Given an array arr[] and an integer k, we need to print k maximum elements of given array. The elements should printed in the order of the input.
Note : k is always less than or equal to n.
Examples:
Input : arr[] = {10 50 30 60 15}
k = 2
Output : 50 60
The top 2 elements are printed
as per their appearance in original
array.
Input : arr[] = {50 8 45 12 25 40 84}
k = 3
Output : 50 45 84
Method 1: We search for the maximum element k times in the given array. Each time we find one maximum element, we print it and replace it with minus infinite (Integer.MIN_VALUE in Java) in the array. Also, the position of all k maximum elements is marked using an array so that with the help of that array we can print the elements in the order given in the original array. The time complexity of this method is O(n*k).
Java
// Java program to find k maximum elements// of array in original orderimport java.util.*;import java.util.Arrays;import java.util.Collections;class GFG{ // Function to print k Maximum elementspublic static void printMax(int arr[], int k, int n){ int[] brr = new int[n]; Arrays.fill(brr, 0); int[] crr = new int[n]; // Copying the array arr // into crr so that it // can be used later for(int i=0;i<n;i++) { crr[i]=arr[i]; } // Iterating for K-times for(int i=0;i<k;i++) { // Finding the maximum element // along with its index int maxi=Integer.MIN_VALUE; int index=0; for(int j=0;j<n;j++) { if(maxi<arr[j]) { maxi=arr[j]; index=j; } } // Assigning 1 in order // to mark the position // of all k maximum numbers brr[index]=1; arr[index]=Integer.MIN_VALUE; } for(int i=0;i<n;i++) { // Printing the k maximum // elements array if(brr[i]==1) System.out.print(crr[i]+" "); }} // Driver Codepublic static void main(String[] args){ int[] arr = { 50, 8, 45, 12, 25, 40, 84 }; int n = arr.length; int k = 3; printMax(arr, k, n);}} |
50 45 84
Time Complexity: O(n*k)
Auxiliary Space: O(n)
Method 2: In this method, we store the original array in a new array and will sort the new array in descending order. After sorting, we iterate the original array from 0 to n and print all those elements that appear in first k elements of new array. For searching, we can do Binary Search.
Java
// Java program to find k maximum // elements of array in original orderimport java.util.Arrays;import java.util.Collections;public class GfG { // Function to print m Maximum elements public static void printMax(int arr[], int k, int n) { // Array to store the copy // of the original array Integer[] brr = new Integer[n]; for (int i = 0; i < n; i++) brr[i] = arr[i]; // Sorting the array in // descending order Arrays.sort(brr, Collections.reverseOrder()); // Traversing through original array and // printing all those elements that are // in first k of sorted array. // Please refer https://goo.gl/uj5RCD // for details of Arrays.binarySearch() for (int i = 0; i < n; ++i) if (Arrays.binarySearch(brr, arr[i], Collections.reverseOrder()) >= 0 && Arrays.binarySearch(brr, arr[i], Collections.reverseOrder()) < k) System.out.print(arr[i]+ " "); } // Driver code public static void main(String args[]) { int arr[] = { 50, 8, 45, 12, 25, 40, 84 }; int n = arr.length; int k = 3; printMax(arr, k, n); }}// This code is contributed by Swetank Modi |
50 45 84
Time Complexity: O(n Log n) for sorting.
Auxiliary Space: O(n)
Please refer complete article on Find k maximum elements of array in original order for more details!
