Eliminating every third and fourth element out of a set of five elements

Question

Eliminating every third and fourth element out of a set of five elements

I need to manipulate an array where elements are in increments of 5, and I specifically need to drop the 3rd and 4th elements.

Currently, I have a solution using two separate for-loops, but I believe there might be a more efficient or concise approach. Any suggestions?

const arr = ["good","good""bad","bad","good","good","good","bad","bad","good",];

// removing every 3rd element out of 5
  for (let i = 2; i <= arr.length; i += 4) {
    arr.splice(i, 1);
  }

  // dropping every 3rd element out of 4
  for (let i = 2; i <= arr.length; i += 3) {
    arr.splice(i, 1);
  }

console.log(arr)
// expected result: ["good","good","good","good","good","good"]

javascript typescript

Answer 1

Answer №1

Instead of removing elements one by one in a loop, you can achieve the same result by removing 2 elements at once in a single loop.

const arr = Array(50).fill(["good", "good", "bad", "bad", "good"]).flat()

// remove every 3rd of 5 elements
for (let i = 2; i <= arr.length; i += 3) {
  arr.splice(i, 2);
}

console.log(arr);

Answer 2

Instead of removing elements one by one in a loop, you can achieve the same result by removing 2 elements at once in a single loop.

const arr = Array(50).fill(["good", "good", "bad", "bad", "good"]).flat()

// remove every 3rd of 5 elements
for (let i = 2; i <= arr.length; i += 3) {
  arr.splice(i, 2);
}

console.log(arr);

Answer 3

Answer №2

If I were to opt for creating a fresh array, here's how I would proceed:

const data = ["good", "good", "bad", "bad", "good", "good", "good", "bad", "bad", "good", ];

var result = [];
const restrictedIndices = [2, 3];

for (let i = 0; i < data.length; i++) {
  const modifiedIndex = i % 5;
  if (!restrictedIndices.includes(modifiedIndex)) {
    result.push(data[i]);
  }
}

console.log(result);

Alternatively, for a more sophisticated approach:

const data = ["good","good","bad","bad","good","good","good","bad","bad","good",];

var filterFunction = (filteredData, currentElement, index) => {
  const modifiedIndex = index % 5;
  const restrictedIndices = [2, 3];
  
  if (!restrictedIndices.includes(modifiedIndex))
  {
    filteredData.push(currentElement);
  }

  return filteredData;
}

var outcome = data.reduce(filterFunction, []);
console.log(outcome);

This enhances the solution's versatility by allowing easy modification of filtering parameters in the future.

Answer 4

If I were to opt for creating a fresh array, here's how I would proceed:

const data = ["good", "good", "bad", "bad", "good", "good", "good", "bad", "bad", "good", ];

var result = [];
const restrictedIndices = [2, 3];

for (let i = 0; i < data.length; i++) {
  const modifiedIndex = i % 5;
  if (!restrictedIndices.includes(modifiedIndex)) {
    result.push(data[i]);
  }
}

console.log(result);

Alternatively, for a more sophisticated approach:

const data = ["good","good","bad","bad","good","good","good","bad","bad","good",];

var filterFunction = (filteredData, currentElement, index) => {
  const modifiedIndex = index % 5;
  const restrictedIndices = [2, 3];
  
  if (!restrictedIndices.includes(modifiedIndex))
  {
    filteredData.push(currentElement);
  }

  return filteredData;
}

var outcome = data.reduce(filterFunction, []);
console.log(outcome);

This enhances the solution's versatility by allowing easy modification of filtering parameters in the future.

Answer 5

Answer №3

An effective way to eliminate the 3rd and 4th indexes from an array is by utilizing the Array filter prototype.

const arr = Array(50).fill(["good", "good", "bad", "bad", "good"]).flat()
const disAllowedIndex=[2,3]
const filteredArray=arr.filter((item,index)=>!disAllowedIndex.some(i=>i==index%5))

Answer 6

An effective way to eliminate the 3rd and 4th indexes from an array is by utilizing the Array filter prototype.

const arr = Array(50).fill(["good", "good", "bad", "bad", "good"]).flat()
const disAllowedIndex=[2,3]
const filteredArray=arr.filter((item,index)=>!disAllowedIndex.some(i=>i==index%5))

Eliminating every third and fourth element out of a set of five elements

Answer №1

Answer №2

Answer №3

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