Explaination of above alogrithm
This code is an algorithm to find the median of an array 'A' in an efficient way, where the median is the middle value of the sorted array. The code divides the problem into smaller subproblems until it can easily find the median.
Let's break down the algorithm step by step:
1. If the array 'A' has 10 or fewer elements:
- The code sorts the array 'A' in ascending order.
- Then it returns the element at the (k-1)th index, which is the kth smallest element in the sorted array.
- This works because in a sorted array, the element at index k-1 will be the kth smallest element.
2. If the array 'A' has more than 10 elements:
- The code partitions the array 'A' into subsets 'S[i]' of five elements each. If there are any remaining elements, they form a subset as well.
- For each subset 'S[i]', the code finds the median by selecting the middle element (element at index 3 when sorted).
3. Next, the code finds the median 'M' of all the medians found in the previous step.
- It creates an array 'x[i]' containing all these medians.
- Then it recursively applies the same algorithm to find the median 'M' of this new array 'x[i]'.
4. The array 'A' is then partitioned into three parts based on the median 'M':
- Elements less than 'M' are put into 'A1'.
- Elements equal to 'M' are put into 'A2'.
- Elements greater than 'M' are put into 'A3'.
5. Finally, the algorithm checks where the desired median 'k' lies relative to the sizes of 'A1' and 'A2':
- If 'k' is less than or equal to the length of 'A1', then the algorithm recursively finds the median in 'A1'.
- If 'k' is greater than the combined length of 'A1' and 'A2', then the algorithm recursively finds the median in 'A3'.
- Otherwise, the median 'M' is returned since it means 'k' lies within 'A2'.
The algorithm keeps breaking down the problem into smaller subproblems until it finds the desired median. By doing so, it ensures that it doesn't have to sort the entire array, making it more efficient than a standard sorting-based approach for finding the median.