# Find m-th smallest value in k sorted arrays

• Difficulty Level : Hard
• Last Updated : 23 Mar, 2023

Given k sorted arrays of possibly different sizes, find m-th smallest value in the merged array.
Examples:

```Input: m = 5
arr[][] = { {1, 3},
{2, 4, 6},
{0, 9, 10, 11}} ;
Output: 4
Explanation The merged array would
be {0 1 2 3 4 6 9 10 11}.  The 5-th
smallest element in this merged
array is 4.

Input: m = 2
arr[][] = { {1, 3, 20},
{2, 4, 6}} ;
Explanation The merged array would
be {1 2 3 4 6 20}. The 2nd smallest element would be 2.
Output: 2```

A simple solution is to create an output array 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(m Log k).
1. Create a min heap of size k and insert 1st element in all the arrays into the heap
2. Repeat following steps m times
…..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 doesnâ€™t have any more elements, then do nothing.
3. Print the last removed item.

## CPP

 `// C++ program to find m-th smallest element``// in the merged arrays.``#include ``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 returns m-th smallest element in``// the array obtained after merging the given``// arrays.``int` `mThLargest(vector > arr, ``int` `m)``{``    ``// Create a min heap with k heap nodes. Every``    ``// heap node has first element of an array``    ``priority_queue, greater > 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``    ``int` `count = 0;``    ``int` `i = 0, j = 0;``    ``while` `(count < m && pq.empty() == ``false``) {``        ``ppi curr = pq.top();``        ``pq.pop();` `        ``// i ==> Array Number``        ``// j ==> Index in the array number``        ``i = curr.second.first;``        ``j = curr.second.second;` `        ``// The next element belongs to same array as``        ``// current.``        ``if` `(j + 1 < arr[i].size())``            ``pq.push({ arr[i][j + 1], { i, j + 1 } });` `        ``count++;``    ``}` `    ``return` `arr[i][j];``}` `// Driver program to test above functions``int` `main()``{``    ``vector > arr{ { 2, 6, 12 },``                              ``{ 1, 9 },``                              ``{ 23, 34, 90, 2000 } };` `    ``int` `m = 4;``    ``cout << mThLargest(arr, m);` `    ``return` `0;``}`

## Java

 `// This function takes an array of arrays as an``//  argument and all arrays are assumed to be``//  sorted. It returns m-th smallest element in``//  the array obtained after merging the given``//  arrays.``import` `java.util.*;``public` `class` `Main {``    ``public` `static` `int` `mThLargest(``int``[][] arr, ``int` `m) {``  ` `        ``// Create a min heap. Every``        ``// heap node has first element of an array``        ``PriorityQueue pq = ``new` `PriorityQueue();``        ``for` `(``int` `i = ``0``; i < arr.length; i++) {``            ``pq.add(``new` `Pair(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``        ``int` `count = ``0``;``        ``int` `i=``0``;``        ``int` `j=``0``;``        ``while` `(count < m && !pq.isEmpty()) {``            ``Pair curr = pq.poll();``  ` `            ``// i ==> Array Number``            ``// j ==> Index in the array number``            ``i = curr.arrayNumber;``            ``j = curr.index;``  ` `            ``// The next element belongs to same array as current.``            ``if` `(j + ``1` `< arr[i].length) {``                ``pq.add(``new` `Pair(arr[i][j + ``1``], i, j + ``1``));``            ``}``            ``count++;``        ``}``        ``return` `arr[i][j];``    ``}``  ` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args) {``        ``int``[][] arr = { { ``2``, ``6``, ``12` `}, { ``1``, ``9` `}, { ``23``, ``34``, ``90``, ``2000` `} };``        ``int` `m = ``4``;``        ``System.out.println(mThLargest(arr, m));``    ``}``}` `//Class to store array number and index``class` `Pair ``implements` `Comparable {``    ``int` `value;``    ``int` `arrayNumber;``    ``int` `index;``  ` `    ``// Constructor``    ``public` `Pair(``int` `v, ``int` `i, ``int` `j) {``        ``value = v;``        ``arrayNumber = i;``        ``index = j;``    ``}``  ` `    ``// Compare two pair according to their values.``    ``public` `int` `compareTo(Pair o) {``        ``return` `this``.value - o.value;``    ``}``}`

## Python3

 `# Python program to find m-th smallest element``# in the merged arrays.``from` `heapq ``import` `*` `#  This function takes an array of arrays as an``#  argument and all arrays are assumed to be``#  sorted. It returns m-th smallest element in``#  the array obtained after merging the given``#  arrays.``def` `mThLargest(arr, m):` `    ``#  Create a min heap. Every``    ``#  heap node has first element of an array``    ``pq ``=` `[]``    ``for` `i ``in` `range``(``len``(arr)):``        ``heappush(pq, (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``    ``count ``=` `0``    ``while` `count < m ``and` `pq:``        ``curr ``=` `heappop(pq)` `        ``#  i ==> Array Number``        ``#  j ==> Index in the array number``        ``i ``=` `curr[``1``][``0``]``        ``j ``=` `curr[``1``][``1``]` `        ``# The next element belongs to same array as current.``        ``if` `j ``+` `1` `< ``len``(arr[i]):``            ``heappush(pq, (arr[i][j ``+` `1``], (i, j ``+` `1``)))``        ``count ``+``=` `1` `    ``return` `arr[i][j]` `# Driver Code``arr ``=` `[[``2``, ``6``, ``12``], [``1``, ``9``], [``23``, ``34``, ``90``, ``2000``]]``m ``=` `4``print``(mThLargest(arr, m))` `# This code is contributed by vivekmaddheshiya205`

Output:

`9`

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