DEV Community: Cesar Del rio

#46 - Sum of a sequence - Codewars Kata (7 kyu)

Cesar Del rio — Sun, 14 Aug 2022 00:43:13 +0000

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Instructions

Task:
our task is to make function, which returns the sum of a sequence of integers.

The sequence is defined by 3 non-negative values: begin, end, step (inclusive).

If begin value is greater than the end, function should returns 0

Examples

2,2,2 --> 2
2,6,2 --> 12 (2 + 4 + 6)
1,5,1 --> 15 (1 + 2 + 3 + 4 + 5)
1,5,3 --> 5 (1 + 4)

My solution:

const sequenceSum = (begin, end, step) => {
  var count = 0;
  for (let i = begin; i <= end; i += step) {
    count += i;
  }
  return count;
};

Explanation

First I made the variable count, in which I'll save the results of the sums in the loop

var count = 0;

Then I used a for loop, in which I started with the value of the begin param, it will iterate until the "i" value is equal to the end value, and in each iteration the "i" value will be summed the "step" param value.
And in each iteration I summed the "i" value to count

  for (let i = begin; i <= end; i += step) {
    count += i;
  }

At the end I just returned count

return count;

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Any advice for Google STEP program?

Cesar Del rio — Mon, 23 May 2022 02:11:12 +0000

Who am I? 🤔

Hi guys, my name is Cesar Del Rio, I'm a 17 year old highschool student from Mexico , I specialize in Javascript with React.js 👨🏽‍💻.

My goal 🎯

I've been investigating about the Google STEP program for first and second year college students, and I'm so excited to apply 💪🏼, I'd be very grateful if you could give me some advice to have more possibility of being selected or telling me a little bit about your experience :)

Please comment what advice would you give me for this 🙌🏼

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#45 - Persistent Bugger - Codewars Kata (6 kyu)

Cesar Del rio — Sat, 21 May 2022 17:25:01 +0000

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Instructions

Task:
Write a function, persistence, that takes in a positive parameter num and returns its multiplicative persistence, which is the number of times you must multiply the digits in num until you reach a single digit.

Example

39 --> 3 (because 3*9 = 27, 2*7 = 14, 1*4 = 4 and 4 has only one digit)
999 --> 4 (because 9*9*9 = 729, 7*2*9 = 126, 1*2*6 = 12, and finally 1*2 = 2)
4 --> 0 (because 4 is already a one-digit number)

My solution:

function persistence(num) {
  let count = 0;
  while(num.toString().length > 1){
    num = num.toString().split('').reduce((acc,el)=>acc *= +el,1)
    count++
  }
  return count
}

Explanation

First I created the variable count, where I'll save the number of times the number had to be multiplied so it gets to only one digit

let count = 0;

Then I used a while loop that will iterate until the length of the number is 1

  while(num.toString().length > 1)

Inside of that while loop, I revalued the variable "num" to save the result of the multiplication of the digits, I did this by splitting the number into an array, and then I just used the reduce function to save the result of the multiplication of the digits.

Note: In the next iteration of the loop the variable "num" will be the result of the multiplication from the last digits

After that I just added one to the count variable

  num = num.toString().split('').reduce((acc,el)=>acc *= +el,1)
  count++

At the end I just returned count

  return count

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#44 - Where is my parent!?(cry) - Codewars Kata (6 kyu)

Cesar Del rio — Sat, 14 May 2022 20:19:57 +0000

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Instructions

Mothers arranged a dance party for the children in school. At that party, there are only mothers and their children. All are having great fun on the dance floor when suddenly all the lights went out. It's a dark night and no one can see each other. But you were flying nearby and you can see in the dark and have ability to teleport people anywhere you want.

Legend:
-Uppercase letters stands for mothers, lowercase stand for their children, i.e. "A" mother's children are "aaaa".
-Function input: String contains only letters, uppercase letters are unique.

Task:
Place all people in alphabetical order where Mothers are followed by their children, i.e. "aAbaBb" => "AaaBbb".

My solution:

function findChildren(str) {
  str = str.toLowerCase()
//   beeeebb
  let arr = str.split('')
//   [ 'b', 'e', 'e', 'e', 'e', 'b', 'b' ]
  var r = '';
  let letters = [...new Set(arr)].sort()
//   ['b','e']

  for(let i = 0; i<letters.length; i++){
//     Will iterate ['b','e']
    for(let j = 0; j<str.length; j++){
//       Will iterate //   [ 'b', 'e', 'e', 'e', 'e', 'b', 'b' ]
      if(letters[i] == arr[j]) r += arr[j]
    }
  }
  return r.split('').map((x,i)=> x !== r[i-1] ? x = x.toUpperCase() : x ).join('')
//   in the map, if the actual letter isn't the same as the last one it'll make it upperCase becuase it is the first one of those group
}

Explanation

First I converted the string to lower case, and then I splitted it into an array, and I made the variable "r" with an empty string, for saving the result there

  str = str.toLowerCase()
//   beeeebb
  let arr = str.split('')
//   [ 'b', 'e', 'e', 'e', 'e', 'b', 'b' ]
  var r = '';

Then I made the variable "letters" I used a new Set() into the array with the letters so I could get the letters that the array contains, and then I sorted it so I could get it in alphabetical order.

    let letters = [...new Set(arr)].sort()
//   ['b','e']

Then I used 2 for loops, the first one to iterate the Ungrouped letters that the string contains and the second one to iterate all the letters that the string contains mixed.
Inside of the second loop I used a conditional, if the letter that the group contains is equal to the letter being iterated on the mixed string, it'll added to "r"

 for(let i = 0; i<letters.length; i++){
//     Will iterate ['b','e']
    for(let j = 0; j<str.length; j++){
//       Will iterate //   [ 'b', 'e', 'e', 'e', 'e', 'b', 'b' ]
      if(letters[i] == arr[j]) r += arr[j]
    }
  }

At the end "r" is equal to a string with the letters in order but without the first letter being a capital letter ('bbbeeee'), and I want it to be ('BbbEeee').
So I splitted "r" into an array and then I mapped it, if the current letter isn't equal to the letter behind it it means it needs to be Capital because it's the first one of the group, so I make it upperCase, if not I just leave as it is.
And I return this result

return r.split('').map((x,i)=> x !== r[i-1] ? x = x.toUpperCase() : x ).join('')
//   in the map, if the actual letter isn't the same as the last one it'll make it upperCase becuase it is the first one of those group

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How can you help?

Cesar Del rio — Mon, 09 May 2022 01:22:14 +0000

Hi 👋🏼, my name is Cesar Del Rio, I'm a 17 year old software developer 👨🏼‍💻, I'm posting a new Codewars Kata Solution everyday, with the explanation of it.

If you like my content and want to support me to continue my software developer journey 👌🏼, I'll be very grateful :)

You can support by buying a coffee ☕️

#43 - If you can read this... - Codewars Kata (6 kyu)

Cesar Del rio — Sun, 08 May 2022 19:45:07 +0000

Instructions

Task
You'll have to translate a string to Pilot's alphabet (NATO phonetic alphabet).

Note:

There are preloaded dictionary you can use, named NATO
The set of used punctuation is ,.!?.
Punctuation should be kept in your return string, but spaces should not.
Xray should not have a dash within.
Every word and punctuation mark should be seperated by a space ' '.
There should be no trailing whitespace

Example

Input:

If, you can read?

Output:

India Foxtrot , Yankee Oscar Uniform Charlie Alfa November Romeo Echo Alfa Delta ?

My solution:

function to_nato(words) {
  let arr = words.split('').filter(x=> x!== ' ')
  let r = ''
  arr.map((x,i)=>{
    let letter = x.toLowerCase()
    let coded = NATO[letter]
    if(i !== arr.length - 1 && coded) r+=coded + ' '
    else if(i == arr.length- 1 && coded) r += coded
    else if(i !== arr.length - 1 && !coded) r += letter + ' '
    else r+=letter
  })
  return r
}

Explanation

First I created a variable with the string splitted into an array, and another variable "r" with an empty string, where I'll save the last result.

  let arr = words.split('').filter(x=> x!== ' ')
  let r = ''

Then I mapped that array with 2 parameters, x: the letter being mapped and i: the index of it.
First in the loop I converted to lower case the letter and after that I made the variable "coded" to save the letter in pilots alphabet with the NATO object.

  arr.map((x,i)=>{
    let letter = x.toLowerCase()
    let coded = NATO[letter]

Then I used 4 conditionals:
1- If the index of the letter isn't equal to the array length - 1, it means that it isn't the last one, and coded isn't undefined it means it exist, so I'll add to the result string the coded plus a space

    if(i !== arr.length - 1 && coded) r+=coded + ' '
    else r+=letter

2- Else if the letter is the last one in the array and coded exist I'll just add coded to the result without the sapce because is the last letter

    else if(i == arr.length- 1 && coded) r += coded

3- Else if it isn't the last letter and there's no coded string it means that it is one of the symbols like .?!, so I just add the letter with an space

    else if(i !== arr.length - 1 && !coded) r += letter + ' '
    else r+=letter

4- If any condition is true it means that it is a symbol and it's the last element of the array, so It doesn't need a space, so I just add the letter

    else r += letter

At the end I just returned r

    return r

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#42 - How many pages in a book? - Codewars Kata (6 kyu)

Cesar Del rio — Thu, 21 Apr 2022 00:02:06 +0000

Instructions

Task
Every book has n pages with page numbers 1 to n. The summary is made by adding up the number of digits of all page numbers.

Task: Given the summary, find the number of pages n the book has.

Example

If the input is summary=25, then the output must be n=17: The numbers 1 to 17 have 25 digits in total: 1234567891011121314151617.

Be aware that you'll get enormous books having up to 100.000 pages.

All inputs will be valid.

My solution:

function amountOfPages(summary){
  if(summary <= 9) return summary
  var count = 0;
  var res = 0;
  for(let i = 1; i<summary; i++){
    count+=i.toString().length
    if(count==summary) res = i
  }
  return res
}

Explanation

First I used a conditional if the summary pages are less than 9 it means that the digits are the same as the original one

if(summary <= 9) return summary

Then I created two variables:

count where I'll store the count of the pages digits.
res where I'll store the last result.

var count = 0;
var res = 0;

I used a for loop so I could go through every page and in every iteration I add to the variable count the total digits of the page number that is being iterated. Until the variable count is equal to summary, that means that I'm located in the exact page number "i", so I set "res" as the variable "i".

At the end I just return res

for(let i = 1; i<summary; i++){
  count+=i.toString().length
  if(count==summary) res = i
}

return res

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#41 - Split Strings Codewars Kata (6 kyu)

Cesar Del rio — Thu, 07 Apr 2022 01:23:39 +0000

Instructions

Task
Complete the solution so that it splits the string into pairs of two characters. If the string contains an odd number of characters then it should replace the missing second character of the final pair with an underscore ('_').

Examples:

'abc' => ['ab', 'c_']

'abcdef' => ['ab', 'cd', 'ef']

My solution:

function solution(str){
   var arr = str.split('')
   var res = []

   for(let i = 0; i<arr.length; i+=2){
     if(arr[i+1]) res.push(arr[i]+arr[i+1])
     else res.push(arr[i]+'_')
   }

  return res
}

Explanation

First I made an array of the string, and an array to save the result

   var arr = str.split('')
   var res = []

After that I used a loop to iterate through the array, and in every iteration I'll add two to the value of "i", so it iterates every two elements.
Inside of this loop I'll check if there is another element next to the one being iterated if it has another one It'll push the current element and the next one to the result array, if not it'll push the current element plus an underscore.

   for(let i = 0; i<arr.length; i+=2){
     if(arr[i+1]) res.push(arr[i]+arr[i+1])
     else res.push(arr[i]+'_')
   }

At the end I just return the result array

   return res

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#40 - Credit Card Mask Kata (7 kyu)

Cesar Del rio — Wed, 06 Apr 2022 05:34:10 +0000

Instructions

Task
Usually when you buy something, you're asked whether your credit card number, phone number or answer to your most secret question is still correct. However, since someone could look over your shoulder, you don't want that shown on your screen. Instead, we mask it.

Your task is to write a function maskify, which changes all but the last four characters into '#'.

Examples

"4556364607935616" --> "############5616"
     "64607935616" -->      "#######5616"
               "1" -->                "1"
                "" -->                 ""

// "What was the name of your first pet?"

"Skippy" --> "##ippy"

"Nananananananananananananananana Batman!"
-->
"####################################man!"

My solution:

function maskify(cc) {
    return cc.split("").map((x,i) => (i<cc.length-4)?x="#":x).join("");
}

Explanation

First I splitted into an array the string, then I mapped it and in every iteration I checked if the index of the element being iterated isn't smaller than the length of the string minus 4, if it isn't small it means that it hasn't reach the last 4 characters, so I change the number for a "#", but if it has reached the last 4 characters I just leave them as they are, and at the end I just join the array.

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#39 - Two Sum CodeWars Kata (6 kyu)

Cesar Del rio — Tue, 22 Mar 2022 03:54:09 +0000

Instructions

Task
Write a function that takes an array of numbers (integers for the tests) and a target number. It should find two different items in the array that, when added together, give the target value. The indices of these items should then be returned in a tuple / list (depending on your language) like so: (index1, index2).

For the purposes of this kata, some tests may have multiple answers; any valid solutions will be accepted.

The input will always be valid (numbers will be an array of length 2 or greater, and all of the items will be numbers; target will always be the sum of two different items from that array).

Based on: http://oj.leetcode.com/problems/two-sum/

Examples:

twoSum [1, 2, 3] 4 === (0, 2)

My solution:

function twoSum(numbers, target) {
  for(let i = 0; i<numbers.length; i++){
    for(let j = i+1; j<numbers.length; j++){
      let res = numbers[i] + numbers[j]
      if(res == target) return [i,j]
    }
  }
}

Explanation

First I made a loop that will iterate through all the elements of the array

for(let i = 0; i<numbers.length; i++)

Inside of this loop I used another loop that will iterate through all the elements that are after the one that is being iterated in the last loop.

for(let j = i+1; j<numbers.length; j++)

In each iteration I'll check if the sum of the element iterated in the first loop (i) plus the one being iterated in the last loop (j) is equal to the target, it'll return their positions

let res = numbers[i] + numbers[j]
if(res == target) return [i,j]

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#38 - Assemble string CodeWars Kata (6 kyu)

Cesar Del rio — Sat, 26 Feb 2022 02:07:06 +0000

Instructions

Task
In this task, you need to restore a string from a list of its copies.

You will receive an array of strings. All of them are supposed to be the same as the original but, unfortunately, they were corrupted which means some of the characters were replaced with asterisks ("*").

You have to restore the original string based on non-corrupted information you have. If in some cases it is not possible to determine what the original character was, use "#" character as a special marker for that.

If the array is empty, then return an empty string.

Examples:

input = [
  "a*cde",
  "*bcde",
  "abc*e"
]
result = "abcde"


input = [
  "a*c**",
  "**cd*",
  "a*cd*"
]
result = "a#cd#"

My solution:

function assembleString(arr){
  let r = []
  for(let i = 0; i<arr.length; i++){
    for(let j = 0; j<arr[i].length; j++){
      if(i==0) r.push(arr[i][j])
      if(r[j] == '*' || arr[i][j] !== '*') r[j] = arr[i][j]
    }
  }
  return r.map(x=>x=='*' ? '#' : x).join('')
}

Explanation

First I made an empty array in the variable "r"

let r = []

After that I used 2 for loops, the first one to iterate through arr, and then the other to iterate into every array inside of it that contained a word, inside of the second loop, first I checked if it is the first word array, I will push it as it is, so I can get it as a base for the other iterations, and I can start from something.

for(let i = 0; i<arr.length; i++){
    for(let j = 0; j<arr[i].length; j++){
      if(i==0) r.push(arr[i][j])

After that I used another conditional that checked if the the element in "r" that is in the same index position we are iterating is a "" and the character that I'm iterating isn't a '' it means that we discovered another character, so I just change "*" to the correct character

if(r[j] == '*' || arr[i][j] !== '*') r[j] = arr[i][j]

At the end I just returned the "r" array, that contained the last result, but first I mapped it to change every "*" to a "#" and to join it into an array

return r.map(x=>x=='*' ? '#' : x).join('')

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#37 - Row of the odd triangle - CodeWars Kata (6 kyu)

Cesar Del rio — Mon, 21 Feb 2022 06:33:58 +0000

Instructions

Given a triangle of consecutive odd numbers:

               1
            3     5
         7     9    11
      13    15    17    19 
   21    23    25    27    29

find the triangle's row knowing its index (the rows are 1-indexed), e.g.:

odd_row(1) == [1]
odd_row(2) == [3, 5]
odd_row(3) == [7, 9, 11]

Note: your code should be optimized to handle big inputs.

My solution:

function oddRow(n) {
  let first = 1;
  for(let i = 1; i<n; i++){
    first+=i*2
  }

  let r = [first];
  for(let i = 1; i<n; i++){
    r.push(first+=2)
  }

  return r
}

Explanation

First I got the first value of the row that I'm searching on, I did this by using a for loop that will iterate the number of rows, and in every iteration I changed the value of the var "first" by adding 2 to it and revaluing it.

  for(let i = 1; i<n; i++){
    first+=i*2
  }

After that with that value I made the variable "r" that will contain the result that will be returned, inside of it I stored the first number of the row

let r = [first];

After that I used another loop and inside of it, I pushed every value after the first plus 2, and at the end I just returned "r"

for(let i = 1; i<n; i++){
    r.push(first+=2)
  }

  return r

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