Given **k** sorted arrays of different length, merge them into a single array such that the merged array is also sorted.

**Examples:**

Input :{{3, 13}, {8, 10, 11} {9, 15}}Output :{3, 8, 9, 10, 11, 13, 15}Input :{{1, 5}, {2, 3, 4}}Output :{1, 2, 3, 4, 5}

Let S be the total number of elements in all the arrays.

**Simple Approach**: A simple approach is to append the arrays one after another and sort them. Time complexity in this case will be **O( S * log(S))**.

**Efficient Approach**: An efficient solution will be to take pairs of arrays at each step. Then merge the pairs using the two pointer technique of merging two sorted arrays. Thus, after merging all the pairs, the number of arrays will reduce by half.

We, will continue this till the number of remaining arrays doesn’t become 1. Thus, the number of steps required will be of the order log(k) and since at each step, we are taking O(S) time to perform the merge operations, the total time complexity of this approach becomes **O(S * log(k))**.

We already have discussed this approach for merging K sorted arrays of same sizes.

Since in this problem the arrays are of different sizes, we will use dynamic arrays ( eg: vector in case of C++ or arraylist in case of Java ) because they reduce the number of lines and work load significantly.

Below is the implementation of the above approach:

`// C++ program to merge K sorted arrays of ` `// different arrays ` ` ` `#include <iostream> ` `#include <vector> ` `using` `namespace` `std; ` ` ` `// Function to merge two arrays ` `vector<` `int` `> mergeTwoArrays(vector<` `int` `> l, vector<` `int` `> r) ` `{ ` ` ` `// array to store the result ` ` ` `// after merging l and r ` ` ` `vector<` `int` `> ret; ` ` ` ` ` `// variables to store the current ` ` ` `// pointers for l and r ` ` ` `int` `l_in = 0, r_in = 0; ` ` ` ` ` `// loop to merge l and r using two pointer ` ` ` `while` `(l_in + r_in < l.size() + r.size()) { ` ` ` `if` `(l_in != l.size() && (r_in == r.size() || l[l_in] < r[r_in])) { ` ` ` `ret.push_back(l[l_in]); ` ` ` `l_in++; ` ` ` `} ` ` ` `else` `{ ` ` ` `ret.push_back(r[r_in]); ` ` ` `r_in++; ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `ret; ` `} ` ` ` `// Function to merge all the arrays ` `vector<` `int` `> mergeArrays(vector<vector<` `int` `> > arr) ` `{ ` ` ` `// 2D-array to store the results of ` ` ` `// a step temporarily ` ` ` `vector<vector<` `int` `> > arr_s; ` ` ` ` ` `// Loop to make pairs of arrays and merge them ` ` ` `while` `(arr.size() != 1) { ` ` ` ` ` `// To clear the data of previous steps ` ` ` `arr_s.clear(); ` ` ` ` ` `for` `(` `int` `i = 0; i < arr.size(); i += 2) { ` ` ` `if` `(i == arr.size() - 1) ` ` ` `arr_s.push_back(arr[i]); ` ` ` ` ` `else` ` ` `arr_s.push_back(mergeTwoArrays(arr[i], ` ` ` `arr[i + 1])); ` ` ` `} ` ` ` ` ` `arr = arr_s; ` ` ` `} ` ` ` ` ` `// Returning the required output array ` ` ` `return` `arr[0]; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `// Input arrays ` ` ` `vector<vector<` `int` `> > arr{ { 3, 13 }, ` ` ` `{ 8, 10, 11 }, ` ` ` `{ 9, 15 } }; ` ` ` `// Merged sorted array ` ` ` `vector<` `int` `> output = mergeArrays(arr); ` ` ` ` ` `for` `(` `int` `i = 0; i < output.size(); i++) ` ` ` `cout << output[i] << ` `" "` `; ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

3 8 9 10 11 13 15

Note that there exist a better solution using heap (or priority queue). The time complexity of the heap based solution is O(N Log k) where N is total number of elements in all K arrays.

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