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

## Recommended Posts:

- Merge k sorted linked lists | Set 2 (Using Min Heap)
- Merge K sorted linked lists | Set 1
- K'th largest element in a stream
- K'th Smallest/Largest Element in Unsorted Array | Set 2 (Expected Linear Time)
- K'th Smallest/Largest Element in Unsorted Array | Set 1
- Binary Heap
- Merge k sorted arrays | Set 1
- HeapSort
- Sort a nearly sorted (or K sorted) array
- Median in a stream of integers (running integers)
- Sliding Window Maximum (Maximum of all subarrays of size k)
- Next Greater Element
- Applications of Heap Data Structure
- k largest(or smallest) elements in an array | added Min Heap method
- Median of two sorted arrays of same size

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 Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.