# k largest(or smallest) elements in an array | added Min Heap method

Question: Write an efficient program for printing k largest elements in an array. Elements in array can be in any order.

For example, if given array is [1, 23, 12, 9, 30, 2, 50] and you are asked for the largest 3 elements i.e., k = 3 then your program should print 50, 30 and 23.

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Method 1 (Use Bubble k times)

Thanks to Shailendra for suggesting this approach.
1) Modify Bubble Sort to run the outer loop at most k times.
2) Print the last k elements of the array obtained in step 1.

Time Complexity: O(nk)

Like Bubble sort, other sorting algorithms like Selection Sort can also be modified to get the k largest elements.

Method 2 (Use temporary array)
K largest elements from arr[0..n-1]

1) Store the first k elements in a temporary array temp[0..k-1].
2) Find the smallest element in temp[], let the smallest element be min.
3-a) For each element x in arr[k] to arr[n-1]. O(n-k)
If x is greater than the min then remove min from temp[] and insert x.
3-b)Then, determine the new min from temp[]. O(k)
4) Print final k elements of temp[]

Time Complexity: O((n-k)*k). If we want the output sorted then O((n-k)*k + klogk)

Thanks to nesamani1822 for suggesting this method.

Method 3(Use Sorting)
1) Sort the elements in descending order in O(nLogn)
2) Print the first k numbers of the sorted array O(k).
Following is the implementation of above.

## C++

 `// C++ code for k largest elements in an array ` `#include ` `using` `namespace` `std; ` ` `  `void` `kLargest(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``// Sort the given array arr in reverse ` `    ``// order. ` `    ``sort(arr, arr + n, greater<``int``>()); ` ` `  `    ``// Print the first kth largest elements ` `    ``for` `(``int` `i = 0; i < k; i++) ` `        ``cout << arr[i] << ``" "``; ` `} ` ` `  `// driver program ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 23, 12, 9, 30, 2, 50 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `k = 3; ` `    ``kLargest(arr, n, k); ` `} ` ` `  `// This article is contributed by Chhavi `

## Java

 `// Java code for k largest elements in an array ` `import` `java.util.Arrays; ` `import` `java.util.Collections; ` ` `  `class` `GFG { ` `    ``public` `static` `void` `kLargest(Integer[] arr, ``int` `k) ` `    ``{ ` `        ``// Sort the given array arr in reverse order ` `        ``// This method doesn't work with primitive data ` `        ``// types. So, instead of int, Integer type ` `        ``// array will be used ` `        ``Arrays.sort(arr, Collections.reverseOrder()); ` ` `  `        ``// Print the first kth largest elements ` `        ``for` `(``int` `i = ``0``; i < k; i++) ` `            ``System.out.print(arr[i] + ``" "``); ` `    ``} ` ` `  `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``Integer arr[] = ``new` `Integer[] { ``1``, ``23``, ``12``, ``9``, ` `                                        ``30``, ``2``, ``50` `}; ` `        ``int` `k = ``3``; ` `        ``kLargest(arr, k); ` `    ``} ` `} ` `// This code is contributed by Kamal Rawal `

## Python

 `''' Python3 code for k largest elements in an array'''` ` `  `def` `kLargest(arr, k): ` `    ``# Sort the given array arr in reverse  ` `    ``# order. ` `    ``arr.sort(reverse ``=` `True``) ` `    ``# Print the first kth largest elements ` `    ``for` `i ``in` `range``(k): ` `        ``print` `(arr[i], end ``=``" "``)  ` ` `  `# Driver program ` `arr ``=` `[``1``, ``23``, ``12``, ``9``, ``30``, ``2``, ``50``] ` `# n = len(arr) ` `k ``=` `3` `kLargest(arr, k) ` ` `  `# This code is contributed by shreyanshi_arun. `

## C#

 `// C# code for k largest elements in an array ` `using` `System; ` ` `  `class` `GFG { ` `    ``public` `static` `void` `kLargest(``int``[] arr, ``int` `k) ` `    ``{ ` `        ``// Sort the given array arr in reverse order ` `        ``// This method doesn't work with primitive data ` `        ``// types. So, instead of int, Integer type ` `        ``// array will be used ` `        ``Array.Sort(arr); ` `        ``Array.Reverse(arr); ` ` `  `        ``// Print the first kth largest elements ` `        ``for` `(``int` `i = 0; i < k; i++) ` `            ``Console.Write(arr[i] + ``" "``); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``int``[] arr = ``new` `int``[] { 1, 23, 12, 9, ` `                                ``30, 2, 50 }; ` `        ``int` `k = 3; ` `        ``kLargest(arr, k); ` `    ``} ` `} ` ` `  `// This code contributed by Rajput-Ji `

## PHP

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Output:

```50 30 23
```

Time complexity: O(nlogn)

Method 4 (Use Max Heap)
1) Build a Max Heap tree in O(n)
2) Use Extract Max k times to get k maximum elements from the Max Heap O(klogn)

Time complexity: O(n + klogn)

Method 5(Use Oder Statistics)
1) Use order statistic algorithm to find the kth largest element. Please see the topic selection in worst-case linear time O(n)
2) Use QuickSort Partition algorithm to partition around the kth largest number O(n).
3) Sort the k-1 elements (elements greater than the kth largest element) O(kLogk). This step is needed only if sorted output is required.

Time complexity: O(n) if we don’t need the sorted output, otherwise O(n+kLogk)

Thanks to Shilpi for suggesting the first two approaches.

Method 6 (Use Min Heap)
This method is mainly an optimization of method 1. Instead of using temp[] array, use Min Heap.

1) Build a Min Heap MH of the first k elements (arr to arr[k-1]) of the given array. O(k)

2) For each element, after the kth element (arr[k] to arr[n-1]), compare it with root of MH.
……a) If the element is greater than the root then make it root and call heapify for MH
……b) Else ignore it.
// The step 2 is O((n-k)*logk)

3) Finally, MH has k largest elements and root of the MH is the kth largest element.

Time Complexity: O(k + (n-k)Logk) without sorted output. If sorted output is needed then O(k + (n-k)Logk + kLogk)

All of the above methods can also be used to find the kth largest (or smallest) element.

## C++

 `#include ` `using` `namespace` `std; ` ` `  `// Swap function to interchange ` `// the value of variables x and y ` `int` `swap(``int``& x, ``int``& y) ` `{ ` `    ``int` `temp = x; ` `    ``x = y; ` `    ``y = temp; ` `} ` ` `  `// Min Heap Class ` `// arr holds reference to an integer  ` `// array size indicate the number of ` `// elements in Min Heap ` `class` `MinHeap { ` ` `  `    ``int` `size; ` `    ``int``* arr; ` ` `  `public``: ` `    ``// Constructor to initialize the size and arr ` `    ``MinHeap(``int` `size, ``int` `input[]); ` ` `  `    ``// Min Heapify function, that assumes that ` `    ``// 2*i+1 and 2*i+2 are min heap and fix the ` `    ``// heap property for i. ` `    ``void` `heapify(``int` `i); ` ` `  `    ``// Build the min heap, by calling heapify ` `    ``// for all non-leaf nodes. ` `    ``void` `buildHeap(); ` `}; ` ` `  `// Constructor to initialize data ` `// members and creating mean heap ` `MinHeap::MinHeap(``int` `size, ``int` `input[]) ` `{ ` `    ``// Initializing arr and size ` ` `  `    ``this``->size = size; ` `    ``this``->arr = input; ` ` `  `    ``// Building the Min Heap ` `    ``buildHeap(); ` `} ` ` `  `// Min Heapify function, that assumes ` `// 2*i+1 and 2*i+2 are min heap and  ` `// fix min heap property for i ` ` `  `void` `MinHeap::heapify(``int` `i) ` `{ ` `    ``// If Leaf Node, Simply return ` `    ``if` `(i >= size / 2) ` `        ``return``; ` ` `  `    ``// variable to store the smallest element ` `    ``// index out of i, 2*i+1 and 2*i+2 ` `    ``int` `smallest; ` ` `  `    ``// Index of left node ` `    ``int` `left = 2 * i + 1; ` ` `  `    ``// Index of right node ` `    ``int` `right = 2 * i + 2; ` ` `  `    ``// Select minimum from left node and  ` `    ``// current node i, and store the minimum ` `    ``// index in smallest variable ` `    ``smallest = arr[left] < arr[i] ? left : i; ` ` `  `    ``// If right child exist, compare and ` `    ``// update the smallest variable ` `    ``if` `(right < size) ` `        ``smallest = arr[right] < arr[smallest] ` `                             ``? right : smallest; ` ` `  `    ``// If Node i violates the min heap  ` `    ``// property, swap  current node i with ` `    ``// smallest to fix the min-heap property  ` `    ``// and recursively call heapify for node smallest. ` `    ``if` `(smallest != i) { ` `        ``swap(arr[i], arr[smallest]); ` `        ``heapify(smallest); ` `    ``} ` `} ` ` `  `// Build Min Heap ` `void` `MinHeap::buildHeap() ` `{ ` `    ``// Calling Heapify for all non leaf nodes ` `    ``for` `(``int` `i = size / 2 - 1; i >= 0; i--) { ` `        ``heapify(i); ` `    ``} ` `} ` ` `  `void` `FirstKelements(``int` `arr[],``int` `size,``int` `k){ ` `    ``// Creating Min Heap for given ` `    ``// array with only k elements ` `    ``MinHeap* m = ``new` `MinHeap(k, arr); ` ` `  `    ``// Loop For each element in array ` `    ``// after the kth element ` `    ``for` `(``int` `i = k; i < size; i++) { ` ` `  `        ``// if current element is smaller  ` `        ``// than minimum element, do nothing  ` `        ``// and continue to next element ` `        ``if` `(arr > arr[i]) ` `            ``continue``; ` ` `  `        ``// Otherwise Change minimum element to  ` `        ``// current element, and call heapify to ` `        ``// restore the heap property ` `        ``else` `{ ` `            ``arr = arr[i]; ` `            ``m->heapify(0); ` `        ``} ` `    ``} ` `    ``// Now min heap contains k maximum ` `    ``// elements, Iterate and print ` `    ``for` `(``int` `i = 0; i < k; i++) { ` `        ``cout << arr[i] << ``" "``; ` `    ``} ` `} ` `// Driver Program ` `int` `main() ` `{ ` ` `  `    ``int` `arr[] = { 11, 3, 2, 1, 15, 5, 4, ` `                           ``45, 88, 96, 50, 45 }; ` ` `  `    ``int` `size = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``// Size of Min Heap ` `    ``int` `k = 3; ` ` `  `    ``FirstKelements(arr,size,k); ` ` `  `    ``return` `0; ` `} ` `// This code is contributed by Ankur Goel `

Output:

```50 88 96
```

Please write comments if you find any of the above explanations/algorithms incorrect, or find better ways to solve the same problem.

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