Problem Statement:-

Design a class to find the k<sup>th</sup> largest element in a stream. Note that it is the k<sup>th</sup> largest element in the sorted order, not the k<sup>th</sup> distinct element.

Implement KthLargest class:

• KthLargest(int k, int[] nums) Initializes the object with the integer k and the stream of integers nums.

• int add(int val) Appends the integer val to the stream and returns the element representing the k<sup>th</sup> largest element in the stream.

Link: https://leetcode.com/problems/kth-largest-element-in-a-stream/description/

Problem Explanation with examples:-

Example 1

Input
["KthLargest", "add", "add", "add", "add", "add"]
[[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]
Output
[null, 4, 5, 5, 8, 8]

Explanation
KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);
kthLargest.add(3);   // return 4
kthLargest.add(5);   // return 5
kthLargest.add(10);  // return 5
kthLargest.add(9);   // return 8
kthLargest.add(4);   // return 8

Constraints

• 1 <= k <= 10<sup>4</sup>

• 0 <= nums.length <= 10<sup>4</sup>

• -10<sup>4</sup> <= nums[i] <= 10<sup>4</sup>

• -10<sup>4</sup> <= val <= 10<sup>4</sup>

• At most 10<sup>4</sup> calls will be made to add.

• It is guaranteed that there will be at least k elements in the array when you search for the k<sup>th</sup> element.

Intuition:-

• Simply add the new element to the array and sort it in descending order. Then return the kth largest element.

Solution:-

• In the constructor, we initialize the array and the value of k.

• In the add function, we append the new element to the array and sort it in descending order. Then we return the kth element from the array.

Code:-

JAVA Solution

class KthLargest {
    private int k;
    private int[] nums;

    public KthLargest(int k, int[] nums) {
        this.k = k;
        this.nums = nums;
    }

    public int add(int val) {
        int[] updatedNums = Arrays.copyOf(nums, nums.length + 1);
        updatedNums[nums.length] = val;
        Arrays.sort(updatedNums);
        nums = updatedNums;
        return nums[nums.length - k];
    }
}

Python Solution

class KthLargest:

    def __init__(self, k: int, nums: List[int]):
        self.k = k
        self.nums = nums


    def add(self, val: int) -> int:
        self.nums.append(val)
        self.nums.sort(reverse = True)
        return self.nums[self.k-1]

Complexity Analysis:-

TIME:-

The time complexity of add method is O(n log n) because sorting the array takes O(n log n) time, where n is the length of the array nums

SPACE:-

The space complexity is O(1) because the additional space used is constant, regardless of the input size.

References:-

• Sorting

Connect with me:-

• Twitter

• Github

• LinkedIn