Explain how to find 100 largest number...

To find the 100 largest numbers out of an array of 1 billion numbers, the most efficient approach is to use a min-heap (also known as a priority queue). This method ensures that the time complexity is optimized compared to a brute force approach like sorting the entire array. Here’s a step-by-step explanation of the process:

Steps to Find the 100 Largest Numbers Using a Min-Heap

1. Initialize a Min-Heap:

   • Create a min-heap of size 100. This heap will store the 100 largest numbers encountered so far.

2. Populate the Heap:

   • Iterate through the first 100 elements of the array and insert them into the min-heap. This step ensures that the heap is initially filled with the first 100 numbers.

3. Process the Remaining Elements:

   • For each of the remaining elements in the array (from the 101st element to the 1 billionth element), compare the element with the root of the min-heap (the smallest element in the heap).
   • If the current element is larger than the root of the min-heap, remove the root and insert the current element into the heap. This step ensures that the heap always contains the 100 largest elements encountered so far.

4. Extract the 100 Largest Elements:

   • After processing all elements, the min-heap will contain the 100 largest elements of the array. You can then extract these elements from the heap.

Example Code in Java

Here is an example implementation in Java:

import java.util.PriorityQueue;

public class LargestNumbers {
    public static void main(String[] args) {
        int[] array = new int[1000000000]; // Example array with 1 billion elements
        // Assume array is populated with data

        int k = 100;
        PriorityQueue<Integer> minHeap = new PriorityQueue<>(k);

        // Populate the heap with the first 100 elements
        for (int i = 0; i < k; i++) {
            minHeap.add(array[i]);
        }

        // Process the remaining elements
        for (int i = k; i < array.length; i++) {
            if (array[i] > minHeap.peek()) {
                minHeap.poll();
                minHeap.add(array[i]);
            }
        }

        // The heap now contains the 100 larg...