move all instances of that target value to the end end of the array

This problem was asked by Intel.

You are given an array and a target value, move all instances of that target value to the end end of the array

Example

function moveElement(arr, val){

};

// testcase
const arr = [2, 1, 2, 2, 2, 3, 4, 2];
console.log(moveElement(arr, 2)); // [4, 1, 3, 2, 2, 2, 2, 2]

Explanation : All the occurences of 2 are moved at the end of the array.

Solution

function moveElement(arr, val) {
  let i = 0,
    j = arr.length - 1;

  while (i < j) {
    while (i < j && arr[j] === val) {
      j--;
    }
    if (arr[i] === val) {
      [arr[i], arr[j]] = [arr[j], arr[i]];
    }
    i++;
  }
  return arr
}

Explanation

• Initialize i and j pointers at both ends of the array.
• Loop while i is less than j:
• Shift j leftwards until we find a non-target value.
• If arr[i] is the target, swap it with arr[j].
• Increment i towards right.
• This continues until i crosses j, which means all target values are at the end.
• So this approach efficiently moves the target values to the end in a single pass. The swap operation keeps the original order intact.