If two or more resistors are connected in parallel, the overall resistance of the circuit reduces. It is possible to calculate the total resistance of a parallel circuit by using this formula:

Create a function that takes an array of parallel resistance values, and calculates the total resistance of the circuit.

Worked Example

parallelResistance([6, 3, 6]) ➞ 1.5

// 1/RTotal = 1/6 + 1/3 + 1/6
// 1/RTotal = 2/3
// RTotal = 3/2 = 1.5

Examples

parallelResistance([6, 3]) ➞ 2

parallelResistance([10, 20, 10]) ➞ 4

parallelResistance([500, 500, 500]) ➞ 166.6
// Round to the nearest decimal place

Notes

Note that you should rearrange to return the Resistance Total, not 1 / Resistance Total.
Round to the nearest decimal place.
All inputs will be valid.

Solution:

function parallelResistance(arr) {
  return parseFloat(
    (1 / arr.reduce((acc, curr) => (acc += 1 / curr), 0)).toFixed(1)
  );
}

Calculate an Earned Run Average [Edabit]

August 31, 2021August 28, 2021 Ketan RamtekeLeave a comment

Edabit

Create a function that returns an Earned Run Average (ERA). An ERA is calculated by multiplying 9 by the quotient of Earned Runs Allowed er divided by ip Innings Pitched.

In baseball statistics, innings are represented with a fractional part of .1 (1/3) or .2 (2/3) to represent the number of outs in an inning. A whole number or a number with a fractional part of .0 represents a full inning with three outs. Check the Resources tab for a deeper explanation.

Examples

era(22, 99) ➞ 2.00

era(23, 99.1) ➞ 2.08

era(24, 99.2) ➞ 2.17

Notes

ERA is represented with a scale of 2: 2.08
For 1/3 and 2/3, use a scale of 2.

Solution:

1
2
3

function era(er, ip) {
	return ((er/ip)*9).toFixed(4).substr(0,4);
}

Majority Vote [Edabit]

August 30, 2021August 26, 2021 Ketan RamtekeLeave a comment

Edabit

Create a function that returns the majority vote in an array. A majority vote is an element that occurs > N/2 times in an array (where N is the length of the array).

Examples

majorityVote(["A", "A", "B"]) ➞ "A"

majorityVote(["A", "A", "A", "B", "C", "A"]) ➞ "A"

majorityVote(["A", "B", "B", "A", "C", "C"]) ➞ null

Notes

The frequency of the majority element must be strictly greater than 1/2.
If there is no majority element, return null.
If the array is empty, return null.

Solution:

function majorityVote(arr) {
  let obj = {};
  if (!arr.length) {
    return null;
  }
  let set = new Set(arr);
  if (set.size === 1) {
    return arr[0];
  }
  for (let vote of arr) {
    if (obj[vote]) {
      obj[vote] += 1;
    } else {
      obj[vote] = 1;
    }
  }
  let sol = Object.entries(obj).sort((a, b) => b[1] - a[1]);
  return sol[0][1] == sol[sol.length - 1][1] ? null : sol[0][0];
}

Multiplying Numbers in a String [Edabit]

August 29, 2021August 29, 2021 Ketan RamtekeLeave a comment

Edabit

Given a string of numbers separated by a comma and space, return the product of the numbers.

Examples

multiplyNums("2, 3") ➞ 6

multiplyNums("1, 2, 3, 4") ➞ 24

multiplyNums("54, 75, 453, 0") ➞ 0

multiplyNums("10, -2") ➞ -20

Notes

Bonus: Try to complete this challenge in one line!

Solution:

1
2
3

function multiplyNums(nums) {
  return nums.split(",").reduce((acc, curr) => (acc *= parseInt(curr)), 1);
}

Spoonerisms [Edabit]

August 29, 2021August 25, 2021 Ketan RamtekeLeave a comment

Edabit

A spoonerism is when the first letters / sounds of two words are transposed onto one another. Create a function that takes a two-word string and performs a spoonerism on the phrase.

Examples

spoonerise("history lecture") ➞ "listory hecture"

spoonerise("loud noises") ➞ "noud loises"

spoonerise("chow mein") ➞ "mow chein"

spoonerise("edabit rules!") ➞ "redabit ules!"

Notes

Only two words will be parsed into the function. Don’t worry about handling more than two.
You won’t always just have to swap the first letters, take care to notice which letters have been switched in the examples (notice the difference between vowel-starting and consonant-starting words).

Solution:

function spoonerise(phrase) {
  let [w1, w2] = phrase.split(" ");
  let i = 0;
  while (!"aeiou".includes(w1[i])) {
    i++;
  }
  let [prew1, postw1] = [w1.substr(0, i), w1.substr(i)];
  i = 0;
  while (!"aeiou".includes(w2[i])) {
    i++;
  }
  let [prew2, postw2] = [w2.substr(0, i), w2.substr(i)];
  return prew2 + postw1 + " " + prew1 + postw2;
}

pigLatin 3.0 [Edabit]

August 28, 2021August 24, 2021 Ketan RamtekeLeave a comment

Edabit

Write a function that converts a sentence into pig latin.

Rules for converting to pig latin:

For words that begin with a vowel (a, e, i, o, u), add “way”.
Otherwise, move all letters before the first vowel to the end and add “ay”.
For simplicity, no punctuation will be present in the inputs.

Examples

pigLatinSentence("this is pig latin") ➞ "isthay isway igpay atinlay"

pigLatinSentence("wall street journal") ➞ "allway eetstray ournaljay"

Notes

All letters will be in lowercase.

Solution:

function pigLatinSentence(sentence) {
  let arr = sentence.split(" ");
  let sol = arr.map((word) => {
    if ("aeiouAEIOU".includes(word[0])) {
      return word + "way";
    } else {
      let i = 0;
      while (!"aeiouAEIOU".includes(word[i])) {
        i++;
      }
      return word.substr(i) + word.substr(0, i) + "ay";
    }
  });
  return sol.join(" ");
}

Making a Box 2.0! [Edabit]

August 27, 2021August 23, 2021 Ketan RamtekeLeave a comment

Edabit

This is based on Helen Yu’s Making a Box challenge. This challenge is the same execpt that instead of an array of strings, your function should output a matrix of characters.

Examples

charBox(1) ➞ [
  ["#"]
]

charBox(4) ➞ [
  ["#", "#", "#", "#"],
  ["#", " ", " ", "#"],
  ["#", " ", " ", "#"],
  ["#", "#", "#", "#"]
]

charBox(2) ➞ [
  ["#", "#"],
  ["#", "#"]
]

Notes

As an added bonus, try making charBox(0) output [[]] and make any strings, non-integers, and negative numbers output -1.

Solution:

function charBox(size) {
  let sol = [];
  if (size == 1) {
    return [["#"]];
  }
  let first = Array.from({ length: size }, () => "#");
  let mid = Array.from({ length: size - 2 }, () => " ");
  mid.unshift("#");
  mid.push("#");
  for (let i = 0; i < size; i++) {
    if (i == 0 || i == size - 1) {
      sol.push(first);
    } else {
      sol.push(mid);
    }
  }
  if (isNaN(parseInt(size)) || size < 0 || Math.round(size) !== size) {
    return -1;
  }
  return size == 0 ? [[]] : sol;
}

Name Count Equality [Edabit]

August 26, 2021August 22, 2021 Ketan RamtekeLeave a comment

Edabit

Create a function that counts the embedded names in the string and determines the equality. The names are embedded in a string of mixed special symbols and characters. The names to be counted to are adjoined with the ampersand symbol & and as the second parameter. See the following examples for more details.

Examples

equalCount("Peter!@#$Paul&*#Peter!--@|#$Paul#$Peter@|Paul$%^^Peter++Paul%$%^Peter++Paul#$#$#Peter@|Paul", "Peter&Paul")
➞ {"Peter":6, "Paul": 6, "equality": true}

equalCount("Elliot!@#$Sam!--@|#$Elliot@|Sam++Elliot$%^Elliot@|Sam!@#Elliot!@#$Sam!--@|#$Elliot", "Sam&Elliot")
➞ {"Sam": 4, "Elliot": 6, "equality": false, "difference": 2}
// "difference" key is added to the object if count is not equal.

equalCount("Tim!@#$Kit&&*#Tim!--@|#$Kit#$%Tim@|Kit$%^^Tim++Kit%$%^Tim++Kit#$#$#Tim@|Kit", "Ken&Tom")
➞ {"Ken": 0, "Tom": 0, "equality": true}

Notes

Make sure to return the result as an object with the corresponding keys seen in the above examples and the difference key when needed.

Solution:

function equalCount(str, names) {
  let strArr = str.replace(/[^a-zA-Z]/g, "").match(/[A-Z][a-z]+/g);
  let obj = {};
  for (let name of strArr) {
    if (obj[name]) {
      obj[name] += 1;
    } else {
      obj[name] = 1;
    }
  }
  //   console.log(obj);
  let [name1, name2] = names.split("&");
  //   console.log(name1, name2);
  let key = obj[name1] === obj[name2] ? "equality" : "difference";
  let sol = {
    [name1]: obj[name1] ? obj[name1] : 0,
    [name2]: obj[name2] ? obj[name2] : 0,
  };

  return key == "equality"
    ? { ...sol, equality: true }
    : {
        ...sol,
        equality: false,
        difference: Math.abs(obj[name1] - obj[name2]),
      };
}

Combine Two Objects Into One, Summing Like Values [Edabit]

August 25, 2021August 20, 2021 Ketan RamtekeLeave a comment

Edabit

Take two objects with similar key values and combine them into a new object summing any values that belong to the same category.

There’s a married couple, Hank and his spouse Sheila. Hank works at a power plant making $70,000 a year, and Sheila is a teacher making $40,000 a year. They both earn rental income from separate rental properties, Hank will get $12,000 and Sheila $10,000. The new object will show their separate income but combine the rental income because it fits the same category.

const user1 = {
  powerPlant: 70000,
  rental: 12000
}

const user2 = {
  teaching: 40000,
  rental: 10000
}

becomes ➞ {
  powerPlant: 70000,
  teaching: 40000,
  rental: 22000   // The rental income is added together.
}

Arguments

user1Income (Object) ⁠— an income object for user1
user2Income (Object) ⁠— an income object for user2
returns: A new object with like values combined

Challenges

They won’t have the same number of key-value pairs.
The return object must return the values ordered from lowest to highest so your answers can match the test answers.

Solution:

function combine(user1Income, user2Income) {
  let sol = { ...user1Income };
  Object.keys(user2Income).forEach((key) => {
    if (sol[key]) {
      sol[key] += user2Income[key];
    } else {
      sol[key] = user2Income[key];
    }
  });
  let temp = Object.keys(sol).map((key) => [key, sol[key]]);
  temp.sort((a, b) => a[1] - b[1]);
  sol = {};
  temp.forEach((arr) => {
    sol = { ...sol, [arr[0]]: arr[1] };
  });
  return sol;
}

Find First Character That Repeats [Edabit]

August 24, 2021August 19, 2021 Ketan RamtekeLeave a comment

Edabit

Create a function that takes a string and returns the first character that repeats. If there is no repeat of a character, return "-1".

Examples

firstRepeat("legolas") ➞ "l"

firstRepeat("Gandalf") ➞ "a"

firstRepeat("Balrog") ➞ "-1"

firstRepeat("Isildur") ➞ "-1"
// Case sensitive "I" not equal to "i"

Notes

Tests are case sensitive.

Solution:

function firstRepeat(chars) {
	let obj={};
	for(let ch of [...chars]){
		if(obj[ch]){
			return ch;
		}else{
			obj[ch]=1;
		}
	}
	return "-1";
}

Ketan Ramteke

Code(Not optimized TBH), Random Rants and Things which I usually don't write on Quora.

Month: August 2021

Sum of Resistance in Parallel Circuits [Edabit]

Worked Example

Examples

Notes

Calculate an Earned Run Average [Edabit]

Examples

Notes

Majority Vote [Edabit]

Examples

Notes

Multiplying Numbers in a String [Edabit]

Examples

Notes

Spoonerisms [Edabit]

Examples

Notes

pigLatin 3.0 [Edabit]

Examples

Notes

Making a Box 2.0! [Edabit]

Examples

Notes

Name Count Equality [Edabit]

Examples

Notes

Combine Two Objects Into One, Summing Like Values [Edabit]

Arguments

Challenges

Find First Character That Repeats [Edabit]

Examples

Notes