Tips for iterating through/range over an array depending on the initial point?

Question

Tips for iterating through/range over an array depending on the initial point?

var ZODIAC = ["RAT", "OX", "TIGER", "RABBIT", "DRAGON", "SNAKE", "HORSE", "SHEEP", "MONKEY", "ROOSTER", "DOG", "PIG"];
var STARTING_ZODIAC = "MONKEY";

Is there a way to output the elements in the array that start with Monkey and end with Sheep?

javascript arrays

Answer 1

Answer №1

To ensure that your index variable cycles back to 0 when it reaches the end of the ZODIAC array, you can utilize the modulo operator:

const ZODIAC = ["RAT", "OX", "TIGER", "RABBIT", "DRAGON", "SNAKE", "HORSE", "SHEEP", "MONKEY", "ROOSTER", "DOG", "PIG"];
const STARTING_ZODIAC = "MONKEY";
const startIndex = ZODIAC.indexOf(STARTING_ZODIAC);

console.log(STARTING_ZODIAC);
for (let i = startIndex + 1; i !== startIndex; i = (i + 1) % ZODIAC.length) {
  console.log(ZODIAC[i]);
}

Alternatively, you can reorder the sections of the array using the slice method to achieve the same result:

const ZODIAC = ["RAT", "OX", "TIGER", "RABBIT", "DRAGON", "SNAKE", "HORSE", "SHEEP", "MONKEY", "ROOSTER", "DOG", "PIG"];
const STARTING_ZODIAC = "MONKEY";
const startIndex = ZODIAC.indexOf(STARTING_ZODIAC);

[
   ...ZODIAC.slice(startIndex),
   ...ZODIAC.slice(0, startIndex)
].forEach(str => console.log(str));

Answer 2

To ensure that your index variable cycles back to 0 when it reaches the end of the ZODIAC array, you can utilize the modulo operator:

const ZODIAC = ["RAT", "OX", "TIGER", "RABBIT", "DRAGON", "SNAKE", "HORSE", "SHEEP", "MONKEY", "ROOSTER", "DOG", "PIG"];
const STARTING_ZODIAC = "MONKEY";
const startIndex = ZODIAC.indexOf(STARTING_ZODIAC);

console.log(STARTING_ZODIAC);
for (let i = startIndex + 1; i !== startIndex; i = (i + 1) % ZODIAC.length) {
  console.log(ZODIAC[i]);
}

Alternatively, you can reorder the sections of the array using the slice method to achieve the same result:

const ZODIAC = ["RAT", "OX", "TIGER", "RABBIT", "DRAGON", "SNAKE", "HORSE", "SHEEP", "MONKEY", "ROOSTER", "DOG", "PIG"];
const STARTING_ZODIAC = "MONKEY";
const startIndex = ZODIAC.indexOf(STARTING_ZODIAC);

[
   ...ZODIAC.slice(startIndex),
   ...ZODIAC.slice(0, startIndex)
].forEach(str => console.log(str));

Answer 3

Answer №2

There are a total of 6 answers provided, however none of them take into consideration the solution that seems obvious to me:

for (let i = 0; i < ZODIAC.length; i++) {
   console.log(ZODIAC[(startIndex + i) % ZODIAC.length]);
}

You can iterate through the loop 12 times and utilize the modulus operator to cycle through the numbers in a seamless manner: 4, 5, 6...10, 11, 0, 1, 2, 3.

Answer 4

There are a total of 6 answers provided, however none of them take into consideration the solution that seems obvious to me:

for (let i = 0; i < ZODIAC.length; i++) {
   console.log(ZODIAC[(startIndex + i) % ZODIAC.length]);
}

You can iterate through the loop 12 times and utilize the modulus operator to cycle through the numbers in a seamless manner: 4, 5, 6...10, 11, 0, 1, 2, 3.

Answer 5

Answer №3

To easily display the zodiac signs, simply print them out starting from the chosen sign and then in reverse order.

function displayZodiacs(chosenSign, zodiacSigns) {
  const startIndex = zodiacSigns.indexOf(chosenSign);

  // Print from start to end of array
  for (let i = startIndex; i < zodiacSigns.length; i++) {
    console.log(zodiacSigns[i]);
  }

  // Print from beginning of array to chosen sign
  for (let i = 0; i < startIndex; i++) {
    console.log(zodiacSigns[i]);
  }
}

displayZodiacs(CHOOSEN_SIGN, ZODIAC_SIGNS);

Answer 6

To easily display the zodiac signs, simply print them out starting from the chosen sign and then in reverse order.

function displayZodiacs(chosenSign, zodiacSigns) {
  const startIndex = zodiacSigns.indexOf(chosenSign);

  // Print from start to end of array
  for (let i = startIndex; i < zodiacSigns.length; i++) {
    console.log(zodiacSigns[i]);
  }

  // Print from beginning of array to chosen sign
  for (let i = 0; i < startIndex; i++) {
    console.log(zodiacSigns[i]);
  }
}

displayZodiacs(CHOOSEN_SIGN, ZODIAC_SIGNS);

Answer 7

Answer №4

Here's another interesting approach:

let zodiacDouble = ZODIAC.concat(ZODIAC);
let initial = ZODIAC.indexOf(START_ZODIAC);

for (let j = 0; j < ZODIAC.length; j++) {
  console.log(zodiacDouble[initial + j]);
}

This method involves duplicating the array and then displaying 12 elements starting from a specified index.

Answer 8

Here's another interesting approach:

let zodiacDouble = ZODIAC.concat(ZODIAC);
let initial = ZODIAC.indexOf(START_ZODIAC);

for (let j = 0; j < ZODIAC.length; j++) {
  console.log(zodiacDouble[initial + j]);
}

This method involves duplicating the array and then displaying 12 elements starting from a specified index.

Answer 9

Answer №5

One of the most direct solutions in my opinion involves a do-while loop that starts at a specific index and loops until it reaches back to that same index. Using a do-block is a straightforward way to handle this particular situation:

const ZODIAC = ["RAT", "OX", "TIGER", "RABBIT", "DRAGON", "SNAKE", "HORSE", "SHEEP", "MONKEY", "ROOSTER", "DOG", "PIG"];
const STARTING_ZODIAC = "MONKEY";

let i = start_ndx = ZODIAC.indexOf(STARTING_ZODIAC);

do {
  console.log(i, ZODIAC[i++])
  i == ZODIAC.length && (i=0)
} while (i!==start_ndx)

Note: In my perspective, other answers appear overly complex by using modulo or inefficient by copying values and creating new arrays. The approach shown above avoids both complexities and inefficiencies, making it easier to maintain.

Answer 10

One of the most direct solutions in my opinion involves a do-while loop that starts at a specific index and loops until it reaches back to that same index. Using a do-block is a straightforward way to handle this particular situation:

const ZODIAC = ["RAT", "OX", "TIGER", "RABBIT", "DRAGON", "SNAKE", "HORSE", "SHEEP", "MONKEY", "ROOSTER", "DOG", "PIG"];
const STARTING_ZODIAC = "MONKEY";

let i = start_ndx = ZODIAC.indexOf(STARTING_ZODIAC);

do {
  console.log(i, ZODIAC[i++])
  i == ZODIAC.length && (i=0)
} while (i!==start_ndx)

Note: In my perspective, other answers appear overly complex by using modulo or inefficient by copying values and creating new arrays. The approach shown above avoids both complexities and inefficiencies, making it easier to maintain.

Tips for iterating through/range over an array depending on the initial point?

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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