Sort an array containing only 0s, 1s, and 2s.

This problem can be efficiently solved using the Dutch National Flag algorithm. This algorithm sorts the array in a single pass with O(n) time complexity and O(1) space complexity. The core idea is to maintain three pointers: low, mid, and high. low points to the beginning of the array, mid is the current element being processed, and high points to the end of the array.

Initialize low and mid to 0, and high to n-1 (where n is the array size).

Iterate while mid <= high:

1. If the element at mid is 0: Swap a[low] and a[mid], then increment both low and mid.
2. If the element at mid is 1: Simply increment mid.
3. If the element at mid is 2: Swap a[mid] and a[high], then decrement high. Do not increment mid in this case, as the swapped element from high needs to be processed.

This process partitions the array into three sections: elements less than low are 0s, elements between low and mid are 1s, and elements greater than high are 2s.