value decrements by one after each iteration.

Inside the loop, each value of i is pushed into result array using the push method. This adds the current value of i to the end of the result array.

After the loop finishes, the result array is returned from the function.

function solution(start, end) {
    const result = []
    for(let i = start; i >= end; i--) {
        result.push(i)
    }
    return result
}

짝수 홀수 개수

Description

정수가 담긴 리스트 num_list가 주어질 때, num_list의 원소 중 짝수와 홀수의 개수를 담은 배열을 return 하도록 solution 함수를 완성해보세요.

제한사항

1 ≤ num_list의 길이 ≤ 100

0 ≤ num_list의 원소 ≤ 1,000

입출력 예

num_list result

[1, 2, 3, 4, 5] [2, 3]

[1, 3, 5, 7] [0, 4]

입출력 예 설명

입출력 예 #1

[1, 2, 3, 4, 5]에는 짝수가 2, 4로 두 개, 홀수가 1, 3, 5로 세 개 있습니다.

입출력 예 #2

[1, 3, 5, 7]에는 짝수가 없고 홀수가 네 개 있습니다.

num_list	result
[1, 2, 3, 4, 5]	[2, 3]
[1, 3, 5, 7]	[0, 4]

Description:

짝수 홀수 개수

Counting odd and even numbers

정수가 담긴 리스트 num_list가 주어질 때,

Given a list num_list of integers,

num_list의 원소 중 짝수와 홀수의 개수를 담은 배열을 return 하도록 solution 함수를 완성해보세요.

Complete the solution function to return an array containing the count of even and odd numbers found in num_list.

제한사항 (constraints):

The num_list should contain a minimum of 1 element and a maximum of 100 elements and each element in num_list must be between 0 and 1,000, inclusive.

입출력 예 (input-output examples):

#1: When num_list = [1, 2, 3, 4, 5], the result should be [2, 3].

입출력 예 설명 (explanation of input-output examples):

Based on the example #1:

[1, 2, 3, 4, 5] has two even numbers: 2 and 4, and three odd numbers: 1, 3, 5.

JavaScript Solution

First, we initialize two variables evenCount and oddCount and set them both to 0. This is where we will store how many odd and even numbers appeared in our num_list array.

Then, we use a for loop to iterate over each element of the num_list. The loop starts with i equal to 0 and continues as long as i is less than the length of num_list. The loop increments i by 1 in each iteration.

Inside the loop, it checks whether the current element at index i is even or odd. This is done by checking if the remainder of dividing the element by 2 is equal to 0. If the current element is even, it increments the evenCount variable by 1. Otherwise, if the element is odd, it increments the oddCount variable by 1.

After the loop finishes iterating over all the elements in the num_list array, it returns an array containing the counts of even and odd numbers, respectively, as [evenCount, oddCount].

function solution(num_list) {
    let evenCount = 0
    let oddCount = 0
    for (let i = 0; i < num_list.length; i++) {
        if(num_list[i] % 2 === 0) evenCount++
        else oddCount++
    }
    return [evenCount, oddCount]
}

배열의 유사도

Description

두 배열이 얼마나 유사한지 확인해보려고 합니다. 문자열 배열 s1과 s2가 주어질 때 같은 원소의 개수를 return하도록 solution 함수를 완성해주세요.

제한사항

1 ≤ s1, s2의 길이 ≤ 100

1 ≤ s1, s2의 원소의 길이 ≤ 10

s1과 s2의 원소는 알파벳 소문자로만 이루어져 있습니다

s1과 s2는 각각 중복된 원소를 갖지 않습니다.

입출력 예

s1 s2 result

["a", "b", "c"] ["com", "b", "d", "p", "c"] 2

["n", "omg"] ["m", "dot"] 0

입출력 예 설명

입출력 예 #1

"b"와 "c"가 같으므로 2를 return합니다.

입출력 예 #2

같은 원소가 없으므로 0을 return합니다.

s1	s2	result
["a", "b", "c"]	["com", "b", "d", "p", "c"]	2
["n", "omg"]	["m", "dot"]	0

Description:

배열의 유사도

Array's similarity

두 배열이 얼마나 유사한지 확인해보려고 합니다.

You want to determine how similar the two arrays are.

문자열 배열 s1과 s2가 주어질 때 같은 원소의 개수를 return하도록 solution 함수를 완성해주세요.

Complete the solution function to return the number of the same elements when given arrays of strings s1 and s2.

제한사항 (constraints):

Arrays s1 and s2 can have a minimum length of 1 and a maximum length of 100, and each string in the arrays can have a minimum length of 1 character and a maximum length of 10 characters. Moreover, the elements of s1 and s2 consist only of lowercase letters of the alphabet and there are no duplicate elements within each array.

입출력 예 (input-output examples):

#1: When s1 = ["a", "b", "c"] and s2 = ["com", "b", "d", "p", "c"], result = 2.

입출력 예 설명 (explanation of input-output examples):

Based on the example #1:

Returns 2 because "b" and "c" appear in both arrays.

JavaScript Solution

First, we initialize a variable called count to keep track of the number of common elements between the arrays. The initial value is set to 0.

Then we use two nested for loops to compare each element of s1 with each element of s2. The outer loop iterates over the elements of s1, and the inner loop iterates over the elements of s2.

Inside the inner loop, it checks if the current element of s1 at index i is equal to the current element of s2 at index j. If they are equal, it means that there is a common element, so it increments the count variable by 1.

After the loops finish iterating over all the elements in s1 and s2, the function returns the final value of count, which represents the number of common elements between the two arrays.

function solution(s1, s2) {
  let count = 0
  for (let i = 0; i < s1.length; i++) {
    for (let j = 0; j < s2.length; j++) {
      if(s1[i] === s2[j]) count++
    }
  }
  return count
}

Wrap-up

And that's a wrap! I hope you've enjoyed today's lesson. Arrays and for loops are super useful tools that developers use all the time. Understanding how they work and how to use them effectively can really level up your coding game. So keep on learning and experimenting, and soon you'll be able to make the most of arrays and for loops in your own projects.

For those who are keen on trying to solve some similar challenges on their own, I have picked some additional challenges related to arrays and for loops:

See you next week!

[#2] KoreanCoding 101: operators, if... else, ternary operator

Ha Anna — Sun, 07 May 2023 01:02:18 +0000

Hello there! Today, we'll explore some advanced concepts related to logic and conditionals in programming. Don't worry if you're not yet confident in these areas - I'll guide you through each step and introduce some Korean technical vocabulary to help you better understand the concepts.

As we dive into these topics, we'll encounter a charming mascot named 머쓱이, who represents Programmers.co.kr. Let's get started!

Operators | 연산자

We already are familiar with arithmetic operators (addition, subtraction, division, multiplication, remainder) but now it's time to get to know comparison operators.

Comparison operators are used to compare two values and return a boolean value (either true (긍정, 참) or false (부정, 거짓)) based on the result of the comparison (비교). If you are not sure how to use them, I recommend going through javascript.info's JavaScript Fundamentals: Comparisons.

`if... else` statement | `if... else` 조건문

We use if... else to make decisions in code. It works by checking a condition (조건) (like whether a number is greater than 1), and then performing different actions based on whether the condition is true (긍정, 참) or false (부정, 거짓).

let number = 1

if(number === 1) {
  console.log('Oho! The number is 1!') // This is the output.
} else {
  console.log('The number is not 1.')
}

If we were to change number to another number, for example, 3:

let number = 3

if(number === 1) {
  console.log('Oho! The number is 1!')
} else {
  console.log('The number is not 1.') // This is the output.
}

Then the console would log out the 'Oh! The number is not 1.' from the else statement.

We can also add multiple conditions (조건) by adding else if (condition) between if and else elements:

let number = 3

if(number === 1) {
  console.log('Oho! Number is 1!')
} else if (number > 1) {
  console.log('Not 1 but bigger than 1.') // This is the output.
} else {
  console.log('Hmm... Number is not 1.')
}

And it is also worth mentioning that else is optional and we can have standalone if statements (if 조건문) or if... if else statements (if... if else 조건문) too:

let number = 1

if(number === 1) {
  console.log('Oho! Number is 1!') // This is the output.
} 

if(number > 0) {
  console.log('Number is bigger than 0!') // This is also the output.
}

Here, both statements evaluate true and get logged out because they are not combined into one statement.

Ternary operator | 삼항 연산자

Now that you are a bit more comfortable with conditional statements, let's take a look at something even cooler - the ternary operator.

The ternary operator, also known as the conditional operator, is a shorthand way of writing a if...else statement. It has the following syntax:

condition ? expression1 : expression2

The condition is evaluated first. If it is true, expression1 is executed. If it is false, expression2 is executed. Here's a real-life example:

let age = 22
let message = age >= 18 ? "You are an adult" : "You are not an adult";
console.log(message) // Output: "You are an adult"

If age is greater than or equal to 18, the value of the message variable will be "You are an adult", and if it is less than 18, the value of the message will be "You are not an adult". The value of the message variable is determined by the result of the ternary operator.

Ternary operators can also be chained together. This technique is used to simplify nested if...else statements or switch cases. It allows you to write a single line of code that performs multiple conditional checks and returns different values based on those conditions.

condition1 ? value1 : condition2 ? value2 : condition3 ? value3 : defaultValue

const temperature = 20;
const isRaining = false;
const message = temperature > 30 ? "It's hot outside" :
                isRaining ? "It's raining outside" :
                "It's a nice day outside";

console.log(message); // Output: It's a nice day outside

The ternary operator can make code more concise and readable in some cases, but it can also make code harder to read if it is overused. I'll show you an example later when solving one of the challenges. When it comes to the ternary operator, it is best to use it only when it improves the clarity of the code.

Vocabulary

Nouns:

변수 - variable

문자열 - string

숫자 - number

불린 - boolean

배수 - multiple (a number that can be divided by another number without a remainder)

짝수 - even (number)

홀수 - odd (number)

if... else 조건문 - if... else conditional statement

삼항 연산자 - ternary operator

조건 연산자 - conditional operator

긍정, 참 - true

부정, 거짓 - false

거짓같은 값 - falsy

참 같은 값 - truthy

논리 연산자 - logical operator (&&, ||, !)

논리곱 연산자 - AND operator (&&)

논리합 연산자 - OR operator (||)

논리부정 연산자 - NOT operator (!)

관계연 산자 - relational operator (==, ===, !=, !==, <, >, <=, >=)

동등 연산자 - equality operator (==)

일치 연산자 - strict equality operator (===)

부등 연산자 - inequality operator (!=)

불일치 연산자 - strict inequality operator (!==)

작음 - less than (<)

큼 - greater than (>)

크거나 같음 - greater or equal to (>=)

작거나 같음 - less or equal to (<=)

for (반복)문 - for loop

코드 블록 - code block

형 변환 - type conversion

예각 - acute angle

직각 - right angle

둔각 - obtuse angle

평각 - straight angle

Verbs:

조건을 평가하다 - to evaluate the condition

만족하다 - to be satisfied (a condition)

실행하다 - to run (a statement)

반환하다 - to return (a value)

비교하다 - to compare

반복하다 - to repeat

Phrases:

condition1을 만족하면 statement1을 실행하다.

if condition1 is satisfied run statement1.

Challenges

n의 배수

Description

수 num과 n이 매개 변수로 주어질 때, num이 n의 배수이면 1을 return n의 배수가 아니라면 0을 return하도록 solution 함수를 완성해주세요.

제한사항

2 ≤ num ≤ 100

2 ≤ n ≤ 9

입출력 예

num n result

98 2 1

34 3 0

입출력 예 설명

입출력 예 #1

98은 2의 배수이므로 1을 return합니다.

입출력 예 #2

32는 3의 배수가 아니므로 0을 return합니다.

num	n	result
98	2	1
34	3	0

Description:

n의 배수

the multiple of n

수 num과 n이 매개 변수로 주어질 때

Integers num and n are given as parameters,

num이 n의 배수이면 1을 return

return 1 if num is a multiple of n

n의 배수가 아니라면 0을 return하도록 solution 함수를 완성해주세요.

and return 0 if it is not to complete the solution function.

제한사항 (constraints):

The constraints section of the challenge informs us that num can have values ranging from 2 to 100, and that n can have values ranging from 2 to 9.

입출력 예 (input-output examples):

#1: When num = 98 and n = 1 the result should be 1.

#2: When num = 32 and n = 3 the result should be 0.

입출력 예 설명 (explanation of input-output examples):

Based on the example #1:

98은 2의 배수이므로 1을 return합니다.

It returns 1 because 98 is a multiple of 2.

Based on the example #2:

32는 3의 배수가 아니므로 0을 return합니다.

It returns 0 because 32 is not a multiple of 3.

JavaScript Solution

If we want to determine whether one number can be divided evenly by another, we have a few options in JavaScript. One popular way is to use an if...else statement (if...else 조건문), which is a great fit for this problem.

To check whether num is divisible by n, we use the modulo operator % to see if there is a remainder. If there is no remainder, we know that num is divisible by n and we return 1. If there is a remainder, we return 0.

Another approach is to use a ternary operator (삼항 연산자), which is a shorthand way to write the same code. This approach is shorter and more concise, which some developers might prefer.

// if... else solution
function solution(num, n) {
  if (num % n === 0) {
    return 1
  } else {
    return 0
  }
}

// ternary operator solution
function solution(num, n) {
  return (num % n === 0) ? 1 : 0
}

각도기

Description

각에서 0도 초과 90도 미만은 예각, 90도는 직각, 90도 초과 180도 미만은 둔각 180도는 평각으로 분류합니다. 각 angle이 매개변수로 주어질 때 예각일 때 1, 직각일 때 2, 둔각일 때 3, 평각일 때 4를 return하도록 solution 함수를 완성해주세요.

예각 : 0 < angle < 90

직각 : angle = 90

둔각 : 90 < angle < 180

평각 : angle = 180

제한사항

0 < angle ≤ 180

angle은 정수입니다.

입출력 예

angle result

70 1

91 3

180 4

입출력 예 설명

입출력 예 #1

angle이 70이므로 예각입니다. 따라서 1을 return합니다.

입출력 예 #2

angle이 91이므로 둔각입니다. 따라서 3을 return합니다.

입출력 예 #2

angle이 180이므로 평각입니다. 따라서 4를 return합니다.

angle	result
70	1
91	3
180	4

Description:

각도기

Protractor

각에서 0도 초과 90도 미만은 예각,

An angle greater than 0 degrees and less than 90 degrees is an acute angle,

90도는 직각,

90 degrees is a right angle,

90도 초과 180도 미만은 둔각

greater than 90 degrees and less than 180 degrees is an obtuse angle,

180도는 평각으로 분류합니다.

and an angle equal to 180 degrees is a straight angle.

각 angle이 매개변수로 주어질 때

When angle is given as a parameter, return

예각일 때 1,

1 for an acute angle,

직각일 때 2,

2 for a right angle,

둔각일 때 3,

3 for an obtuse angle,

평각일 때 4를 return하도록 solution 함수를 완성해주세요.

and 4 for a straight angle to complete the solution function.

예각 : 0 < angle < 90

acute angle: 0 < angle < 90

직각 : angle = 90

right angle: angle = 90

둔각: 90 < angle < 180

obtuse angle: 90 < angle < 180

평각 : angle = 180

straight angle: angle = 180

제한사항 (constraints):

The constraints section of the challenge informs us that angle is an integer (정수) and it can have values ranging from 0 to 180.

입출력 예 (input-output examples):

#1: When angle = 70 the result should be 1.

#2: When angle = 91 the result should be 3.

#3: When angle = 180 the result should be 4.

입출력 예 설명 (explanation of input-output examples):

Based on the example #1:

angle이 70이므로 예각입니다. 따라서 1을 return합니다.

Since angle is 70, it is an acute angle. So it returns 1.

Based on the example #2:

angle이 91이므로 둔각입니다. 따라서 3을 return합니다.

Since angle is 91, it is an obtuse angle. So it returns 3.

Based on the example #3:

angle이 180이므로 평각입니다. 따라서 4를 return합니다.

Since angle is 180, it is a straight angle. So it returns 4.

JavaScript Solution

Here we will need two more else if statements added to our if... else statement. We also need to make sure we use relational operators (관계연 산자) correctly.

This can be also solved in two ways, more or less descriptive. The edge cases will also be handled differently.

The first solution provides more specific information on how the function is handling different angles within the range of 0 to 180. It breaks down the range into four distinct cases, including angles between 0 and 90, 90, angles between 90 and 180, and 180. This approach allows for more precise identification of the angle's location and how it's being handled, which can make debugging and maintenance easier in the long run.

The second solution is more concise and combines the second and fourth cases, resulting in only three cases. While this approach is still functional, it is slightly less descriptive because it doesn't provide as much information about how the function is handling different angles. As a result, it may be less clear to other developers or maintainers who are reading or working with the code.

Overall, the choice between these solutions comes down to a balance between the need for precise information and the desire for simplicity and conciseness. In general, it's best to strive for a balance between these two factors, creating code that is both readable and efficient.

// #1: more descriptive solution
function solution(angle) {
    if(0 < angle && angle < 90) return 1
    else if (angle === 90) return 2
    else if (90 < angle && angle < 180) return 3
    else if (angle === 180) return 4
    else return
}

// #2: a slightly less descriptive solution
function solution(angle) {
    if(0 < angle && angle < 90) return 1
    else if (angle === 90) return 2
    else if (90 < angle && angle < 180) return 3
    else return 4
}

숫자 비교하기

Description

정수 num1과 num2가 매개변수로 주어집니다. 두 수가 같으면 1 다르면 -1을 return하도록 solution 함수를 완성해주세요.

제한사항

0 ≤ num1 ≤ 10,000

0 ≤ num2 ≤ 10,000

입출력 예

num1 num2 result

2 3 -1

11 11 1

7 99 -1

입출력 예 설명

입출력 예 설명 #1

num1이 2이고 num2가 3이므로 다릅니다. 따라서 -1을 return합니다.

입출력 예 설명 #2

num1이 11이고 num2가 11이므로 같습니다. 따라서 1을 return합니다.

입출력 예 설명 #3

num1이 7이고 num2가 99이므로 다릅니다. 따라서 -1을 return합니다.

num1	num2	result
2	3	-1
11	11	1
7	99	-1

Description:

숫자 비교하기

Comparing numbers

정수 num1과 num2가 매개변수로 주어집니다.

Integers num1 and num2 are given as parameters.

두 수가 같으면 1 다르면 -1을 return하도록 solution 함수를 완성해주세요.

Complete the solution function to return 1 if the two numbers are the same and -1 if they are different.

제한사항 (constraints):

The constraints section of the challenge informs us that both num1 and num2 can have values ranging from 0 to 10,000.

입출력 예 (input-output examples):

#1: When num1 = 2 and num2 = 3 the result should be -1.

#2: When num1 = 11 and num2 = 11 the result should be 1.

입출력 예 설명 (explanation of input-output examples):

Based on the example #1:

num1이 2이고 num2가 3이므로 다릅니다. 따라서 -1을 return합니다.

Return -1 because num1 is 2 and num2 is 3, which means they are not equal.

Based on the example #2:

num1이 11이고 num2가 11이므로 같습니다. 따라서 1을 return합니다.

Return 1 because num1 is 11 and num2 is 11, which means they are equal.

JavaScript Solution

When it comes to comparing two numbers in JavaScript, we have a couple of options. We can use an if...else statement (if...else 조건문) or a ternary operator (삼항 연산자). Both of these methods will work just fine.

One important thing to keep in mind is that we want to use a strict equality operator === (일치 연산자) to compare the numbers. This ensures that we are comparing the numbers' values and types, so we don't run into any unexpected issues.

// if... else solution
function solution(num1, num2) {
    if (num1 === num2) {
        return 1
    } else {
        return -1
    }
}

// ternary operator solution
function solution(num1, num2) {
    return (num1 === num2) ? 1 : -1
}

옷가게 할인 받기

Description

머쓱이네 옷가게는 10만 원 이상 사면 5%, 30만 원 이상 사면 10%, 50만 원 이상 사면 20%를 할인해줍니다.

구매한 옷의 가격 price가 주어질 때, 지불해야 할 금액을 return 하도록 solution 함수를 완성해보세요.

제한사항

10 ≤ price ≤ 1,000,000

price는 10원 단위로(1의 자리가 0) 주어집니다.

소수점 이하를 버린 정수를 return합니다.

입출력 예

price result

150,000 142,500

580,000 464,000

입출력 예 설명

입출력 예 #1

150,000원에서 5%를 할인한 142,500원을 return 합니다.

입출력 예 #2

580,000원에서 20%를 할인한 464,000원을 return 합니다.

price	result
150,000	142,500
580,000	464,000

Description:

옷가게 할인 받기

Clothing store discount

머쓱이네 옷가게는 10만 원 이상 사면 5%,

If 머쓱이 spends more than 100,000 won in the clothing store, a 5% discount is given.

30만 원 이상 사면 10%,

A 10% discount is given above 300,000 won,

50만 원 이상 사면 20%를 할인해줍니다.

and a 20% discount is given above 500,000 won.

구매한 옷의 가격 price가 주어질 때,

Given the price,

지불해야 할 금액을 return 하도록 solution 함수를 완성해보세요.

return the discounted price to complete the solution function.

제한사항 (constraints):

The constraints section of the challenge informs us that price can have values ranging from 10 to 1,000,000.

price는 10원 단위로(1의 자리가 0) 주어집니다.

price is given in increments of 10 won (starting from 0).

소수점 이하를 버린 정수를 return합니다.

Returns an integer without the decimal point. (10000 not 10,000)

입출력 예 (input-output examples):

#1: When price = 150,000 the result should be 142,500.

#2: When price = 580,000 the result should be 464,000.

입출력 예 설명 (explanation of input-output examples):

Based on the example #1:

150,000원에서 5%를 할인한 142,500원을 return 합니다.

Returns 142,500 won, which is a 5% discount from 150,000 won.

Based on the example #2:

580,000원에서 20%를 할인한 464,000원을 return 합니다.

Returns 464,000 won, a 20% discount from 580,000 won.

JavaScript Solution

With this example, let's talk about when to use ternary operators (삼항 연산자) versus if...else statements (if...else 조건문).

In this example, we apply discounts based on the original price in three different ranges. If the price falls within a certain range, a discount percentage is applied to it, and we get a new discounted price:

If price is between 100,000 and 299,999, the discount applied is 5% and the new discounted price is calculated by multiplying price by 0.95.
If price is between 300,000 and 499,999, the discount applied is 10% and the new discounted price is calculated by multiplying price by 0.9.
If price is greater than or equal to 500,000, the discount applied is 20% and the new discounted price is calculated by multiplying price by 0.8.
If price is less than 100,000, no discount is applied and the original price is returned.

At the end of the function, we also have to use the Math.floor() method to round down the discounted price to the nearest whole number.

We can write this function using an if...else statement (if...else 조건문), which is easy to read and understand. However, if we try to use a ternary operator (삼항 연산자) to write this function, we will need to chain multiple ternary operators together, resulting in a long and hard-to-read code (Have you ever heard of spaghetti code?).

// if... else statement solution
function solution(price) {
  let discount = 0

  if (price >= 100000 && price < 300000) {
    discount = price * 0.95
  } else if (price >= 300000 && price < 500000) {
    discount = price * 0.9
  } else if (price >= 500000) {
    discount = price * 0.8
  } else {
    discount = price
  }

  return Math.floor(discount)
}

// ternary operator solution
function solution(price) {
  let discount = (price >= 100000 && price < 300000) ? price * 0.95 : (price >= 300000 && price < 500000) ? price * 0.9 : (price >= 500000) ? price * 0.8 : price
  return Math.floor(discount)
}

Wrap-up

Congratulations, you did it! 머쓱이 and I are both incredibly proud of you!

By learning the important vocabulary related to else...if statements, logical operators, and ternary operators, you've taken another step forward towards fluency in technical Korean.

And if you want to support this series and the work I put into it, you can do it by buying me a coffee here:

%%[bmc]

Thanks!

I hope you have an amazing weekend, and I look forward to seeing you next week!

[#1] KoreanCoding 101: Basic math (기초 수학)

Ha Anna — Sun, 07 May 2023 00:57:53 +0000

Let's start with a short announcement: A new post for this series will be published every Friday, and depending on the complexity of the challenges, I'll aim to cover around 2-4 of them.

Now, let's get into it! 💫

Do you enjoy math? If not, don't worry! The math we'll be covering today is something that most of us encounter in our daily lives.

Vocabulary

Nouns:

덧셈 - addition

뺄셈 - subtraction

곱셈 - multiplication

나눗셈 - division

나머지 - remainder (modulo)

합 - sum, total

차 - difference

곱 - product (multiplication)

정수 - integer (a whole number, e.g. 3)

분수 - fraction (a part of a whole, e.g. 0.5)

몫 - quotient (the integer part of a division)

함수 - function

매개변수 - parameter

제한사항 - constraints

입출력 예 - input-output examples

입출력 예 설명 - explanation of input-output examples

Verbs:

2에 1을 더하다 - to add 1 to 2

2에 1을 추가하다 - to add 1 to 2

2에 1을 빼다 - to subtract 1 from 2

2에 1를 곱하다 - to multiply 2 by 1

1을 2로 나누다 - to divide 1 by 2

Phrases:

[something]을/를 return 하도록 solution 함수를 완성해주세요.

Complete the solution function to return [something].

정수 num1과 num2가 주어질 때...

given two integers num1 and num2...

num1과 num2의 합

the sum of num1 and num2

num1에서 num2를 뺀 값

the value obtained by subtracting num2 from num1

num1과 num2를 곱한 값

the value obtained by multiplying num1 by num2

num1을 num2로 나눈 값

the value obtained by dividing num1 by num2

num1을 num2로 나눈 몫

the result of dividing num1 by num2

num1를 num2로 나눈 나머지

the remainder (modulo) of dividing num1 by num2

Challenges

두 수의 합

Description

정수 num1과 num2가 주어질 때, num1과 num2의 합을 return하도록 solution 함수를 완성해주세요.

제한사항

-50,000 ≤ num1 ≤ 50,000

-50,000 ≤ num2 ≤ 50,000

입출력 예

num1 num2 result

2 3 5

100 2 102

입출력 예 설명

입출력 예 #1

num1이 2이고 num2가 3이므로 2 + 3 = 5를 return합니다.

입출력 예 #2

num1이 100이고 num2가 2이므로 100 + 2 = 102를 return합니다.

num1	num2	result
2	3	5
100	2	102

Description:

When we look at the description of this challenge, we can already see some familiar words and phrases. Typically, coding challenges start by outlining the parameters that will be provided.

두 수의 합

The sum of two numbers

정수 num1과 num2가 주어질 때

Given two integers num1 and num2,

num1과 num2의 합을 return하도록 solution 함수를 완성해주세요.

complete the solution function to return the sum of num1 and num2.

제한사항 (constraints):

The constraints section of the challenge informs us that both num1 and num2 can have values ranging from -50,000 to 50,000.

입출력 예 (input-output examples):

Examples are a valuable tool for understanding the challenge description. In this case, we are provided with two input-output examples, which illustrate the given integers and the expected output from the solution function.

When num1 = 2 and num2 = 3 result should be 5.

입출력 예 설명 (explanation of input-output examples):

This section provides a detailed explanation of how the input-output process works for each example. For instance, in the given example where num1 = 2 and num2 = 3, we can read:

num1이 2이고 num2가 3이므로 2 + 3 = 5를 return합니다.

Since num1 is 2 and num2 is 3, the function returns 2 + 3 = 5.

JavaScript Solution

And now we know exactly what to do and we can complete our solution function.

function solution(num1, num2) {
    return num1 + num2
}

두 수의 차

Description

정수 num1과 num2가 주어질 때, num1에서 num2를 뺀 값을 return하도록 solution 함수를 완성해주세요.

제한사항

-50000 ≤ num1 ≤ 50000

-50000 ≤ num2 ≤ 50000

입출력 예

num1 num2 result

2 3 -1

100 2 98

입출력 예 설명

입출력 예 #1

num1이 2이고 num2가 3이므로 2 - 3 = -1을 return합니다.

입출력 예 #2

num1이 100이고 num2가 2이므로 100 - 2 = 98을 return합니다.

num1	num2	result
2	3	-1
100	2	98

Description:

This challenge is pretty similar to the last one, so let's dive straight into it!

두 수의 차

The difference between the two numbers

정수 num1과 num2가 주어질 때,

Given two integers num1 and num2,

num1에서 num2를 뺀 값을 return하도록 solution 함수를 완성해주세요.

complete the solution function to return the value obtained by subtracting num2 from num1.

제한사항 (constraints):

The constraints section of the challenge informs us that both num1 and num2 can have values ranging from -50,000 to 50,000.

입출력 예 (input-output examples):

When num1 = 2 and num2 = 3 result should be -1.

입출력 예 설명 (explanation of input-output examples):

Based on the example #1:

num1이 2이고 num2가 3이므로 2 - 3 = -1을 return합니다.

Since num1 is 2 and num2 is 3, the function returns 2 - 3 = -1.

JavaScript Solution

function solution(num1, num2) {
    return num1 - num2
}

두 수의 곱

Description

정수 num1, num2가 매개변수 주어집니다. num1과 num2를 곱한 값을 return 하도록 solution 함수를 완성해주세요.

제한사항

0 ≤ num1 ≤ 100

0 ≤ num2 ≤ 100

입출력 예

num1 num2 result

3 4 12

27 19 513

입출력 예 설명

입출력 예 #1

num1이 3, num2가 4이므로 3 * 4 = 12를 return합니다.

입출력 예 #2

num1이 27, num2가 19이므로 27 * 19 = 513을 return합니다.

num1	num2	result
3	4	12
27	19	513

Description:

두 수의 곱

The product of two numbers

정수 num1, num2가 매개변수 주어집니다.

Integers num1, num2 are given as parameters.

num1과 num2를 곱한 값을 return 하도록 solution 함수를 완성해주세요.

Complete the solution function to return the value obtained by multiplying num1 by num2.

제한사항 (constraints):

The constraints section of the challenge informs us that both num1 and num2 can have values ranging from 0 to 100.

입출력 예 (input-output examples):

When num1 = 3 and num2 = 4 result should be 12.

입출력 예 설명 (explanation of input-output examples):

Based on the example #1:

num1이 3, num2가 4이므로 3 * 4 = 12를 return합니다.

Since num1 is 3 and num2 is 4, the function returns 3 * 4 = 12.

JavaScript Solution

function solution(num1, num2) {
    return num1 * num2
}

두 수의 나눗셈

Description

정수 num1과 num2가 매개변수로 주어질 때, num1을 num2로 나눈 값에 1,000을 곱한 후 정수 부분을 return 하도록 solution 함수를 완성해주세요.

제한사항

0 < num1 ≤ 100

0 < num2 ≤ 100

입출력 예

num1 num2 result

3 2 1500

7 3 2333

1 16 62

입출력 예 설명

입출력 예 #1

num1이 3, num2가 2이므로 3 / 2 = 1.5에 1,000을 곱하면 1500이 됩니다.

입출력 예 #2

num1이 7, num2가 3이므로 7 / 3 = 2.33333...에 1,000을 곱하면 2333.3333.... 이 되며, 정수 부분은 2333입니다.

입출력 예 #3

num1이 1, num2가 16이므로 1 / 16 = 0.0625에 1,000을 곱하면 62.5가 되며, 정수 부분은 62입니다.

num1	num2	result
3	2	1500
7	3	2333
1	16	62

Description:

두 수의 나눗셈

The division of two numbers

정수 num1과 num2가 매개변수로 주어질 때,

Integers num1 and num2 are given as parameters,

num1을 num2로 나눈 값에 1,000을 곱한 후 정수 부분을 return 하도록 solution 함수를 완성해주세요.

First divide num1 by num2 and then multiply the resulting value by 1000. To complete the solution function, the returned value should always be an integer.

제한사항 (constraints):

The constraints section of the challenge informs us that both num1 and num2 can have values ranging from 0 to 100.

입출력 예 (input-output examples):

Example #1:

When num1 = 3 and num2 = 2 result should be 1500.

Example #2:

When num1 = 7 and num2 = 3 result should be 2333.

입출력 예 설명 (explanation of input-output examples):

Based on the example #1:

num1이 3, num2가 2이므로 3 / 2 = 1.5에 1,000을 곱하면 1500이 됩니다.

Since num1 is 3 and num2 is 2, 3 / 2 = 1.5 multiplied by 1,000 is 1500.

Based on the example #2:

num1이 7, num2가 3이므로 7 / 3 = 2.33333...에 1,000을 곱하면 2333.3333.... 이 되며, 정수 부분은 2333입니다.

Since num1 is 7 and num2 is 3, 7/3 = 2.33333... Multiplied by 1,000 it is 2333.333... The integer part is 2333.

JavaScript Solution

In this challenge, simply dividing and multiplying will not be enough since it will give us a repeating decimal when we are required to return just the integer. To return the correct solution, we can wrap (num1/num2)*1000 in Math.floor(), which will return the largest integer that is less than or equal to a given number.

function solution(num1, num2) {
    return Math.floor((num1/num2) * 1000)
}

나머지 구하기

Description

정수 num1, num2가 매개변수로 주어질 때, num1를 num2로 나눈 나머지를 return 하도록 solution 함수를 완성해주세요.

제한사항

0 < num1 ≤ 100

0 < num2 ≤ 100

입출력 예

num1 num2 result

3 2 1

10 5 0

입출력 예 설명

입출력 예 #1

num1이 3, num2가 2이므로 3을 2로 나눈 나머지 1을 return 합니다.

입출력 예 #2

num1이 10, num2가 5이므로 10을 5로 나눈 나머지 0을 return 합니다.

num1	num2	result
3	2	1
10	5	0

Description:

나머지 구하기

Finding the rest

정수 num1, num2가 매개변수로 주어질 때,

Given two integers num1 and num2,

num1를 num2로 나눈 나머지를 return 하도록 solution 함수를 완성해주세요.

Complete the solution function to return the remainder (modulo) of num1 by num2.

제한사항 (constraints):

The constraints section of the challenge informs us that both num1 and num2 can have values ranging from 0 to 100.

입출력 예 (input-output examples):

When num1 = 3 and num2 = 2 result should be 1.

입출력 예 설명 (explanation of input-output examples):

Based on the example #1:

num1이 3, num2가 2이므로 3을 2로 나눈 나머지 1을 return 합니다.

Since num1 is 3 and num2 is 2, 3 is divided by 2 and the remainder is 1.

JavaScript Solution

If you don't remember how the remainder works, you can read all about it here.

We will use the % symbol in our solution function, as it represents the remainder in computer languages.

function solution(num1, num2) {
    return num1 % num2
}

몫 구하기

Description

정수 num1, num2가 매개변수로 주어질 때, num1을 num2로 나눈 몫을 return 하도록 solution 함수를 완성해주세요.

제한사항

0 < num1 ≤ 100

0 < num2 ≤ 100

입출력 예

num1 num2 result

10 5 2

7 2 3

입출력 예 설명

입출력 예 #1

num1이 10, num2가 5이므로 10을 5로 나눈 몫 2를 return 합니다.

입출력 예 #2

num1이 7, num2가 2이므로 7을 2로 나눈 몫 3을 return 합니다.

num1	num2	result
10	5	2
7	2	3

Description:

몫 구하기

Finding quotient (the integer part of a division)

정수 num1, num2가 매개변수로 주어질 때,

Integers num1, num2 are given as parameters,

num1을 num2로 나눈 몫을 return 하도록 solution 함수를 완성해주세요.

Complete the solution function by returning the integer part of a division of num1 by num2.

제한사항 (constraints):

The constraints section of the challenge informs us that both num1 and num2 can have values ranging from 0 to 100.

입출력 예 (input-output examples):

When num1 = 10 and num2 = 5 result should be 2.

입출력 예 설명 (explanation of input-output examples):

For example #1:

num1이 10, num2가 5이므로 10을 5로 나눈 몫 2를 return 합니다.

Since num1 is 10 and num2 is 5, 10 divided by 5 returns the quotient of 2.

JavaScript Solution

Here again, since some of the divisions will give us a repeating decimal. To return just an integer, we can simply use Math.floor().

function solution(num1, num2) {
    return Math.floor(num1/num2)
}

Wrap-up

Thank you for following along with this lesson! I hope you found it helpful. Remember, it's important to take things at your own pace and gradually build your skills when it comes to solving coding challenges, especially if you are also learning a new language. Building a strong technical Korean foundation is essential for tackling more complex challenges with ease.

Next week, we'll dive into topics that go beyond basic math, such as if/else statements, booleans, ternary operators, and more! I look forward to continuing this learning journey with you.

In the meantime, I hope you all have a fantastic weekend!

I'll see you next Friday!

KoreanCoding 101: Programmers.co.kr Account registration

Ha Anna — Sun, 07 May 2023 00:56:22 +0000

Are you trying to improve your technical Korean skills and looking to get started with some challenges in Korean? If yes, then you've come to the right place! In this post, we'll go through the process of registering on programmers.co.kr, a website that offers a variety of coding challenges and exercises.

How to register?

To start, head over to programmers.co.kr and click on the 회원가입 (membership registration) button, located in the upper right corner of the website.

From there, you can sign up using your name (이름), email (이메일), and password (비밀번호) or choose to connect your Google, Kakao, Github, or Facebook account.

Once you've created an account, you'll be prompted to accept the website's terms of use (이용약관), allow the collection and use of your personal information (프로그래머스 개인정보 수집 및 이용 동의), and confirm that you're 14 years or older (만 14세 이상입니다). You'll also have the option to agree to receive marketing emails (마케팅 활용 동의 및 광고 수신 동의).

After completing the initial registration, you'll be asked to provide additional information, such as your desired job (희망직무) and main technology (주요기술). For your desired job, choose your area of interest, and for your main technology, select the programming language you use the most.

Once you've provided all the necessary information, you're almost ready to start exploring the website. But first, you'll need to verify your account by clicking on the link sent to your email.

What now?

Now that you're registered and verified, it's time to start exploring the website. I recommend filling out your profile with relevant information and trying out some coding challenges. You can find the coding challenges under 코딩테스트 연습 (coding test practice) in the website's navbar. If you're new to coding challenges, I suggest starting with the beginner-friendly challenge, 두 수의 합 (no translation this time, try to figure out what to do by just looking at input and output!).

If you have any questions or need assistance, feel free to leave a comment. I'm always happy to help! In the next post, we'll tackle our first challenge together, so stay tuned.

Thanks for reading!

KoreanCoding 101: Introduction

Ha Anna — Sun, 07 May 2023 00:51:14 +0000

Hi everyone, my name is Anna and I'm excited to share my new blog series, "KoreanCoding 101: Technical Korean for Non-Korean Speakers", with you. As a non-native Korean speaker living in South Korea and trying to find a job as a Frontend Developer in a Korean company, I've realized how important it is to have a strong grasp of Technical Korean for coding challenges.

I've been studying Korean for a while now and even hold a TOPIK level 4, but I found that my technical Korean wasn't quite up to par for the coding challenges and tests required during job interviews. So, I started practicing on programmers.co.kr, a popular platform used by many Korean tech companies.

As I worked through these challenges, I realized that it would be helpful to have a resource that explains the technical Korean terms commonly used in coding challenges. That's why I decided to start this blog series, where I'll walk you through each challenge step-by-step, breaking down the technical terms and explaining my thought process along the way. While I'll be using JavaScript to tackle these challenges, the principles and concepts we'll cover can be applied to any programming language. So even if you prefer working with Python or C, you can still follow along and apply these techniques to your language of choice.

My hope is that this series will be helpful to other non-Korean speakers who are interested in joining the Korean tech industry but may be struggling with the language barrier in coding challenges. Join me as we explore Technical Korean and work towards becoming more confident and proficient in our coding skills!

Edit:

Someone kindly pointed out that it would be great if I shared a bit more about my background and explained my interest in seeking employment with Korean companies.

Let me give you some context: I've been living in South Korea since 2021, but my ties with this country actually started back in 2016. I have family here, and I plan on continuing to call Korea my home for many years to come. So, it only made sense to explore job opportunities in Korea.

But my reasons for seeking employment here go beyond just familial ties. For one, there are so many exciting startups and companies in Korea that I can see myself being a part of. I'm drawn to the innovation and creativity that I've seen in the Korean business scene. Plus, I prefer working face-to-face with colleagues. As a freelancer, I often work remotely and don't get much social interaction. So, the idea of being a part of a team in person is really appealing to me.

Lastly, I should mention that I genuinely love my life here in Korea. I appreciate the culture, the food, and the people. It's a beautiful country with so much to offer, and I feel fortunate to have the opportunity to be here.

So, that's a bit about me and my reasons for seeking employment with Korean companies. I hope this explanation helps to clarify my background and motivations.

Github: How to fix broken image paths

Ha Anna — Sun, 03 Jul 2022 11:54:13 +0000

Are you trying to deploy your page to Github, and suddenly, after everything was uploaded, your images stop showing up?
No worries, many people encounter this issue!

^{As an active member of the Scrimba discord community, I have noticed many beginners struggle when deploying their pages in the beginning. So I am writing this article in hopes of helping more people learn how to troubleshoot their code.}

To show how to troubleshoot this issue step by step, I created a folder that consists of index.html, style.css and a cat.jpg.
See Live Page
See Github Repository

We need to check two things:

Our image's location in the Github repository
Our HTML/CSS code

Depending on our Github repository's structure, we will have to tweak HTML/CSS code to make the file paths work again.

File position case scenarios

All files in one folder

Here, we can see that all files are together, which makes them siblings.
In this case, your image tag should look like this:

 <img src="cat.jpg" alt="Kitty">

Image files in another folder

Often, images are in a separate folder, and the image path has to be structured differently.

Here, the images folder is a sibling of index.html and style.css; however, the cat.jpg is a child of the images folder.

In this case, your image tag should look like this:

<img src="images/cat.jpg" alt="">

I recommend using inspector on your HTML/CSS. This way, you can check the image paths and change them without having to make any commits (but remember to edit your code after finding the correct solution!):

And that's it!
Happy coding!

Additional useful links:

Inspect Element: How to Temporarily Edit Any Webpage by Zapier.com
GitHub Pages Tutorial by Happycoding.io

DEV Community: Ha Anna

[#3] KoreanCoding 101: for loops and arrays

Arrays | 배열

for loop | for 문

Vocabulary

Challenges

Wrap-up

[#2] KoreanCoding 101: operators, if... else, ternary operator

Operators | 연산자

if... else statement | if... else 조건문

Ternary operator | 삼항 연산자

Vocabulary

Challenges

Wrap-up

[#1] KoreanCoding 101: Basic math (기초 수학)

Vocabulary

Challenges

Wrap-up

KoreanCoding 101: Programmers.co.kr Account registration

How to register?

What now?

KoreanCoding 101: Introduction

Github: How to fix broken image paths

We need to check two things:

File position case scenarios

All files in one folder

Image files in another folder

`for` loop | `for` 문

`if... else` statement | `if... else` 조건문