Given an unsorted array. The array has this property that every element in array is at most k distance from its position in sorted array where k is a positive integer smaller than size of array. Which sorting algorithm can be easily modified for sorting this array and what is the obtainable time complexity?a)Insertion Sort with time complexity O(kn)b)Heap Sort with time complexity O(nLogk)c)Quick Sort with time complexity O(kLogk)d)Merge Sort with time complexity O(kLogk)Correct answer is option 'B'. Can you explain this answer?

Question

Answer

1) to sort the array firstly create a min-heap with first k+1 elements and a separate array as resultant array.

2) because elements are at most k distance apart from original position so, it is guranteed that the smallest element will be in this K+1 elements.

3) remove the smallest element from the min-heap(extract min) and put it in the result array.

4) Now,insert another element from the unsorted array into the mean-heap, now,the second smallest element will be in this, perform extract min and continue this process until no more elements are in the unsorted array.finally, use simple heap sort for the remaining elements Time Complexity ------------------------ 1) O(k) to build the initial min-heap 2) O((n-k)logk) for remaining elements... 3) 0(1) for extract min so overall O(k) + O((n-k)logk) + 0(1) = O(nlogk)

Answer

1) to sort the array firstly create a min-heap with first k+1 elements and a separate array as resultant array.

2) because elements are at most k distance apart from original position so, it is guranteed that the smallest element will be in this K+1 elements.

3) remove the smallest element from the min-heap(extract min) and put it in the result array.

4) Now,insert another element from the unsorted array into the mean-heap, now,the second smallest element will be in this, perform extract min and continue this process until no more elements are in the unsorted array.finally, use simple heap sort for the remaining elements Time Complexity ------------------------ 1) O(k) to build the initial min-heap 2) O((n-k)logk) for remaining elements... 3) 0(1) for extract min so overall O(k) + O((n-k)logk) + 0(1) = O(nlogk)

Answer

Understanding the Problem
The given unsorted array has the unique property that each element is at most k positions away from where it would be in a fully sorted array. This characteristic is pivotal in choosing an efficient sorting algorithm.
Why Heap Sort?
Heap Sort can be effectively modified to handle this scenario. Here's why:
- Min-Heap Structure: We can use a min-heap (or priority queue) of size k+1. This allows us to efficiently access the smallest element among the k nearest elements to their sorted position.
- Operation: For each element in the array, we can:
- Add the element to the min-heap.
- Remove the smallest element from the heap and place it in the sorted position.
- Efficiency: Since the heap maintains a size of k+1, insertion and extraction operations will be O(log(k)).
Time Complexity Explanation
- Total Operations: We perform this operation for all n elements in the array.
- Final Complexity: The overall time complexity becomes O(n log(k)), where:
- n is the number of elements in the array.
- log(k) is the time taken for operations on the heap.
Comparison with Other Algorithms
- Insertion Sort: While it can work, its time complexity in the worst case is O(n^2), which is inefficient compared to Heap Sort.
- Quick Sort and Merge Sort: Both algorithms generally have O(n log(n)) complexity, but they do not leverage the k-distance property effectively.
Conclusion
Thus, Heap Sort is the most suitable choice for sorting an array with the given constraints, providing an efficient O(n log(k)) time complexity. This makes it ideal for scenarios where k is significantly smaller than n.

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