In data structures and algorithms, write and explain an algorithm to separate all ones and zeroes in a binary array so that all the zero elements come before all the one elements.

Introduction / Context:
This interview style data structures and algorithms question asks you to design an efficient way to separate all ones and zeroes in a binary array so that all zero elements appear before all one elements. It tests your understanding of in place partitioning techniques, time complexity analysis, and how to translate an informal requirement into a clear step by step algorithm that can be implemented in a programming language such as C, C++, Java, or Python.

Given Data / Assumptions:

• The input is an array that contains only binary values: 0 and 1.
• The array can contain any number of zeroes and ones in any initial order.
• The goal is to rearrange the same array so that all 0s come before all 1s.
• The relative order among zeroes or among ones does not matter.
• We prefer an in place algorithm with O(1) extra space and O(n) time.

Concept / Approach:
A very natural and efficient approach is to use a two pointer partition algorithm, similar to the partition step in quick sort when the pivot is 0 or 1. You maintain two indices: one that scans from the left and one that scans from the right. The left pointer moves forward until it finds a misplaced 1, and the right pointer moves backward until it finds a misplaced 0. When both pointers have stopped on misplaced elements, you swap those elements. This continues until the pointers cross, at which point all zeroes are on the left side and all ones are on the right side. This design ensures a single linear scan and constant extra memory usage.

Step-by-Step Solution:
Step 1: Let left = 0 (start index) and right = n - 1 (last index), where n is the length of the array. Step 2: While left is less than right, move left forward while array[left] is equal to 0, because 0 is already on the correct side. Step 3: Similarly, move right backward while array[right] is equal to 1, because 1 is already on the correct side. Step 4: When you stop, if left is still less than right, you have array[left] equal to 1 and array[right] equal to 0, so swap these two elements. Step 5: Increment left and decrement right, then repeat the scan process until left is greater than or equal to right, which means the array is partitioned into zeroes followed by ones.

Verification / Alternative check:
You can verify correctness by observing invariants. At any time, all elements strictly to the left of left are guaranteed to be 0, and all elements strictly to the right of right are guaranteed to be 1. The loop terminates when left passes right, at which point every position has been checked or corrected, so no misplaced 1 exists on the left side and no misplaced 0 exists on the right side. An alternative method is to count how many zeros exist with a first pass and then fill that many zeros followed by ones in a second pass, but that requires two passes through the array, while the two pointer approach completes in a single pass.

Why Other Options Are Wrong:
Option B uses a nested loop, which degrades to O(n^2) time and is inefficient compared to a linear solution. Option C suggests a comparison based sorting algorithm, which is unnecessary overhead because the array contains only 0s and 1s and can be partitioned more simply. Option D using a balanced tree complicates the solution and adds extra memory overhead. Option E copies data to extra arrays and uses more memory than needed when a simple in place partition is perfectly adequate.

Common Pitfalls:
A frequent mistake is ignoring in place requirements and unnecessarily using extra arrays. Another pitfall is incorrectly updating pointers so that the algorithm gets stuck in an infinite loop or performs out of bounds access. Some candidates also attempt to use general sorting functions without recognizing that a specialized partition algorithm can solve this specific binary separation problem more efficiently. Taking a moment to think about invariants and pointer movements helps avoid these errors.

Final Answer:
The most efficient approach is to use a two pointer partition algorithm that scans from both ends, swapping misplaced 1s on the left with 0s on the right in a single O(n) in place pass so that all zeroes precede all ones in the final array.