# Kth Largest Element in an Array

## Question

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

**Notice**

You can swap elements in the array

**Example**

In array \[9,3,2,4,8], the 3rd largest element is 4.

In array \[1,2,3,4,5], the 1st largest element is 5, 2nd largest element is 4, 3rd largest element is 3 and etc.

**Challenge**

O(n) time, O(1) extra memory.

## Analysis

### Sort Array

* O(nlogn) time

### Max Heap

* O(nlogk) running time + O(k) memory

在数字集合中寻找第k大，可以考虑用Max Heap，将数组遍历一遍，加入到一个容量为k的PriorityQueue，最后poll() k-1次，那么最后剩下在堆顶的就是kth largest的数字了。

另外此题用quick sort中的 [quick select](https://en.wikipedia.org/wiki/Quickselect) 的思路来解，更优化，参考：[http://www.jiuzhang.com/solutions/kth-largest-element/](http://www.jiuzhang.com/solutions/kth-largest-element/)

### Quick Select

* Time Complexity: average = O(n); worst case O(n^2), O(1) space

注意事项：

* partition的主要思想：将比pivot小的元素放到pivot左边，比pivot大的放到pivot右边
* pivot的选取决定了partition所得结果的效率，可以选择left pointer，更好的选择是在left和right范围内随机生成一个；

Time Complexity O(n)来自于O(n) + O(n/2) + O(n/4) + ... \~ O(2n)，此时每次partition的pivot大约将区间对半分。

Source: [https://discuss.leetcode.com/topic/15256/4-c-solutions-using-partition-max-heap-priority\_queue-and-multiset-respectively](https://discuss.leetcode.com/topic/15256/4-c-solutions-using-partition-max-heap-priority_queue-and-multiset-respectively)

> So, in the **average** sense, the problem is reduced to approximately half of its original size, giving the recursion `T(n) = T(n/2) + O(n)` in which `O(n)` is the time for partition. This recursion, once solved, gives `T(n) = O(n)` and thus we have a linear time solution. Note that since we only need to consider **one half** of the array, the time complexity is `O(n)`. If we need to consider both the two halves of the array, like quicksort, then the recursion will be `T(n) = 2T(n/2) + O(n)` and the complexity will be `O(nlogn)`.
>
> Of course, `O(n)` is the average time complexity. In the worst case, the recursion may become `T(n) = T(n - 1) + O(n)` and the complexity will be `O(n^2)`.

## Solution

### Sort Array

O(nlogn)

```java
public class Solution {
    public int findKthLargest(int[] nums, int k) {
            final int N = nums.length;
            Arrays.sort(nums);
            return nums[N - k];
    }
}
```

### Max Heap

O(N lg K) running time + O(K) memory

```java
class Solution {
    /*
     * @param k : description of k
     * @param nums : array of nums
     * @return: description of return
     */
    public int kthLargestElement(int k, int[] nums) {
        if (nums == null || nums.length == 0 || k == 0) {
            return -1;
        }
        PriorityQueue<Integer> heap = new PriorityQueue<Integer>(k, new Comparator<Integer>() {
            @Override
            public int compare(Integer o1, Integer o2) {
                return o2 - o1;
            }
        });
        for (int i = 0; i < nums.length; i++) {
            heap.offer(nums[i]);
        }
        for (int j = 0; j < k - 1; j++) {
            heap.poll();
        }
        return heap.peek();
    }
};
```

### Min Heap

O(N lg K) running time + O(K) memory

```java
public int findKthLargest(int[] nums, int k) {

    final PriorityQueue<Integer> pq = new PriorityQueue<>();
    for(int val : nums) {
        pq.offer(val);

        if(pq.size() > k) {
            pq.poll();
        }
    }
    return pq.peek();
}
```

#### Quick Select: O(N) best case / O(N^2) worst case running time + O(1) memory

```java
    public int findKthLargest(int[] nums, int k) {

        k = nums.length - k;
        int lo = 0;
        int hi = nums.length - 1;
        while (lo < hi) {
            final int j = partition(nums, lo, hi);
            if(j < k) {
                lo = j + 1;
            } else if (j > k) {
                hi = j - 1;
            } else {
                break;
            }
        }
        return nums[k];
    }

    private int partition(int[] a, int lo, int hi) {

        int i = lo;
        int j = hi + 1;
        while(true) {
            while(i < hi && less(a[++i], a[lo]));
            while(j > lo && less(a[lo], a[--j]));
            if(i >= j) {
                break;
            }
            exch(a, i, j);
        }
        exch(a, lo, j);
        return j;
    }

    private void exch(int[] a, int i, int j) {
        final int tmp = a[i];
        a[i] = a[j];
        a[j] = tmp;
    }

    private boolean less(int v, int w) {
        return v < w;
    }
```

#### Use Shuffle : O(N) guaranteed running time + O(1) space

```java
    public int findKthLargest(int[] nums, int k) {

        shuffle(nums);
        k = nums.length - k;
        int lo = 0;
        int hi = nums.length - 1;
        while (lo < hi) {
            final int j = partition(nums, lo, hi);
            if(j < k) {
                lo = j + 1;
            } else if (j > k) {
                hi = j - 1;
            } else {
                break;
            }
        }
        return nums[k];
    }

    private void shuffle(int a[]) {

        final Random random = new Random();
        for(int ind = 1; ind < a.length; ind++) {
            final int r = random.nextInt(ind + 1);
            exch(a, ind, r);
        }
    }
```

### Quick Select (Partition with two pointers)

O(N) best case / O(N^2) worst case running time + O(1) memory

```java
class Solution {
    /*
     * @param k : description of k
     * @param nums : array of nums
     * @return: description of return
     */
    public int kthLargestElement(int k, int[] nums) {
        if (nums == null || nums.length == 0 || k <= 0 || k > nums.length) {
            return 0;
        }

        return select(nums, 0, nums.length - 1, nums.length - k);

    }

    public int select(int[] nums, int left, int right, int k) {
        if (left == right) {
            return nums[left];
        }

        int pivotIndex = partition(nums, left, right);
        if (pivotIndex == k) {
            return nums[pivotIndex];
        } else if (pivotIndex < k) {
            return select(nums, pivotIndex + 1, right, k);
        }  else {
            return select(nums, left, pivotIndex - 1, k);
        }
    }

    public int partition(int[] nums, int left, int right) {

        // Init pivot, better to be random
        int pivot = nums[left];

        // Begin partition
        while (left < right) {
            while (left < right && nums[right] >= pivot) { // skip nums[i] that equals pivot
                right--;
            }
            nums[left] = nums[right];
            while (left < right && nums[left] <= pivot) { // skip nums[i] that equals pivot
                left++;
            }
            nums[right] = nums[left];
        }

        // Recover pivot to array
        nums[left] = pivot;
        return left;
    }
}
```

### Quick Select (with random pivot)

Source: [wikipedia: QuickSelect](https://en.wikipedia.org/wiki/Quickselect)

Animation:

* [geekviewpoint: quickselect](http://www.geekviewpoint.com/java/search/quickselect)
* [csanimated: Quicksort](http://www.csanimated.com/animation.php?t=Quicksort)

```java
import java.util.Random;

class Solution {
    /*
     * @param k : description of k
     * @param nums : array of nums
     * @return: description of return
     */
    public int kthLargestElement(int k, int[] nums) {
        if (nums == null || nums.length == 0 || k <= 0 || k > nums.length) {
            return 0;
        }

        return select(nums, 0, nums.length - 1, nums.length - k);

    }

    public int select(int[] nums, int left, int right, int k) {
        if (left == right) {
            return nums[left];
        }

        int pivotIndex = partition(nums, left, right);
        if (pivotIndex == k) {
            return nums[pivotIndex];
        } else if (pivotIndex < k) {
            return select(nums, pivotIndex + 1, right, k);
        }  else {
            return select(nums, left, pivotIndex - 1, k);
        }
    }

    public void swap(int[] nums, int x, int y) {
        int tmp = nums[x];
        nums[x] = nums[y];
        nums[y] = tmp;
    }

    public int partition(int[] nums, int left, int right) {

        Random rand = new Random();
        int pivotIndex = rand.nextInt((right - left) + 1) + left;
        // Init pivot
        int pivotValue = nums[pivotIndex];

        swap(nums, pivotIndex, right);

        // First index that nums[firstIndex] > pivotValue
        int firstIndex = left;

        for (int i = left; i <= right - 1; i++) {
            if (nums[i] < pivotValue) {
                swap(nums, firstIndex, i);
                firstIndex++;
            }
        }

        // Recover pivot to array
        swap(nums, right, firstIndex);
        return firstIndex;
    }

    public static void main(String[] args) {
        System.out.println("kth Largest Element: Quick Select");
        int[] A = {21, 3, 34, 5, 13, 8, 2, 55, 1, 19};
        Solution search = new Solution();
        int expResult[] = {1, 2, 3, 5, 8, 13, 19, 21, 34, 55};
        int k = expResult.length;
        int err = 0;
        for (int exp : expResult) {
            if (exp != search.kthLargestElement(k--, A)) {
                System.out.println("Test failed: " + k);
                err++;
            }
        }
        System.out.println("Test finished");
    }
}
```

#### LeetCode Quick Select

```java
import java.util.Random;
class Solution {
  int [] nums;

  public void swap(int a, int b) {
    int tmp = this.nums[a];
    this.nums[a] = this.nums[b];
    this.nums[b] = tmp;
  }


  public int partition(int left, int right, int pivot_index) {
    int pivot = this.nums[pivot_index];
    // 1. move pivot to end
    swap(pivot_index, right);
    int store_index = left;

    // 2. move all smaller elements to the left
    for (int i = left; i <= right; i++) {
      if (this.nums[i] < pivot) {
        swap(store_index, i);
        store_index++;
      }
    }

    // 3. move pivot to its final place
    swap(store_index, right);

    return store_index;
  }

  public int quickselect(int left, int right, int k_smallest) {
    /*
    Returns the k-th smallest element of list within left..right.
    */

    if (left == right) // If the list contains only one element,
      return this.nums[left];  // return that element

    // select a random pivot_index
    Random random_num = new Random();
    int pivot_index = left + random_num.nextInt(right - left); 

    pivot_index = partition(left, right, pivot_index);

    // the pivot is on (N - k)th smallest position
    if (k_smallest == pivot_index)
      return this.nums[k_smallest];
    // go left side
    else if (k_smallest < pivot_index)
      return quickselect(left, pivot_index - 1, k_smallest);
    // go right side
    return quickselect(pivot_index + 1, right, k_smallest);
  }

  public int findKthLargest(int[] nums, int k) {
    this.nums = nums;
    int size = nums.length;
    // kth largest is (N - k)th smallest
    return quickselect(0, size - 1, size - k);
  }
}
```

## Reference

LeetCode:

* [Solution explained](https://discuss.leetcode.com/topic/14597/solution-explained)
* [Solutions using Partition, Max-Heap, priority\_queue and multiset respectively](https://discuss.leetcode.com/topic/15256/4-c-solutions-using-partition-max-heap-priority_queue-and-multiset-respectively)
* \[Concise JAVA solution based on Quick Select]\([https://leetcode.com/problems/kth-largest-element-in-an-array/discuss/60333/Concise-JAVA-solution-based-on-Quick-Select\\](https://leetcode.com/problems/kth-largest-element-in-an-array/discuss/60333/Concise-JAVA-solution-based-on-Quick-Select\)/)&#x20;

Source: [wikipedia: QuickSelect](https://en.wikipedia.org/wiki/Quickselect)

Animation:

* [geekviewpoint: quickselect](http://www.geekviewpoint.com/java/search/quickselect)
* [csanimated: Quicksort](http://www.csanimated.com/animation.php?t=Quicksort)
