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(nLogn) sorting algorithm. This approach takes O(nkLognk) time.

We can merge arrays in O(nk*Logk) time using Min Heap. 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 a 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<iostream> #include<limits.h> 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[0]; }       // to replace root with new node x and heapify() new root     void replaceMin(MinHeapNode x) { harr[0] = 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][0]; // 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[0]);       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).