Below is my code for trying to understand the median of medians algorithm (using blocks of size 5). I understand how to get medians of the input, but I'm not sure how to code the block to keep recursing the input until I just have the median. Then after getting that median, I'm not sure how to use it as a pivot to throw away the useless information to partition the input. getMediansArray
returns an array of size ceil(input.length/5) and getMedians
just returns the median from an array (only used on arrays of length <= 5).
public static int[] findKthElement(int[] input, int k) {
int numOfMedians = (int) Math.ceil(input.length/5.0);
int[] medians = new int[numOfMedians];
medians = getMediansArray(input, medians)
// (1) This only gets the first iteration of medians of the
// input. How do I recurse on this until I just have one median?
// (2) how should I partition about the pivot once I get it?
}
public static int[] getMediansArray(int[] input, int[] medians) {
int numOfMedians = (int) Math.ceil(input.length/5.0);
int[] five = new int[5];
for (int i = 0; i < numOfMedians; i++) {
if (i != numOfMedians - 1) {
for (int j = 0; j < 5; j++) {
five[j] = input[(i*5)+j];
}
medians[i] = getMedian(five);
} else {
int numOfRemainders = input.length % 5;
int[] remainder = new int[numOfRemainders];
for (int j = 0; j < numOfRemainders; j++) {
remainder[j] = input[(i*5)+j];
}
medians[i] = getMedian(five);
}
}
return medians;
}
public static int getMedian(int[] input) {
Arrays.sort(input);
if (input.length % 2 == 0) {
return (input[input.length/2] + input[input.length/2 - 1]) / 2;
}
return input[input.length/2];
}
Complexity Analysis Time Complexity: The worst-case time complexity for this algorithm is O(n²), but it can be improved if we choose the pivot element randomly. If we randomly select the pivot, the expected time complexity would be linear, O(n). Space Complexity: O(logn) average for recursion call stack (Think!)
K'th smallest element in an unsorted array using sorting: Sort the given array and return the element at index K-1 in the sorted array. Follow the given steps to solve the problem: Sort the input array in the increasing order. Return the element at the K-1 index (0 – Based indexing) in the sorted array.
The simplest solution is to sort the array and return the kth element. This solution has a time complexity of O(n*logn).
Median of medians is basically just the quick-select algorithm (http://en.wikipedia.org/wiki/Quickselect) improved. While quick-select has O(n) average time complexity, it can slow down to O(n^2) for tricky input.
What you do after finding a median of medians is nothing but an iteration of quick-select algorithm. Median of medians has a nice property that it will be always larger than 30% of elements and smaller than 30% of elements. This guarantees that quick-select using median of medians for a pivot will run in worst time complexity of O(n). Refer to: http://en.wikipedia.org/wiki/Median_of_medians
I suggest you start by implementing quick-select. Once you do that, you can use code you already have to select pivot in each step of quick-select.
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