Tuesday, May 26, 2015

[LeetCode] Kth Largest Element in an Array

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.
Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.
Solution: A naive way is to first sort the array in the decreasing (non-increasing) order, and then return the kth element. This takes O(n log n) time at best, where n = array's length.
Quickselect is a selection algorithm to efficiently find the kth largest/smallest element in an unsorted array. Below is an implementation of the quickselect algorithm, both recursive and iterative.
Further reading: Selection algorithm.
Complexity: O(n) on average, where n = array's length.
Code in Java:
Recursive version.
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public class Solution {
    private void swap(int[] nums, int i, int j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
        return;
    }
    
    private int partition(
        int[] nums, 
        int low, 
        int high, 
        int pivotIdx
    ) {
        int pivotVal = nums[pivotIdx];
        swap(nums, pivotIdx, high);
        int storeIdx = low;
        for (int i = low; i < high; i++) {
            if (nums[i] > pivotVal) {
                swap(nums, storeIdx, i);
                storeIdx++;
            }
        }
        swap(nums, storeIdx, high);
        return storeIdx;
    }
    
    private int findKthLargest(
        int[] nums, 
        int low, 
        int high, 
        int k
    ) {
        if (low == high) {
            return nums[low];
        }
        int offset = new java.util.Random().nextInt(
                         high - low + 1
                     );
        int pivotIdx = low + offset;
        pivotIdx = partition(nums, low, high, pivotIdx);
        if (pivotIdx == k) {
            return nums[k];
        }
        if (pivotIdx > k) {
            return findKthLargest(nums, low, pivotIdx - 1, k);
        } else {
            return findKthLargest(nums, pivotIdx + 1, high, k);
        }
    }
    
    public int findKthLargest(int[] nums, int k) {
        return findKthLargest(nums, 0, nums.length - 1, k - 1);
    }
}

Iterative version.
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public class Solution {
    private void swap(int[] nums, int i, int j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
        return;
    }
    
    private int partition(
        int[] nums, 
        int low, 
        int high, 
        int pivotIdx
    ) {
        int pivotVal = nums[pivotIdx];
        swap(nums, pivotIdx, high);
        int storeIdx = low;
        for (int i = low; i < high; i++) {
            if (nums[i] > pivotVal) {
                swap(nums, storeIdx, i);
                storeIdx++;
            }
        }
        swap(nums, storeIdx, high);
        return storeIdx;
    }
    
    private int findKthLargest(
        int[] nums, 
        int low, 
        int high, 
        int k
    ) {
        if (low == high) {
            return nums[low];
        }
        while (true) {
            int offset = new java.util.Random().nextInt(
                             high - low + 1
                         );
            int pivotIdx = low + offset;
            pivotIdx = partition(nums, low, high, pivotIdx);
            if (pivotIdx == k) {
                return nums[k];
            }
            if (pivotIdx > k) {
                high = pivotIdx - 1;
            } else {
                low = pivotIdx + 1;
            }            
        }
    }
    
    public int findKthLargest(int[] nums, int k) {
        return findKthLargest(nums, 0, nums.length - 1, k - 1);
    }
}
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