# Merge k sorted arrays | Set 1

Given k sorted arrays of size n each, merge them and print the sorted output.

Example:

Input:
k = 3, n = 4
arr[][] = { {1, 3, 5, 7},
{2, 4, 6, 8},
{0, 9, 10, 11}} ;

Output: 0 1 2 3 4 5 6 7 8 9 10 11
Explanation: The output array is a sorted array that contains all the elements of the input matrix.

Input:
k = 3, n = 4
arr[][] = { {1, 5, 6, 8},
{2, 4, 10, 12},
{3, 7, 9, 11},
{13, 14, 15, 16}} ;

Output: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Explanation: The output array is a sorted array that contains all the elements of the input matrix.

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

Naive Approach: The very naive method is to create an output array of size n * k and then copy all the elements into the output array followed by sorting.

• Algorithm:
1. Create a output array of size n * k.
2. Traverse the matrix from start to end and insert all the elements in output array.
3. Sort and print the output array.
• Implementation:

 `// C++ program to merge k sorted arrays of size n each. ` `#include ` `using` `namespace` `std; ` `#define n 4 ` ` `  ` `  `// A utility function to print array elements ` `void` `printArray(``int` `arr[], ``int` `size) ` `{ ` `   ``for` `(``int` `i=0; i < size; i++) ` `       ``cout << arr[i] << ``" "``; ` `} ` ` `  `// This function takes an array of arrays as an argument and ` `// All arrays are assumed to be sorted. It merges them together ` `// and prints the final sorted output. ` `void` `mergeKArrays(``int` `arr[][n], ``int` `a, ``int` `output[]) ` `{ ` `    ``int` `c=0; ` `     `  `    ``//traverse the matrix ` `    ``for``(``int` `i=0; i

Output:

```Merged array is
1 2 6 9 12 20 23 34 34 90 1000 2000
```
• Complexity Analysis:

• Time Complexity :O(n*k*log(n*k)).
since resulting array is of N*k size.
• Space Complexity :O(n*k), The output array is of size n*k.

Efficient Approach The process might begin with merging arrays into groups of two. After the first merge, we have k/2 arrays. Again merge arrays in groups, now we have k/4 arrays. This is similar to merge sort. Divide k arrays into two halves containing an equal number of arrays until there are two arrays in a group. This is followed by merging the arrays in a bottom-up manner.

• Algorithm:
1. Create a recursive function which takes k arrays and returns the output array.
2. In the recursive function, if the value of k is 1 then return the array else if the value of k is 2 then merge the two arrays in linear time and return the array.
3. If the value of k is greater than 2 then divide the group of k elements into two equal halves and recursively call the function, i.e 0 to k/2 array in one recursive function and k/2 to k array in another recursive function.
4. Print the output array.
• Implementation:

 `// C++ program to merge k sorted arrays of size n each. ` `#include ` `using` `namespace` `std; ` `#define n 4 ` ` `  `// Merge arr1[0..n1-1] and arr2[0..n2-1] into  ` `// arr3[0..n1+n2-1]  ` `void` `mergeArrays(``int` `arr1[], ``int` `arr2[], ``int` `n1,  ` `                             ``int` `n2, ``int` `arr3[])  ` `{  ` `    ``int` `i = 0, j = 0, k = 0;  ` `   `  `    ``// Traverse both array  ` `    ``while` `(i

Output:

```Merged array is
1 2 6 9 12 20 23 34 34 90 1000 2000
```
• Complexity Analysis:

• Time Complexity: O( n * k * log k).
There are log k levels as in each level the k arrays are divided in half and at each level the k arrays are traversed. So time Complexity is O( n * k ).
• Space Complexity: O( n * k * log k).
In each level O( n*k ) space is required So Space Complexity is O( n * k * log k).

Alternative Efficient Approach: The idea is to use Min Heap. This MinHeap based solution has the same time complexity which is O(NK log K). But for a different and particular sized array, this solution works much better. The process must start with creating a MinHeap and inserting the first element of all the k arrays. Remove the root element of Minheap and put it in the output array and insert the next element from the array of removed element. To get the result the step must continue until there is no element in the MinHeap left.
MinHeap: A Min-Heap is a complete binary tree in which the value in each internal node is smaller than or equal to the values in the children of that node. Mapping the elements of a heap into an array is trivial: if a node is stored at index k, then its left child is stored at index 2k + 1 and its right child at index 2k + 2.

• Algorithm:
1. Create a min Heap and insert the first element of all k arrays.
2. Run a loop until the size of MinHeap is greater than zero.
3. Remove the top element of the MinHeap and print the element.
4. Now insert the next element from the same array in which the removed element belonged.
5. If the array doesn’t have any more elements, then replace root with infinite.After replacing the root, heapify the tree.
• Implementation:

## C++

 `// C++ program to merge k sorted  ` `// arrays of size n each. ` `#include ` `using` `namespace` `std; ` ` `  `#define n 4 ` ` `  `// A min-heap node ` `struct` `MinHeapNode ` `{ ` `// The element to be stored ` `    ``int` `element; ` ` `  `// index of the array from which the element is taken ` `    ``int` `i; ` ` `  `// index of the next element to be picked from the array  ` `    ``int` `j; ` `}; ` ` `  `// Prototype of a utility function to swap two min-heap nodes ` `void` `swap(MinHeapNode *x, MinHeapNode *y); ` ` `  `// A class for Min Heap ` `class` `MinHeap ` `{ ` ` `  `// pointer to array of elements in heap ` `    ``MinHeapNode *harr;  ` ` `  `// size of min heap ` `    ``int` `heap_size;  ` `public``: ` `    ``// Constructor: creates a min heap of given size ` `    ``MinHeap(MinHeapNode a[], ``int` `size); ` ` `  `    ``// to heapify a subtree with root at given index ` `    ``void` `MinHeapify(``int` `); ` ` `  `    ``// to get index of left child of node at index i ` `    ``int` `left(``int` `i) { ``return` `(2*i + 1); } ` ` `  `    ``// to get index of right child of node at index i ` `    ``int` `right(``int` `i) { ``return` `(2*i + 2); } ` ` `  `    ``// to get the root ` `    ``MinHeapNode getMin() { ``return` `harr; } ` ` `  `    ``// to replace root with new node x and heapify() new root ` `    ``void` `replaceMin(MinHeapNode x) { harr = x;  MinHeapify(0); } ` `}; ` ` `  `// This function takes an array of arrays as an argument and ` `// All arrays are assumed to be sorted. It merges them together ` `// and prints the final sorted output. ` `int` `*mergeKArrays(``int` `arr[][n], ``int` `k) ` `{ ` ` `  `// To store output array ` `    ``int` `*output = ``new` `int``[n*k];   ` ` `  `    ``// Create a min heap with k heap nodes.  ` `    ``// Every heap node has first element of an array ` `    ``MinHeapNode *harr = ``new` `MinHeapNode[k]; ` `    ``for` `(``int` `i = 0; i < k; i++) ` `    ``{ ` ` `  `// Store the first element ` `        ``harr[i].element = arr[i];  ` ` `  `// index of array ` `        ``harr[i].i = i; ` ` `  ` ``// Index of next element to be stored from the array ` `        ``harr[i].j = 1;  ` `    ``} ` ` `  `// Create the heap ` `    ``MinHeap hp(harr, k);  ` ` `  `    ``// Now one by one get the minimum element from min ` `    ``// heap and replace it with next element of its array ` `    ``for` `(``int` `count = 0; count < n*k; count++) ` `    ``{ ` `        ``// Get the minimum element and store it in output ` `        ``MinHeapNode root = hp.getMin(); ` `        ``output[count] = root.element; ` ` `  `        ``// Find the next elelement that will replace current ` `        ``// root of heap. The next element belongs to same ` `        ``// array as the current root. ` `        ``if` `(root.j < n) ` `        ``{ ` `            ``root.element = arr[root.i][root.j]; ` `            ``root.j += 1; ` `        ``} ` `        ``// If root was the last element of its array ` `// INT_MAX is for infinite         ` `else` `root.element =  INT_MAX;  ` ` `  `        ``// Replace root with next element of array ` `        ``hp.replaceMin(root); ` `    ``} ` ` `  `    ``return` `output; ` `} ` ` `  `// FOLLOWING ARE IMPLEMENTATIONS OF  ` `// STANDARD MIN HEAP METHODS FROM CORMEN BOOK ` `// Constructor: Builds a heap from a given  ` `// array a[] of given size ` `MinHeap::MinHeap(MinHeapNode a[], ``int` `size) ` `{ ` `    ``heap_size = size; ` `    ``harr = a;  ``// store address of array ` `    ``int` `i = (heap_size - 1)/2; ` `    ``while` `(i >= 0) ` `    ``{ ` `        ``MinHeapify(i); ` `        ``i--; ` `    ``} ` `} ` ` `  `// A recursive method to heapify a  ` `// subtree with root at given index.  ` `// This method assumes that the subtrees  ` `// are already heapified ` `void` `MinHeap::MinHeapify(``int` `i) ` `{ ` `    ``int` `l = left(i); ` `    ``int` `r = right(i); ` `    ``int` `smallest = i; ` `    ``if` `(l < heap_size && harr[l].element < harr[i].element) ` `        ``smallest = l; ` `    ``if` `(r < heap_size && harr[r].element < harr[smallest].element) ` `        ``smallest = r; ` `    ``if` `(smallest != i) ` `    ``{ ` `        ``swap(&harr[i], &harr[smallest]); ` `        ``MinHeapify(smallest); ` `    ``} ` `} ` ` `  `// A utility function to swap two elements ` `void` `swap(MinHeapNode *x, MinHeapNode *y) ` `{ ` `    ``MinHeapNode temp = *x;  *x = *y;  *y = temp; ` `} ` ` `  `// A utility function to print array elements ` `void` `printArray(``int` `arr[], ``int` `size) ` `{ ` `   ``for` `(``int` `i=0; i < size; i++) ` `       ``cout << arr[i] << ``" "``; ` `} ` ` `  `// Driver program to test above functions ` `int` `main() ` `{ ` `    ``// Change n at the top to change number of elements ` `    ``// in an array ` `    ``int` `arr[][n] =  {{2, 6, 12, 34}, ` `                     ``{1, 9, 20, 1000}, ` `                     ``{23, 34, 90, 2000}}; ` `    ``int` `k = ``sizeof``(arr)/``sizeof``(arr); ` ` `  `    ``int` `*output = mergeKArrays(arr, k); ` ` `  `    ``cout << ``"Merged array is "` `<< endl; ` `    ``printArray(output, n*k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to merge k sorted  ` `// arrays of size n each. ` ` `  `// A min heap node ` `class` `MinHeapNode ` `{ ` `    ``int` `element; ``// The element to be stored ` `     `  `     ``// index of the array from  ` `     ``// which the element is taken ` `    ``int` `i; ` `     `  `    ``// index of the next element  ` `    ``// to be picked from array ` `    ``int` `j;  ` ` `  `    ``public` `MinHeapNode(``int` `element, ``int` `i, ``int` `j) ` `    ``{ ` `        ``this``.element = element; ` `        ``this``.i = i; ` `        ``this``.j = j; ` `    ``} ` `}; ` ` `  `// A class for Min Heap ` `class` `MinHeap ` `{ ` `    ``MinHeapNode[] harr; ``// Array of elements in heap ` `    ``int` `heap_size; ``// Current number of elements in min heap ` ` `  `    ``// Constructor: Builds a heap from  ` `    ``// a given array a[] of given size ` `    ``public` `MinHeap(MinHeapNode a[], ``int` `size) ` `    ``{ ` `        ``heap_size = size; ` `        ``harr = a; ` `        ``int` `i = (heap_size - ``1``)/``2``; ` `        ``while` `(i >= ``0``) ` `        ``{ ` `            ``MinHeapify(i); ` `            ``i--; ` `        ``} ` `    ``} ` ` `  `    ``// A recursive method to heapify a subtree  ` `    ``// with the root at given index This method ` `    ``// assumes that the subtrees are already heapified ` `    ``void` `MinHeapify(``int` `i) ` `    ``{ ` `        ``int` `l = left(i); ` `        ``int` `r = right(i); ` `        ``int` `smallest = i; ` `        ``if` `(l < heap_size && harr[l].element < harr[i].element) ` `            ``smallest = l; ` `        ``if` `(r < heap_size && harr[r].element < harr[smallest].element) ` `            ``smallest = r; ` `        ``if` `(smallest != i) ` `        ``{ ` `            ``swap(harr, i, smallest); ` `            ``MinHeapify(smallest); ` `        ``} ` `    ``} ` ` `  `    ``// to get index of left child of node at index i ` `    ``int` `left(``int` `i) { ``return` `(``2``*i + ``1``); } ` ` `  `    ``// to get index of right child of node at index i ` `    ``int` `right(``int` `i) { ``return` `(``2``*i + ``2``); } ` ` `  `    ``// to get the root ` `    ``MinHeapNode getMin()  ` `    ``{ ` `        ``if``(heap_size <= ``0``)  ` `        ``{ ` `            ``System.out.println(``"Heap underflow"``); ` `            ``return` `null``; ` `        ``} ` `        ``return` `harr[``0``]; ` `    ``} ` ` `  `    ``// to replace root with new node  ` `    ``// "root" and heapify() new root ` `    ``void` `replaceMin(MinHeapNode root) { ` `        ``harr[``0``] = root; ` `        ``MinHeapify(``0``); ` `    ``} ` ` `  `    ``// A utility function to swap two min heap nodes ` `    ``void` `swap(MinHeapNode[] arr, ``int` `i, ``int` `j) { ` `        ``MinHeapNode temp = arr[i]; ` `        ``arr[i] = arr[j]; ` `        ``arr[j] = temp; ` `    ``} ` ` `  `    ``// A utility function to print array elements ` `    ``static` `void` `printArray(``int``[] arr) { ` `        ``for``(``int` `i : arr) ` `            ``System.out.print(i + ``" "``); ` `        ``System.out.println(); ` `    ``} ` ` `  `    ``// This function takes an array of  ` `    ``// arrays as an argument and All  ` `    ``// arrays are assumed to be sorted.  ` `    ``// It merges them together and  ` `    ``// prints the final sorted output. ` `    ``static` `void` `mergeKSortedArrays(``int``[][] arr, ``int` `k)  ` `    ``{ ` `        ``MinHeapNode[] hArr = ``new` `MinHeapNode[k]; ` `        ``int` `resultSize = ``0``; ` `        ``for``(``int` `i = ``0``; i < arr.length; i++)  ` `        ``{ ` `            ``MinHeapNode node = ``new` `MinHeapNode(arr[i][``0``],i,``1``); ` `            ``hArr[i] = node; ` `            ``resultSize += arr[i].length; ` `        ``} ` ` `  `        ``// Create a min heap with k heap nodes. Every heap node ` `        ``// has first element of an array ` `        ``MinHeap mh = ``new` `MinHeap(hArr, k); ` ` `  `        ``int``[] result = ``new` `int``[resultSize];     ``// To store output array ` ` `  `        ``// Now one by one get the minimum element from min ` `        ``// heap and replace it with next element of its array ` `        ``for``(``int` `i = ``0``; i < resultSize; i++)  ` `        ``{ ` ` `  `            ``// Get the minimum element and store it in result ` `            ``MinHeapNode root = mh.getMin(); ` `            ``result[i] = root.element; ` ` `  `            ``// Find the next element that will replace current ` `            ``// root of heap. The next element belongs to same ` `            ``// array as the current root. ` `            ``if``(root.j < arr[root.i].length) ` `                ``root.element = arr[root.i][root.j++]; ` `            ``// If root was the last element of its array ` `            ``else` `                ``root.element = Integer.MAX_VALUE; ` ` `  `            ``// Replace root with next element of array  ` `            ``mh.replaceMin(root); ` `        ``} ` ` `  `        ``printArray(result); ` ` `  `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]){ ` `        ``int``[][] arr= {{``2``, ``6``, ``12``, ``34``}, ` `                ``{``1``, ``9``, ``20``, ``1000``}, ` `                ``{``23``, ``34``, ``90``, ``2000``}}; ` ` `  `        ``System.out.println(``"Merged array is :"``); ` ` `  `        ``mergeKSortedArrays(arr,arr.length); ` `    ``} ` `}; ` ` `  `// This code is contributed by shubham96301 `

## Python3

 `""" ` `Python3 program to merge k sorted arrays of size n each ` `"""` `import` `sys ` `from` `typing ``import` `List``, Optional ` `Matrix ``=` `List``[``List``[``int``]] ` ` `  ` `  `class` `MinHeapNode: ` `    ``def` `__init__(``self``, el, i, j): ` `        ``self``.element ``=` `el ``# the element to be sorted ` `        ``self``.i ``=` `i         ``# index of array from which element is taken ` `        ``self``.j ``=` `j         ``# index of next element to be picked from array ` ` `  ` `  `class` `MinHeap: ` `    ``def` `__init__(``self``, ar: ``List``[MinHeapNode], size: ``int``): ` `        ``self``.heap_size ``=` `size ` `        ``self``.heap_arr ``=` `ar ` `        ``i ``=` `(``self``.heap_size ``-` `1``) ``/``/` `2` `        ``while` `i >``=` `0``: ` `            ``self``.min_heapify(i) ` `            ``i ``-``=` `1` ` `  `    ``""" ` `    ``A recursive method to heapify a subtree ` `    ``with the root at given index. This method ` `    ``assumes that the subtree are already heapified ` `    ``"""` `    ``def` `min_heapify(``self``, i): ` `        ``l ``=` `left(i) ` `        ``r ``=` `right(i) ` `        ``smallest ``=` `i ` `        ``if` `l < ``self``.heap_size ``and` `self``.heap_arr[l].element < ``self``.heap_arr[i].element: ` `            ``smallest ``=` `l ` `        ``if` `r < ``self``.heap_size ``and` `self``.heap_arr[r].element < ``self``.heap_arr[smallest].element: ` `            ``smallest ``=` `r ` `        ``if` `smallest !``=` `i: ` `            ``swap(``self``.heap_arr, i, smallest) ` `            ``self``.min_heapify(smallest) ` ` `  `    ``def` `get_min(``self``) ``-``> Optional: ` `        ``if` `self``.heap_size <``=` `0``: ` `            ``print``(``'Heap underflow'``) ` `            ``return` `None` `        ``return` `self``.heap_arr[``0``] ` ` `  `    ``# Replace root with new root ` `    ``def` `replace_min(``self``, root): ` `        ``self``.heap_arr[``0``] ``=` `root ` `        ``self``.min_heapify(``0``) ` ` `  ` `  `def` `left(i): ` `    ``return` `2` `*` `i ``+` `1` ` `  ` `  `def` `right(i): ` `    ``return` `2` `*` `i ``+` `2` ` `  ` `  `def` `swap(arr: ``List``[MinHeapNode], i, j): ` `    ``temp ``=` `arr[i] ` `    ``arr[i] ``=` `arr[j] ` `    ``arr[j] ``=` `temp ` ` `  ` `  `def` `merge_k_sorted_arrays(arr: Matrix, k: ``int``): ` `    ``h_arr ``=` `[] ` `    ``result_size ``=` `0` `    ``for` `i ``in` `range``(``len``(arr)): ` `        ``node ``=` `MinHeapNode(arr[i][``0``], i, ``1``) ` `        ``h_arr.append(node) ` `        ``result_size ``+``=` `len``(arr[i]) ` ` `  `    ``min_heap ``=` `MinHeap(h_arr, k) ` `    ``result ``=` `[``0``]``*``result_size ` `    ``for` `i ``in` `range``(result_size): ` `        ``root ``=` `min_heap.get_min() ` `        ``result[i] ``=` `root.element ` `        ``if` `root.j < ``len``(arr[root.i]): ` `            ``root.element ``=` `arr[root.i][root.j] ` `            ``root.j ``+``=` `1` `        ``else``: ` `            ``root.element ``=` `sys.maxsize ` `        ``min_heap.replace_min(root) ` `    ``for` `x ``in` `result: ` `        ``print``(x, end``=``' '``) ` `    ``print``() ` ` `  ` `  `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[ ` `        ``[``2``, ``6``, ``12``, ``34``], ` `        ``[``1``, ``9``, ``20``, ``1000``], ` `        ``[``23``, ``34``, ``90``, ``2000``] ` `    ``] ` `    ``print``(``'Merged Array is:'``) ` `    ``merge_k_sorted_arrays(arr, ``len``(arr)) ` ` `  `# The code is contributed by Rajat Srivastava `

## C#

 `// C# program to merge k sorted  ` `// arrays of size n each. ` `using` `System; ` ` `  `// A min heap node ` `public` `class` `MinHeapNode ` `{ ` `    ``public` `int` `element; ``// The element to be stored ` `     `  `    ``// index of the array from  ` `    ``// which the element is taken ` `    ``public` `int` `i; ` `     `  `    ``// index of the next element  ` `    ``// to be picked from array ` `    ``public` `int` `j;  ` ` `  `    ``public` `MinHeapNode(``int` `element, ``int` `i, ``int` `j) ` `    ``{ ` `        ``this``.element = element; ` `        ``this``.i = i; ` `        ``this``.j = j; ` `    ``} ` `}; ` ` `  `// A class for Min Heap ` `public` `class` `MinHeap ` `{ ` `    ``MinHeapNode[] harr; ``// Array of elements in heap ` `    ``int` `heap_size; ``// Current number of elements in min heap ` ` `  `    ``// Constructor: Builds a heap from  ` `    ``// a given array a[] of given size ` `    ``public` `MinHeap(MinHeapNode []a, ``int` `size) ` `    ``{ ` `        ``heap_size = size; ` `        ``harr = a; ` `        ``int` `i = (heap_size - 1) / 2; ` `        ``while` `(i >= 0) ` `        ``{ ` `            ``MinHeapify(i); ` `            ``i--; ` `        ``} ` `    ``} ` ` `  `    ``// A recursive method to heapify a subtree  ` `    ``// with the root at given index This method ` `    ``// assumes that the subtrees are already heapified ` `    ``void` `MinHeapify(``int` `i) ` `    ``{ ` `        ``int` `l = left(i); ` `        ``int` `r = right(i); ` `        ``int` `smallest = i; ` `        ``if` `(l < heap_size &&  ` `            ``harr[l].element < harr[i].element) ` `            ``smallest = l; ` `        ``if` `(r < heap_size &&  ` `            ``harr[r].element < harr[smallest].element) ` `            ``smallest = r; ` `        ``if` `(smallest != i) ` `        ``{ ` `            ``swap(harr, i, smallest); ` `            ``MinHeapify(smallest); ` `        ``} ` `    ``} ` ` `  `    ``// to get index of left child of node at index i ` `    ``int` `left(``int` `i) { ``return` `(2 * i + 1); } ` ` `  `    ``// to get index of right child of node at index i ` `    ``int` `right(``int` `i) { ``return` `(2 * i + 2); } ` ` `  `    ``// to get the root ` `    ``MinHeapNode getMin()  ` `    ``{ ` `        ``if``(heap_size <= 0)  ` `        ``{ ` `            ``Console.WriteLine(``"Heap underflow"``); ` `            ``return` `null``; ` `        ``} ` `        ``return` `harr; ` `    ``} ` ` `  `    ``// to replace root with new node  ` `    ``// "root" and heapify() new root ` `    ``void` `replaceMin(MinHeapNode root) ` `    ``{ ` `        ``harr = root; ` `        ``MinHeapify(0); ` `    ``} ` ` `  `    ``// A utility function to swap two min heap nodes ` `    ``void` `swap(MinHeapNode[] arr, ``int` `i, ``int` `j)  ` `    ``{ ` `        ``MinHeapNode temp = arr[i]; ` `        ``arr[i] = arr[j]; ` `        ``arr[j] = temp; ` `    ``} ` ` `  `    ``// A utility function to print array elements ` `    ``static` `void` `printArray(``int``[] arr)  ` `    ``{ ` `        ``foreach``(``int` `i ``in` `arr) ` `            ``Console.Write(i + ``" "``); ` `        ``Console.WriteLine(); ` `    ``} ` ` `  `    ``// This function takes an array of  ` `    ``// arrays as an argument and All  ` `    ``// arrays are assumed to be sorted.  ` `    ``// It merges them together and  ` `    ``// prints the final sorted output. ` `    ``static` `void` `mergeKSortedArrays(``int``[,] arr, ``int` `k)  ` `    ``{ ` `        ``MinHeapNode[] hArr = ``new` `MinHeapNode[k]; ` `        ``int` `resultSize = 0; ` `        ``for``(``int` `i = 0; i < arr.GetLength(0); i++)  ` `        ``{ ` `            ``MinHeapNode node = ``new` `MinHeapNode(arr[i, 0], i, 1); ` `            ``hArr[i] = node; ` `            ``resultSize += arr.GetLength(1); ` `        ``} ` ` `  `        ``// Create a min heap with k heap nodes.  ` `        ``// Every heap node has first element of an array ` `        ``MinHeap mh = ``new` `MinHeap(hArr, k); ` ` `  `        ``int``[] result = ``new` `int``[resultSize];     ``// To store output array ` ` `  `        ``// Now one by one get the minimum element  ` `        ``// from min heap and replace it with  ` `        ``// next element of its array ` `        ``for``(``int` `i = 0; i < resultSize; i++)  ` `        ``{ ` ` `  `            ``// Get the minimum element and ` `            ``// store it in result ` `            ``MinHeapNode root = mh.getMin(); ` `            ``result[i] = root.element; ` ` `  `            ``// Find the next element that will  ` `            ``// replace current root of heap.  ` `            ``// The next element belongs to same ` `            ``// array as the current root. ` `            ``if``(root.j < arr.GetLength(1)) ` `                ``root.element = arr[root.i,root.j++]; ` `                 `  `            ``// If root was the last element of its array ` `            ``else` `                ``root.element = ``int``.MaxValue; ` ` `  `            ``// Replace root with next element of array  ` `            ``mh.replaceMin(root); ` `        ``} ` `        ``printArray(result); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String []args) ` `    ``{ ` `        ``int``[,] arr = {{2, 6, 12, 34}, ` `                      ``{1, 9, 20, 1000}, ` `                      ``{23, 34, 90, 2000}}; ` ` `  `        ``Console.WriteLine(``"Merged array is :"``); ` ` `  `        ``mergeKSortedArrays(arr, arr.GetLength(0)); ` `    ``} ` `}; ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```Merged array is
1 2 6 9 12 20 23 34 34 90 1000 2000
```
• Complexity Analysis:

• Time Complexity :O( n * k * log k), Insertion and deletion in a Min Heap requires log k time. So the Overall time compelxity is O( n * k * log k)
• Space Complexity :O(k), If Output is not stored then the only space required is the Min-Heap of k elements. So space Comeplexity is O(k).

Merge k sorted arrays | Set 2 (Different Sized Arrays)

Thanks to vignesh for suggesting this problem and initial solution. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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