# 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 ```

A simple solution is to create an output array of size n*k and one by one copy all arrays to it. Finally, sort the output array using any O(n Log n) sorting algorithm. This approach takes O(nk Log nk) time.

One efficient solution is to first merge arrays into groups of 2. After first merging, we have k/2 arrays. We again merge arrays in groups, now we have k/4 arrays. We keep doing it unit we have one array left. The time complexity of this solution would O(nk Log k). How? Every merging in first iteration would take 2n time (merging two arrays of size n). Since there are total k/2 merging, total time in first iteration would be O(nk). Next iteration would also take O(nk). There will be total O(Log k) iterations, hence time complexity is O(nk Log k)

Another efficient solution is to use Min Heap. This Min Heap based solution has same time complexity which is O(nk Log k). But for different sized arrays, this solution works much better.

Following is detailed algorithm.
1. Create an output array of size n*k.
2. Create a min heap of size k and insert 1st element in all the arrays into the heap
3. Repeat following steps n*k times.
a) Get minimum element from heap (minimum is always at root) and store it in output array.
b) Replace heap root with next element from the array from which the element is extracted. If the array doesn’t have any more elements, then replace root with infinite. After replacing the root, heapify the tree.

Following is C++ implementation of the above algorithm.

 `// C++ program to merge k sorted arrays of size n each. ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `#define n 4 ` ` `  `// A min heap node ` `struct` `MinHeapNode ` `{ ` `    ``int` `element; ``// The element to be stored ` `    ``int` `i; ``// index of the array from which the element is taken ` `    ``int` `j; ``// index of the next element to be picked from array ` `}; ` ` `  `// Prototype of a utility function to swap two min heap nodes ` `void` `swap(MinHeapNode *x, MinHeapNode *y); ` ` `  `// A class for Min Heap ` `class` `MinHeap ` `{ ` `    ``MinHeapNode *harr; ``// pointer to array of elements in heap ` `    ``int` `heap_size; ``// size of min heap ` `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) ` `{ ` `    ``int` `*output = ``new` `int``[n*k];  ``// To store output array ` ` `  `    ``// 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++) ` `    ``{ ` `        ``harr[i].element = arr[i]; ``// Store the first element ` `        ``harr[i].i = i;  ``// index of array ` `        ``harr[i].j = 1;  ``// Index of next element to be stored from array ` `    ``} ` `    ``MinHeap hp(harr, k); ``// Create the heap ` ` `  `    ``// 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 ` `        ``else` `root.element =  INT_MAX; ``//INT_MAX is for infinite ` ` `  `        ``// 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; ` `} `

Output:

```Merged array is
1 2 6 9 12 20 23 34 34 90 1000 2000```

Time Complexity: The main step is 3rd step, the loop runs n*k times. In every iteration of loop, we call heapify which takes O(Logk) time. Therefore, the time complexity is O(nk Logk).

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

This article is attributed to GeeksforGeeks.org

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