We use a slow pointer to track where the next non-zero element should be placed and a fast pointer to find the next non-zero element.

1. l = 0.
2. For r from 0 to len(nums)-1:
   • If nums[r] != 0:
     • Swap nums[l] and nums[r].
     • l += 1.
3. The remaining elements from l to end are already zeroes (or will be swapped into place).

• Time Complexity: O(N).
• Space Complexity: O(1).

def move_zeroes(nums):
    l = 0
    for r in range(len(nums)):
        if nums[r] != 0:
            nums[l], nums[r] = nums[r], nums[l]
            l += 1