When provided with an index, rearrange the elements of the array by swapping

Question

When provided with an index, rearrange the elements of the array by swapping

My goal is to create a behavior similar to that of a radio group, where only one option can be selected at a time. For instance, if we start with the array [0, 1, 1, 1, 1, 1], the elements should be rearranged as follows:

index	array
`0`	`[0, 1, 1, 1, 1, 1]`
`1`	`[1, 0, 1, 1, 1, 1]`
`2`	`[1, 1, 0, 1, 1, 1]`
`3`	`[1, 1, 1, 0, 1, 1]`
`4`	`[1, 1, 1, 1, 0, 1]`
`5`	`[1, 1, 1, 1, 1, 0]`

I have developed a solution for this, but I believe it involves "extra work" in certain scenarios due to unnecessary loops.

function rearrange(array: number[], idx: number) {
  let arr = array.slice();
  let l = arr.length;
  if (arr.indexOf(0) === idx) return arr;

  while (arr.indexOf(0) !== idx) {
    let swap;
    for (let i = 0; i < l; i++) {
      if (arr[i] === 0 || arr[i + 1] === 0) {
        swap = arr[i];
        if (i + 1 < l) {
          arr[i] = arr[i + 1];
          arr[i + 1] = swap;
        }
        if (i + 1 > l) {
          arr[i] = arr[i - 1];
          arr[i - 1] = swap;
        }
      }
    }
  }

  return arr;
}

If you have any ideas on how to simplify or improve this process, please share them.

arrays typescript performance ecmascript-6

Answer 1

Answer №1

Simply use the findIndex method to locate the previous 0, replace it with 1, and then assign 0 to the index idx.

function rearrange(array: number[], idx: number) {
  const arr = [...array];
  arr[arr.indexOf(0)] = 1;
  arr[idx] = 0;
  return arr;
}

Here's another way to go about it...

const rearrange = (array: number[], idx: number) => (
  array.map((_, i) => i === idx ? 0 : 1)
);

Answer 2

Simply use the findIndex method to locate the previous 0, replace it with 1, and then assign 0 to the index idx.

function rearrange(array: number[], idx: number) {
  const arr = [...array];
  arr[arr.indexOf(0)] = 1;
  arr[idx] = 0;
  return arr;
}

Here's another way to go about it...

const rearrange = (array: number[], idx: number) => (
  array.map((_, i) => i === idx ? 0 : 1)
);

Answer 3

Answer №2

If you want to avoid swapping pairs, try rotating a subarray in a single step by using slice and splice methods.

Check out this handy function that modifies the original array instead of creating a new one. Here are some examples to illustrate its usage. It's designed to keep values connected as if they were moved like with a "moving" zero:

function reposition(array, index) {
  let zeroIndex = array.indexOf(0);
  if (zeroIndex < index) {
      arr.splice(zeroIndex, index - zeroIndex + 1, ...arr.slice(zeroIndex + 1, index + 1), 0);
  } else {
      arr.splice(index, zeroIndex - index + 1, 0, ...arr.slice(index, zeroIndex));
  }
}

arr = [0,1,2,3,4,5];

reposition(arr, 3);
console.log(...arr);
reposition(arr, 1);
console.log(...arr);
reposition(arr, 5);
console.log(...arr);
reposition(arr, 1);
console.log(...arr);

Answer 4

If you want to avoid swapping pairs, try rotating a subarray in a single step by using slice and splice methods.

Check out this handy function that modifies the original array instead of creating a new one. Here are some examples to illustrate its usage. It's designed to keep values connected as if they were moved like with a "moving" zero:

function reposition(array, index) {
  let zeroIndex = array.indexOf(0);
  if (zeroIndex < index) {
      arr.splice(zeroIndex, index - zeroIndex + 1, ...arr.slice(zeroIndex + 1, index + 1), 0);
  } else {
      arr.splice(index, zeroIndex - index + 1, 0, ...arr.slice(index, zeroIndex));
  }
}

arr = [0,1,2,3,4,5];

reposition(arr, 3);
console.log(...arr);
reposition(arr, 1);
console.log(...arr);
reposition(arr, 5);
console.log(...arr);
reposition(arr, 1);
console.log(...arr);

When provided with an index, rearrange the elements of the array by swapping

Answer №1

Answer №2

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