215. Kth Largest Element in an Array

Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.

For example,
Given [3,2,1,5,6,4] and k = 2, return 5.

Note: 
You may assume k is always valid, 1 ≤ k ≤ array's length.

Solution:

Method 1: find kth smallest

kth largest number in array nums, is the (nums.length - k + 1)th smallest umber in the array.

And its absolute position in an sorted order is nums.length - k.

Therefore, we recursively call 3-way partition until nums[nums.length - k] is finalized.



Code:

public class Solution {
    public int findKthLargest(int[] nums, int k) {
        if (nums == null || nums.length == 0 || k <= 0) {
            return -1;
        }
        
        k = nums.length - k;
        
        int lo = 0;
        int hi = nums.length - 1;
        partition(nums, lo, hi, k);
        
        return nums[k];
    }
    
    public void partition(int[] nums, int lo, int hi, int pos) {
        if (hi <= lo) {
            return;
        }
        
        int v = nums[lo];
        int lt = lo;
        int gt = hi;
        int i = lo;
        
        while (i <= gt) {
            if (nums[i] < v) {
                exch(nums, lt++, i++);
            }
            else if (nums[i] > v) {
                exch(nums, gt--, i);
            }
            else {
                i++;
            }
        }
        
        if (lt < pos) {
            partition(nums, lt + 1, hi, pos);
        }
        if (lt > pos) {
            partition(nums, lo, lt - 1, pos);
        }
    }
    
    public void exch(int[] nums, int i, int j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    }
}

Method 2: find kth largest

Modify the 3-way partition to put larger number before pivot and smaller number after pivot.

The kth largest number is nums[k - 1] after partition.



Code:

public class Solution {
    public int findKthLargest(int[] nums, int k) {
        if (nums == null || nums.length == 0 || k <= 0) {
            return -1;
        }
        
        int lo = 0;
        int hi = nums.length - 1;
        partition(nums, lo, hi, k - 1);
        
        return nums[k - 1];
    }
    
    public void partition(int[] nums, int lo, int hi, int pos) {
        if (hi <= lo) {
            return;
        }
        
        int v = nums[lo];
        int gt = lo;
        int lt = hi;
        int i = lo;
        
        while (i <= lt) {
            if (nums[i] > v) {
                exch(nums, gt++, i++);
            }
            else if (nums[i] < v) {
                exch(nums, lt--, i);
            }
            else {
                i++;
            }
        }
        
        if (gt < pos) {
            partition(nums, gt + 1, hi, pos);
        }
        if (gt > pos) {
            partition(nums, lo, gt - 1, pos);
        }
    }
    
    public void exch(int[] nums, int i, int j) {
        int swap = nums[i];
        nums[i] = nums[j];
        nums[j] = swap;
    }
}