Below is the implementation of the stream data processor:
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152/** * Brute Force Approach * Time Complexity: O(n) for add operation * Space Complexity: O(n) for storing all elements */class KthLargestBruteForce { /** * @param {number} k * @param {number[]} nums */ constructor(k, nums) { this.k = k; this.nums = nums.sort((a, b) => b - a); // Sort in descending order } /** * @param {number} val * @return {number} */ add(val) { // Insert val into the correct position to maintain sorted order let i = 0; while (i < this.nums.length && this.nums[i] > val) { i++; } this.nums.splice(i, 0, val); // Return the kth largest element return this.nums[this.k - 1]; }} /** * Min Heap Approach * Time Complexity: O(log k) for add operation * Space Complexity: O(k) for storing k elements * * Note: JavaScript doesn't have a built-in heap, so we'll implement a simple min heap */class MinHeap { constructor() { this.heap = []; } size() { return this.heap.length; } peek() { return this.heap[0]; } add(val) { this.heap.push(val); this.bubbleUp(this.heap.length - 1); } poll() { if (this.heap.length === 0) { return null; } const min = this.heap[0]; const last = this.heap.pop(); if (this.heap.length > 0) { this.heap[0] = last; this.bubbleDown(0); } return min; } bubbleUp(index) { while (index > 0) { const parentIndex = Math.floor((index - 1) / 2); if (this.heap[parentIndex] <= this.heap[index]) { break; } // Swap parent and current element [this.heap[parentIndex], this.heap[index]] = [this.heap[index], this.heap[parentIndex]]; index = parentIndex; } } bubbleDown(index) { const lastIndex = this.heap.length - 1; while (true) { const leftChildIndex = 2 * index + 1; const rightChildIndex = 2 * index + 2; let smallestIndex = index; if (leftChildIndex <= lastIndex && this.heap[leftChildIndex] < this.heap[smallestIndex]) { smallestIndex = leftChildIndex; } if (rightChildIndex <= lastIndex && this.heap[rightChildIndex] < this.heap[smallestIndex]) { smallestIndex = rightChildIndex; } if (smallestIndex === index) { break; } // Swap current element with the smallest child [this.heap[index], this.heap[smallestIndex]] = [this.heap[smallestIndex], this.heap[index]]; index = smallestIndex; } }} class KthLargest { /** * @param {number} k * @param {number[]} nums */ constructor(k, nums) { this.k = k; this.heap = new MinHeap(); // Add all elements from nums to the heap for (const num of nums) { this.add(num); } } /** * @param {number} val * @return {number} */ add(val) { this.heap.add(val); // If heap size exceeds k, remove the smallest element if (this.heap.size() > this.k) { this.heap.poll(); } // Return the kth largest element (the root of the heap) return this.heap.peek(); }} // Test caseconst kthLargest = new KthLargest(3, [4, 5, 8, 2]);console.log(kthLargest.add(3)); // 4console.log(kthLargest.add(5)); // 5console.log(kthLargest.add(10)); // 5console.log(kthLargest.add(9)); // 8console.log(kthLargest.add(4)); // 8
Let's break down the implementation:
Implement the stream data processor solution in different programming languages.
Below is the implementation of the stream data processor in different programming languages. Select a language tab to view the corresponding code.
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152/** * Brute Force Approach * Time Complexity: O(n) for add operation * Space Complexity: O(n) for storing all elements */class KthLargestBruteForce { /** * @param {number} k * @param {number[]} nums */ constructor(k, nums) { this.k = k; this.nums = nums.sort((a, b) => b - a); // Sort in descending order } /** * @param {number} val * @return {number} */ add(val) { // Insert val into the correct position to maintain sorted order let i = 0; while (i < this.nums.length && this.nums[i] > val) { i++; } this.nums.splice(i, 0, val); // Return the kth largest element return this.nums[this.k - 1]; }} /** * Min Heap Approach * Time Complexity: O(log k) for add operation * Space Complexity: O(k) for storing k elements * * Note: JavaScript doesn't have a built-in heap, so we'll implement a simple min heap */class MinHeap { constructor() { this.heap = []; } size() { return this.heap.length; } peek() { return this.heap[0]; } add(val) { this.heap.push(val); this.bubbleUp(this.heap.length - 1); } poll() { if (this.heap.length === 0) { return null; } const min = this.heap[0]; const last = this.heap.pop(); if (this.heap.length > 0) { this.heap[0] = last; this.bubbleDown(0); } return min; } bubbleUp(index) { while (index > 0) { const parentIndex = Math.floor((index - 1) / 2); if (this.heap[parentIndex] <= this.heap[index]) { break; } // Swap parent and current element [this.heap[parentIndex], this.heap[index]] = [this.heap[index], this.heap[parentIndex]]; index = parentIndex; } } bubbleDown(index) { const lastIndex = this.heap.length - 1; while (true) { const leftChildIndex = 2 * index + 1; const rightChildIndex = 2 * index + 2; let smallestIndex = index; if (leftChildIndex <= lastIndex && this.heap[leftChildIndex] < this.heap[smallestIndex]) { smallestIndex = leftChildIndex; } if (rightChildIndex <= lastIndex && this.heap[rightChildIndex] < this.heap[smallestIndex]) { smallestIndex = rightChildIndex; } if (smallestIndex === index) { break; } // Swap current element with the smallest child [this.heap[index], this.heap[smallestIndex]] = [this.heap[smallestIndex], this.heap[index]]; index = smallestIndex; } }} class KthLargest { /** * @param {number} k * @param {number[]} nums */ constructor(k, nums) { this.k = k; this.heap = new MinHeap(); // Add all elements from nums to the heap for (const num of nums) { this.add(num); } } /** * @param {number} val * @return {number} */ add(val) { this.heap.add(val); // If heap size exceeds k, remove the smallest element if (this.heap.size() > this.k) { this.heap.poll(); } // Return the kth largest element (the root of the heap) return this.heap.peek(); }} // Test caseconst kthLargest = new KthLargest(3, [4, 5, 8, 2]);console.log(kthLargest.add(3)); // 4console.log(kthLargest.add(5)); // 5console.log(kthLargest.add(10)); // 5console.log(kthLargest.add(9)); // 8console.log(kthLargest.add(4)); // 8
First, understand that we need to design a class that finds the kth largest element in a stream of numbers.
Decide which approach to use: brute force (maintain a sorted list) or min heap (maintain a heap of size k).
Initialize the class with the given k and nums array, setting up the data structure (sorted list or min heap).
Add the new element to the data structure and return the kth largest element.
Consider edge cases such as empty arrays, k = 1, or when the number of elements is less than k.
Consider optimizations such as using a min heap for O(log k) time complexity instead of maintaining a sorted list.
Verify the solution with the provided examples and additional test cases.
Modify the code to implement an alternative approach and test it with the same examples.
Implement a function that solves the stream data processor problem using a different approach than shown above.
If the input array is empty, we initialize the data structure with no elements and add elements as they come.
If k = 1, we need to find the maximum element in the stream.
If the number of elements is less than k, we need to handle this case carefully. The problem guarantees that there will be at least k elements when we search for the kth element.
The stream may contain duplicate elements, which should be treated as separate elements.
Below is the implementation of the stream data processor:
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152/** * Brute Force Approach * Time Complexity: O(n) for add operation * Space Complexity: O(n) for storing all elements */class KthLargestBruteForce { /** * @param {number} k * @param {number[]} nums */ constructor(k, nums) { this.k = k; this.nums = nums.sort((a, b) => b - a); // Sort in descending order } /** * @param {number} val * @return {number} */ add(val) { // Insert val into the correct position to maintain sorted order let i = 0; while (i < this.nums.length && this.nums[i] > val) { i++; } this.nums.splice(i, 0, val); // Return the kth largest element return this.nums[this.k - 1]; }} /** * Min Heap Approach * Time Complexity: O(log k) for add operation * Space Complexity: O(k) for storing k elements * * Note: JavaScript doesn't have a built-in heap, so we'll implement a simple min heap */class MinHeap { constructor() { this.heap = []; } size() { return this.heap.length; } peek() { return this.heap[0]; } add(val) { this.heap.push(val); this.bubbleUp(this.heap.length - 1); } poll() { if (this.heap.length === 0) { return null; } const min = this.heap[0]; const last = this.heap.pop(); if (this.heap.length > 0) { this.heap[0] = last; this.bubbleDown(0); } return min; } bubbleUp(index) { while (index > 0) { const parentIndex = Math.floor((index - 1) / 2); if (this.heap[parentIndex] <= this.heap[index]) { break; } // Swap parent and current element [this.heap[parentIndex], this.heap[index]] = [this.heap[index], this.heap[parentIndex]]; index = parentIndex; } } bubbleDown(index) { const lastIndex = this.heap.length - 1; while (true) { const leftChildIndex = 2 * index + 1; const rightChildIndex = 2 * index + 2; let smallestIndex = index; if (leftChildIndex <= lastIndex && this.heap[leftChildIndex] < this.heap[smallestIndex]) { smallestIndex = leftChildIndex; } if (rightChildIndex <= lastIndex && this.heap[rightChildIndex] < this.heap[smallestIndex]) { smallestIndex = rightChildIndex; } if (smallestIndex === index) { break; } // Swap current element with the smallest child [this.heap[index], this.heap[smallestIndex]] = [this.heap[smallestIndex], this.heap[index]]; index = smallestIndex; } }} class KthLargest { /** * @param {number} k * @param {number[]} nums */ constructor(k, nums) { this.k = k; this.heap = new MinHeap(); // Add all elements from nums to the heap for (const num of nums) { this.add(num); } } /** * @param {number} val * @return {number} */ add(val) { this.heap.add(val); // If heap size exceeds k, remove the smallest element if (this.heap.size() > this.k) { this.heap.poll(); } // Return the kth largest element (the root of the heap) return this.heap.peek(); }} // Test caseconst kthLargest = new KthLargest(3, [4, 5, 8, 2]);console.log(kthLargest.add(3)); // 4console.log(kthLargest.add(5)); // 5console.log(kthLargest.add(10)); // 5console.log(kthLargest.add(9)); // 8console.log(kthLargest.add(4)); // 8
Let's break down the implementation:
Implement the stream data processor solution in different programming languages.
Below is the implementation of the stream data processor in different programming languages. Select a language tab to view the corresponding code.
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152/** * Brute Force Approach * Time Complexity: O(n) for add operation * Space Complexity: O(n) for storing all elements */class KthLargestBruteForce { /** * @param {number} k * @param {number[]} nums */ constructor(k, nums) { this.k = k; this.nums = nums.sort((a, b) => b - a); // Sort in descending order } /** * @param {number} val * @return {number} */ add(val) { // Insert val into the correct position to maintain sorted order let i = 0; while (i < this.nums.length && this.nums[i] > val) { i++; } this.nums.splice(i, 0, val); // Return the kth largest element return this.nums[this.k - 1]; }} /** * Min Heap Approach * Time Complexity: O(log k) for add operation * Space Complexity: O(k) for storing k elements * * Note: JavaScript doesn't have a built-in heap, so we'll implement a simple min heap */class MinHeap { constructor() { this.heap = []; } size() { return this.heap.length; } peek() { return this.heap[0]; } add(val) { this.heap.push(val); this.bubbleUp(this.heap.length - 1); } poll() { if (this.heap.length === 0) { return null; } const min = this.heap[0]; const last = this.heap.pop(); if (this.heap.length > 0) { this.heap[0] = last; this.bubbleDown(0); } return min; } bubbleUp(index) { while (index > 0) { const parentIndex = Math.floor((index - 1) / 2); if (this.heap[parentIndex] <= this.heap[index]) { break; } // Swap parent and current element [this.heap[parentIndex], this.heap[index]] = [this.heap[index], this.heap[parentIndex]]; index = parentIndex; } } bubbleDown(index) { const lastIndex = this.heap.length - 1; while (true) { const leftChildIndex = 2 * index + 1; const rightChildIndex = 2 * index + 2; let smallestIndex = index; if (leftChildIndex <= lastIndex && this.heap[leftChildIndex] < this.heap[smallestIndex]) { smallestIndex = leftChildIndex; } if (rightChildIndex <= lastIndex && this.heap[rightChildIndex] < this.heap[smallestIndex]) { smallestIndex = rightChildIndex; } if (smallestIndex === index) { break; } // Swap current element with the smallest child [this.heap[index], this.heap[smallestIndex]] = [this.heap[smallestIndex], this.heap[index]]; index = smallestIndex; } }} class KthLargest { /** * @param {number} k * @param {number[]} nums */ constructor(k, nums) { this.k = k; this.heap = new MinHeap(); // Add all elements from nums to the heap for (const num of nums) { this.add(num); } } /** * @param {number} val * @return {number} */ add(val) { this.heap.add(val); // If heap size exceeds k, remove the smallest element if (this.heap.size() > this.k) { this.heap.poll(); } // Return the kth largest element (the root of the heap) return this.heap.peek(); }} // Test caseconst kthLargest = new KthLargest(3, [4, 5, 8, 2]);console.log(kthLargest.add(3)); // 4console.log(kthLargest.add(5)); // 5console.log(kthLargest.add(10)); // 5console.log(kthLargest.add(9)); // 8console.log(kthLargest.add(4)); // 8
First, understand that we need to design a class that finds the kth largest element in a stream of numbers.
Decide which approach to use: brute force (maintain a sorted list) or min heap (maintain a heap of size k).
Initialize the class with the given k and nums array, setting up the data structure (sorted list or min heap).
Add the new element to the data structure and return the kth largest element.
Consider edge cases such as empty arrays, k = 1, or when the number of elements is less than k.
Consider optimizations such as using a min heap for O(log k) time complexity instead of maintaining a sorted list.
Verify the solution with the provided examples and additional test cases.
Modify the code to implement an alternative approach and test it with the same examples.
Implement a function that solves the stream data processor problem using a different approach than shown above.
If the input array is empty, we initialize the data structure with no elements and add elements as they come.
If k = 1, we need to find the maximum element in the stream.
If the number of elements is less than k, we need to handle this case carefully. The problem guarantees that there will be at least k elements when we search for the kth element.
The stream may contain duplicate elements, which should be treated as separate elements.