hy Not Iterate to high in One Step? Now, let's address your alternative code, where you propose iterating up to j <= high and performing the swap directly without the final swap step. Here’s the key issue with that approach:

Final Position of Pivot: The crucial step in quicksort is ensuring that the pivot ends up in its correct position after partitioning. When the loop ends, i will be pointing to the last element that is smaller than or equal to the pivot. However, the pivot itself still sits at the last position (array[high]).

If you swap array[i] with array[j] when j == high, you'll be swapping the pivot with the last element that was smaller than or equal to the pivot. This would cause the pivot to land in a position where it may not be correctly ordered in relation to other elements. This would break the partitioning logic. Final Swap Corrects the Position: By performing the final swap (swap(&array[i + 1], &array[high])), you ensure that the pivot is placed exactly in the position where all the smaller elements are on the left and all the larger elements are on the right. This guarantees that the pivot is in the correct sorted position, which is crucial for the recursive calls to continue working.

To Summarize: Why iterate up to j < high in the loop?

We stop the loop at j < high because the pivot is located at array[high], and we don’t want to compare the pivot to itself during the partitioning step. Why the final swap?

After the loop, the element at array[i] is the last element smaller than or equal to the pivot. The pivot itself is at array[high]. The final swap ensures that the pivot gets placed in its correct sorted position, i.e., just after the last element that is smaller than or equal to it. What would happen if you didn’t do the final swap?

The pivot wouldn’t end up in its correct sorted position, which would result in incorrect behavior during the recursive sorting steps, potentially causing the algorithm to not sort the array properly. Code with Your Suggested Approach (For Comparison): If you were to follow your suggested approach, i.e., iterate to j <= high, the final swap becomes unnecessary, but the pivot wouldn't end up in its correct position, and you would likely end up with an unsorted array.