Given k sorted arrays of possibly different sizes, merge them and print the sorted output.

Examples:

Input: k = 3 arr[][] = { {1, 3}, {2, 4, 6}, {0, 9, 10, 11}} ; Output: 0 1 2 3 4 6 9 10 11 Input: k = 2 arr[][] = { {1, 3, 20}, {2, 4, 6}} ; Output: 1 2 3 4 6 20

We have discussed a solution that works for all arrays of same size in Merge k sorted arrays | Set 1.

A **simple solution** is to create an output array and and one by one copy all arrays to it. Finally, sort the output array using. This approach takes O(N Logn N) time where N is count of all elements.

An **efficient solution** is to use heap data structure. The time complexity of heap based solution is O(N Log k).

1. Create an output array.

2. Create a min heap of size k and insert 1st element in all the arrays into the heap

3. Repeat following steps while priority queue is not empty.

…..a) Remove minimum element from heap (minimum is always at root) and store it in output array.

…..b) Insert next element from the array from which the element is extracted. If the array doesn’t have any more elements, then do nothing.

// C++ program to merge k sorted arrays // of size n each. #include <bits/stdc++.h> using namespace std; // A pair of pairs, first element is going to // store value, second element index of array // and third element index in the array. typedef pair<int, pair<int, int> > ppi; // 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. vector<int> mergeKArrays(vector<vector<int> > arr) { vector<int> output; // Create a min heap with k heap nodes. Every // heap node has first element of an array priority_queue<ppi, vector<ppi>, greater<ppi> > pq; for (int i = 0; i < arr.size(); i++) pq.push({ arr[i][0], { i, 0 } }); // Now one by one get the minimum element // from min heap and replace it with next // element of its array while (pq.empty() == false) { ppi curr = pq.top(); pq.pop(); // i ==> Array Number // j ==> Index in the array number int i = curr.second.first; int j = curr.second.second; output.push_back(curr.first); // The next element belongs to same array as // current. if (j + 1 < arr[i].size()) pq.push({ arr[i][j + 1], { i, j + 1 } }); } return output; } // Driver program to test above functions int main() { // Change n at the top to change number // of elements in an array vector<vector<int> > arr{ { 2, 6, 12 }, { 1, 9 }, { 23, 34, 90, 2000 } }; vector<int> output = mergeKArrays(arr); cout << "Merged array is " << endl; for (auto x : output) cout << x << " "; return 0; }

**Output:**

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

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.