Merge k sorted arrays

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 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.

code link




// 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

};




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;

}

// 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--;

    }

}




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;

}

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).