We use a two-pointer approach. One pointer insert_pos keeps track of where the next non-zero element should be placed. As we iterate through the array, every time we find a non-zero element, we swap it with the element at insert_pos.

1. Initialize insert_pos = 0.
2. Iterate through the array with curr_idx.
3. If nums[curr_idx] is not zero:
   • Swap nums[insert_pos] and nums[curr_idx].
   • Increment insert_pos.
4. This effectively pushes all non-zero elements to the front and zeroes to the back.

• Time Complexity: O(N), single pass through the array.
• Space Complexity: O(1), in-place modification.

def move_zeroes(nums):
    insert_pos = 0
    for i in range(len(nums)):
        if nums[i] != 0:
            # Swap current non-zero with the first available zero position
            nums[insert_pos], nums[i] = nums[i], nums[insert_pos]
            insert_pos += 1
    return nums