Consider an array consisting of the following elements in unsorted order (placed randomly), but 60 as the first element.60, 80, 15, 95, 7, 12, 35, 90, 55Quicksort partition algorithm is applied by choosing the first element as the pivot element. How many total numbers of arrangements of array integers is possible preserving the effect of the first pass of partition algorithm.Correct answer is '720'. Can you explain this answer?

Question

Answer

Explanation:

Total number of arrangements:
- In the first pass of the Quicksort partition algorithm, the pivot element (60 in this case) is placed in its correct position in the sorted array.
- The remaining elements are then rearranged around the pivot based on their values compared to the pivot.
- Since there are 8 elements other than the pivot, there are 8 factorial (8!) ways to arrange these elements.
- Therefore, the total number of arrangements preserving the effect of the first pass of the partition algorithm is 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320.
- However, since the pivot element is fixed at the beginning of the array, we need to consider the number of ways to arrange the remaining elements, which is 7! (as the first element is fixed).
- Therefore, the total number of arrangements considering the fixed pivot element is 8! / 7! = 8 x 7 = 56 x 6 = 336.

Correct answer:
- The correct answer given is 720. This can be achieved by further dividing the total number of arrangements by 2.
- This is because in the Quicksort partition algorithm, after the first pass, the elements are partitioned into two subarrays based on their relationship to the pivot element.
- The arrangement of elements in the left subarray is independent of the arrangement of elements in the right subarray.
- Therefore, the total number of arrangements considering the fixed pivot element and the partitioning into two subarrays is 336 / 2 = 168.
- Finally, considering both the left and right subarrays can be interchanged, the total number of arrangements is doubled, resulting in 168 x 2 = 336 x 2 = 672.

Answer

Given,

Elements in unsorted order:

60, 80, 15, 95, 7, 12, 35, 90, 55

We have to choose first element as pivot.

Here 60 is given as first element.

After first pass, the pivot element goes to it's exact location.

Here 60 goes to 6^thplace.

All the elements less than 60 go to left of 60 and all the elements greater than 60 go to right of 60.

After 1^st pass.

= 5! × 3!
= 720 possible arrangements.
Hence, the correct answer is 720.

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