Kth Largest Element in a Stream

Problem statement

Design a class to find the kth largest element in a stream. Note that it is the kth largest element in the sorted order, not the kth 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 kth largest element in the stream.

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

Constraints:

• 1 <= k <= 104
• 0 <= nums.length <= 104
• -104 <= nums[i] <= 104
• -104 <= val <= 104
• At most 104 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 kth element.

My solution

/**
 * @param {number} k
 * @param {number[]} nums
 */
var KthLargest = function(k, nums) {
    this.k = k;
    this.nums = nums.sort((a, b) => a - b);
    
};

/** 
 * @param {number} val
 * @return {number}
 */
KthLargest.prototype.add = function(val) {
    let left = 0;
    let right = this.nums.length - 1;
    let indexToInsert = left;
    while(left <= right) {
        let mid = Math.floor(left + (right - left) / 2);
        if (val > this.nums[mid]) {
            left = mid + 1;
            indexToInsert = mid + 1;
        } else if (val < this.nums[mid]) {
            right = mid - 1;
            indexToInsert = mid;
        } else {
            indexToInsert = mid
            break;
        }
    }
    
    this.nums.splice(indexToInsert, 0, val);
    return this.nums[this.nums.length - this.k];
};

/** 
 * Your KthLargest object will be instantiated and called as such:
 * var obj = new KthLargest(k, nums)
 * var param_1 = obj.add(val)
 */