# Find k maximum elements of array in original order

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 repalce it with minus infinite (INT_MIN in C) in the array. The time complexity of this method is O(n*k).

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.

## C++

 `// CPP program to find k maximum elements  ` `// of array in original order ` `#include ` `using` `namespace` `std; ` ` `  `// Function to print m Maximum elements ` `void` `printMax(``int` `arr[], ``int` `k, ``int` `n) ` `{ ` `    ``// vector to store the copy of the ` `    ``// original array ` `    ``vector<``int``> brr(arr, arr + n); ` ` `  `    ``// Sorting the vector in descending ` `    ``// order. Please refer below link for ` `    ``// details ` `    ``// https://www.geeksforgeeks.org/sort-c-stl/ ` `    ``sort(brr.begin(), brr.end(), greater<``int``>()); ` ` `  `    ``// Traversing through original array and ` `    ``// printing all those elements that are ` `    ``// in first k of sorted vector. ` `    ``// Please refer https://goo.gl/44Rwgt ` `    ``// for details of binary_search() ` `    ``for` `(``int` `i = 0; i < n; ++i) ` `        ``if` `(binary_search(brr.begin(), ` `                 ``brr.begin() + k, arr[i],  ` `                        ``greater<``int``>())) ` `            ``cout << arr[i] << ``" "``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 50, 8, 45, 12, 25, 40, 84 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `k = 3; ` `    ``printMax(arr, k, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find k maximum   ` `// elements of array in original order ` `import` `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 `

Output :

```50 45 84
```

Time Complexity : O(n Log n) for sorting.
Auxiliary Space : O(n)

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