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Find smallest range containing elements from k lists
• Difficulty Level : Hard
• Last Updated : 12 Jun, 2021

Given k sorted lists of integers of size n each, find the smallest range that includes at least element from each of the k lists. If more than one smallest ranges are found, print any one of them.

Example:

```Input: K = 3
arr1[] : [4, 7, 9, 12, 15]
arr2[] : [0, 8, 10, 14, 20]
arr3[] : [6, 12, 16, 30, 50]
Output:
The smallest range is [6 8]

Explanation: Smallest range is formed by
number 7 from the first list, 8 from second
list and 6 from the third list.

Input: k = 3
arr1[] : [4, 7]
arr2[] : [1, 2]
arr3[] : [20, 40]
Output:
The smallest range is [2 20]

Explanation:The range [2, 20] contains 2, 4, 7, 20
which contains element from all the three arrays.```

Naive Approach: Given K sorted list, find a range where there is at least one element from every list. The idea to solve the problem is very simple, keep k pointers which will constitute the elements in the range, by taking the min and max of the k elements the range can be formed. Initially, all the pointers will point to the start of all the k arrays. Store the range max to min. If the range has to be minimised then either the minimum value has to be increased or maximum value has to be decreased. The maximum value cannot be decreased as the array is sorted but the minimum value can be increased. To continue increasing the minimum value, increase the pointer of the list containing the minimum value and update the range until one of the lists exhausts.

• Algorithm:
1. Create an extra space ptr of length k to store the pointers and a variable minrange initialized to a maximum value.
2. Initially the index of every list is 0, therefore initialize every element of ptr[0..k] to 0, the array ptr will store the index of the elements in the range.
3. Repeat the following steps until atleast one list exhausts:
1. Now find the minimum and maximum value among the current elements of all the list pointed by the ptr[0…k] array.
2. Now update the minrange if current (max-min) is less than minrange.
3. increment the pointer pointing to current minimum element.
• Implementation:

## C++

 `// C++ program to finds out smallest range that includes``// elements from each of the given sorted lists.``#include ` `using` `namespace` `std;` `#define N 5` `// array for storing the current index of list i``int` `ptr;` `// This function takes an k sorted lists in the form of``// 2D array as an argument. It finds out smallest range``// that includes elements from each of the k lists.``void` `findSmallestRange(``int` `arr[][N], ``int` `n, ``int` `k)``{``    ``int` `i, minval, maxval, minrange, minel, maxel, flag, minind;` `    ``// initializing to 0 index;``    ``for` `(i = 0; i <= k; i++)``        ``ptr[i] = 0;` `    ``minrange = INT_MAX;` `    ``while` `(1) {``        ``// for maintaining the index of list containing the minimum element``        ``minind = -1;``        ``minval = INT_MAX;``        ``maxval = INT_MIN;``        ``flag = 0;` `        ``// iterating over all the list``        ``for` `(i = 0; i < k; i++) {``            ``// if every element of list[i] is traversed then break the loop``            ``if` `(ptr[i] == n) {``                ``flag = 1;``                ``break``;``            ``}``            ``// find minimum value among all the list elements pointing by the ptr[] array``            ``if` `(ptr[i] < n && arr[i][ptr[i]] < minval) {``                ``minind = i; ``// update the index of the list``                ``minval = arr[i][ptr[i]];``            ``}``            ``// find maximum value among all the list elements pointing by the ptr[] array``            ``if` `(ptr[i] < n && arr[i][ptr[i]] > maxval) {``                ``maxval = arr[i][ptr[i]];``            ``}``        ``}` `        ``// if any list exhaust we will not get any better answer, so break the while loop``        ``if` `(flag)``            ``break``;` `        ``ptr[minind]++;` `        ``// updating the minrange``        ``if` `((maxval - minval) < minrange) {``            ``minel = minval;``            ``maxel = maxval;``            ``minrange = maxel - minel;``        ``}``    ``}` `    ``printf``(``"The smallest range is [%d, %d]\n"``, minel, maxel);``}` `// Driver program to test above function``int` `main()``{``    ``int` `arr[][N] = {``        ``{ 4, 7, 9, 12, 15 },``        ``{ 0, 8, 10, 14, 20 },``        ``{ 6, 12, 16, 30, 50 }``    ``};` `    ``int` `k = ``sizeof``(arr) / ``sizeof``(arr);` `    ``findSmallestRange(arr, N, k);` `    ``return` `0;``}``// This code is contributed by Aditya Krishna Namdeo`

## Java

 `// Java program to finds out smallest range that includes``// elements from each of the given sorted lists.``class` `GFG {` `    ``static` `final` `int` `N = ``5``;` `    ``// array for storing the current index of list i``    ``static` `int` `ptr[] = ``new` `int``[``501``];` `    ``// This function takes an k sorted lists in the form of``    ``// 2D array as an argument. It finds out smallest range``    ``// that includes elements from each of the k lists.``    ``static` `void` `findSmallestRange(``int` `arr[][], ``int` `n, ``int` `k)``    ``{``        ``int` `i, minval, maxval, minrange, minel = ``0``, maxel = ``0``, flag, minind;` `        ``// initializing to 0 index;``        ``for` `(i = ``0``; i <= k; i++) {``            ``ptr[i] = ``0``;``        ``}` `        ``minrange = Integer.MAX_VALUE;` `        ``while` `(``true``) {``            ``// for maintaining the index of list containing the minimum element``            ``minind = -``1``;``            ``minval = Integer.MAX_VALUE;``            ``maxval = Integer.MIN_VALUE;``            ``flag = ``0``;` `            ``// iterating over all the list``            ``for` `(i = ``0``; i < k; i++) {``                ``// if every element of list[i] is traversed then break the loop``                ``if` `(ptr[i] == n) {``                    ``flag = ``1``;``                    ``break``;``                ``}``                ``// find minimum value among all the list elements pointing by the ptr[] array``                ``if` `(ptr[i] < n && arr[i][ptr[i]] < minval) {``                    ``minind = i; ``// update the index of the list``                    ``minval = arr[i][ptr[i]];``                ``}``                ``// find maximum value among all the list elements pointing by the ptr[] array``                ``if` `(ptr[i] < n && arr[i][ptr[i]] > maxval) {``                    ``maxval = arr[i][ptr[i]];``                ``}``            ``}` `            ``// if any list exhaust we will not get any better answer, so break the while loop``            ``if` `(flag == ``1``) {``                ``break``;``            ``}` `            ``ptr[minind]++;` `            ``// updating the minrange``            ``if` `((maxval - minval) < minrange) {``                ``minel = minval;``                ``maxel = maxval;``                ``minrange = maxel - minel;``            ``}``        ``}``        ``System.out.printf(``"The smallest range is [%d, %d]\n"``, minel, maxel);``    ``}` `    ``// Driver program to test above function``    ``public` `static` `void` `main(String[] args)``    ``{` `        ``int` `arr[][] = {``            ``{ ``4``, ``7``, ``9``, ``12``, ``15` `},``            ``{ ``0``, ``8``, ``10``, ``14``, ``20` `},``            ``{ ``6``, ``12``, ``16``, ``30``, ``50` `}``        ``};` `        ``int` `k = arr.length;` `        ``findSmallestRange(arr, N, k);``    ``}``}``// this code contributed by Rajput-Ji`

## Python

 `# Python3 program to finds out``# smallest range that includes``# elements from each of the``# given sorted lists.` `N ``=` `5` `# array for storing the``# current index of list i``ptr ``=` `[``0` `for` `i ``in` `range``(``501``)]` `# This function takes an k sorted``# lists in the form of 2D array as``# an argument. It finds out smallest``# range that includes elements from``# each of the k lists.``def` `findSmallestRange(arr, n, k):` `    ``i, minval, maxval, minrange, minel, maxel, flag, minind ``=` `0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``, ``0``        ` `    ``# initializing to 0 index``    ``for` `i ``in` `range``(k ``+` `1``):``        ``ptr[i] ``=` `0` `    ``minrange ``=` `10``*``*``9``        ` `    ``while``(``1``):   ``        ` `            ``# for maintaining the index of list``            ``# containing the minimum element``        ``minind ``=` `-``1``        ``minval ``=` `10``*``*``9``        ``maxval ``=` `-``10``*``*``9``        ``flag ``=` `0` `        ``# iterating over all the list``        ``for` `i ``in` `range``(k):``            ` `                ``# if every element of list[i] is``                ``# traversed then break the loop``            ``if``(ptr[i] ``=``=` `n):``                ``flag ``=` `1`   `                ``break` `            ``# find minimum value among all the list``            ``# elements pointing by the ptr[] array``            ``if``(ptr[i] < n ``and` `arr[i][ptr[i]] < minval):``                ``minind ``=` `i ``# update the index of the list``                ``minval ``=` `arr[i][ptr[i]]``            ` `            ``# find maximum value among all the``            ``# list elements pointing by the ptr[] array``            ``if``(ptr[i] < n ``and` `arr[i][ptr[i]] > maxval):``                ``maxval ``=` `arr[i][ptr[i]]``            ` `        `  `        ``# if any list exhaust we will``        ``# not get any better answer,``        ``# so break the while loop``        ``if``(flag):``            ``break` `        ``ptr[minind] ``+``=` `1` `        ``# updating the minrange``        ``if``((maxval``-``minval) < minrange):``            ``minel ``=` `minval``            ``maxel ``=` `maxval``            ``minrange ``=` `maxel ``-` `minel``    ` `    ``print``(``"The smallest range is ["``, minel, maxel, ``"]"``)` `# Driver code``arr ``=` `[``    ``[``4``, ``7``, ``9``, ``12``, ``15``],``    ``[``0``, ``8``, ``10``, ``14``, ``20``],``    ``[``6``, ``12``, ``16``, ``30``, ``50``]``    ``]` `k ``=` `len``(arr)` `findSmallestRange(arr, N, k)` `# This code is contributed by mohit kumar`

## C#

 `// C# program to finds out smallest``// range that includes elements from``// each of the given sorted lists.``using` `System;` `class` `GFG {` `    ``static` `int` `N = 5;` `    ``// array for storing the current index of list i``    ``static` `int``[] ptr = ``new` `int``;` `    ``// This function takes an k sorted``    ``// lists in the form of 2D array as``    ``// an argument. It finds out smallest range``    ``// that includes elements from each of the k lists.``    ``static` `void` `findSmallestRange(``int``[, ] arr,``                                  ``int` `n, ``int` `k)``    ``{``        ``int` `i, minval, maxval, minrange,``            ``minel = 0, maxel = 0, flag, minind;` `        ``// initializing to 0 index;``        ``for` `(i = 0; i <= k; i++) {``            ``ptr[i] = 0;``        ``}` `        ``minrange = ``int``.MaxValue;` `        ``while` `(``true``) {``            ``// for maintining the index of``            ``// list containing the minimum element``            ``minind = -1;``            ``minval = ``int``.MaxValue;``            ``maxval = ``int``.MinValue;``            ``flag = 0;` `            ``// iterating over all the list``            ``for` `(i = 0; i < k; i++) {``                ``// if every element of list[i]``                ``// is traversed then break the loop``                ``if` `(ptr[i] == n) {``                    ``flag = 1;``                    ``break``;``                ``}` `                ``// find minimum value among all the``                ``// list elements pointing by the ptr[] array``                ``if` `(ptr[i] < n && arr[i, ptr[i]] < minval) {``                    ``minind = i; ``// update the index of the list``                    ``minval = arr[i, ptr[i]];``                ``}` `                ``// find maximum value among all the``                ``// list elements pointing by the ptr[] array``                ``if` `(ptr[i] < n && arr[i, ptr[i]] > maxval) {``                    ``maxval = arr[i, ptr[i]];``                ``}``            ``}` `            ``// if any list exhaust we will``            ``// not get any better answer,``            ``// so break the while loop``            ``if` `(flag == 1) {``                ``break``;``            ``}` `            ``ptr[minind]++;` `            ``// updating the minrange``            ``if` `((maxval - minval) < minrange) {``                ``minel = minval;``                ``maxel = maxval;``                ``minrange = maxel - minel;``            ``}``        ``}``        ``Console.WriteLine(``"The smallest range is"``                              ``+ ``"[{0}, {1}]\n"``,``                          ``minel, maxel);``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{` `        ``int``[, ] arr = {``            ``{ 4, 7, 9, 12, 15 },``            ``{ 0, 8, 10, 14, 20 },``            ``{ 6, 12, 16, 30, 50 }``        ``};` `        ``int` `k = arr.GetLength(0);` `        ``findSmallestRange(arr, N, k);``    ``}``}` `// This code has been contributed by 29AjayKumar`

## Javascript

 ``

Output:

`The smallest range is [6 8]`
• Complexity Analysis:
• Time complexity : O(n * k2), to find the maximum and minimum in an array of length k the time required is O(k), and to traverse all the k arrays of length n (in worst case), the time complexity is O(n*k), then the total time complexity is O(n*k2).
• Space complexity: O(k), an extra array is required of length k so the space complexity is O(k)

Efficient approach: The approach remains the same but the time complexity can be reduced by using min-heap or priority queue. Min heap can be used to find the maximum and minimum value in logarithmic time or log k time instead of linear time. Rest of the approach remains the same.

• Algorithm:
1. create an Min heap to store k elements, one from each array and a variable minrange initialized to a maximum value and also keep a variable max to store the maximum integer.
2. Initially put the first element of each element from each list and store the maximum value in max.
3. Repeat the following steps until atleast one list exhausts :
1. To find the minimum value or min, use the top or root of the Min heap which is the minimum element.
2. Now update the minrange if current (max-min) is less than minrange.
3. remove the top or root element from priority queue and insert the next element from the list which contains the min element and update the max with the new element inserted.
• Implementation:

## C++

 `// C++ program to finds out smallest range that includes``// elements from each of the given sorted lists.``#include ``using` `namespace` `std;` `#define N 5` `// A min heap node``struct` `MinHeapNode {``    ``// The element to be stored``    ``int` `element;` `    ``// index of the list from which the element is taken``    ``int` `i;` `    ``// index of the next element to be picked from list``    ``int` `j;``};` `// Prototype of a utility function to swap two min heap nodes``void` `swap(MinHeapNode* x, MinHeapNode* y);` `// A class for Min Heap``class` `MinHeap {` `    ``// pointer to array of elements in heap``    ``MinHeapNode* harr;` `    ``// size of min heap``    ``int` `heap_size;` `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; }` `    ``// to replace root with new node x and heapify() new root``    ``void` `replaceMin(MinHeapNode x)``    ``{``        ``harr = x;``        ``MinHeapify(0);``    ``}``};` `// 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--;``    ``}``}` `// A recursive method to heapify a subtree with root at``// given index. This method assumes that the subtrees``// are already heapified``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);``    ``}``}` `// This function takes an k sorted lists in the form of``// 2D array as an argument. It finds out smallest range``// that includes elements from each of the k lists.``void` `findSmallestRange(``int` `arr[][N], ``int` `k)``{``    ``// Create a min heap with k heap nodes. Every heap node``    ``// has first element of an list``    ``int` `range = INT_MAX;``    ``int` `min = INT_MAX, max = INT_MIN;``    ``int` `start, end;` `    ``MinHeapNode* harr = ``new` `MinHeapNode[k];``    ``for` `(``int` `i = 0; i < k; i++) {``        ``// Store the first element``        ``harr[i].element = arr[i];` `        ``// index of list``        ``harr[i].i = i;` `        ``// Index of next element to be stored``        ``// from list``        ``harr[i].j = 1;` `        ``// store max element``        ``if` `(harr[i].element > max)``            ``max = harr[i].element;``    ``}` `    ``// Create the heap``    ``MinHeap hp(harr, k);` `    ``// Now one by one get the minimum element from min``    ``// heap and replace it with next element of its list``    ``while` `(1) {``        ``// Get the minimum element and store it in output``        ``MinHeapNode root = hp.getMin();` `        ``// update min``        ``min = hp.getMin().element;` `        ``// update range``        ``if` `(range > max - min + 1) {``            ``range = max - min + 1;``            ``start = min;``            ``end = max;``        ``}` `        ``// Find the next element that will replace current``        ``// root of heap. The next element belongs to same``        ``// list as the current root.``        ``if` `(root.j < N) {``            ``root.element = arr[root.i][root.j];``            ``root.j += 1;` `            ``// update max element``            ``if` `(root.element > max)``                ``max = root.element;``        ``}` `        ``// break if we have reached end of any list``        ``else``            ``break``;` `        ``// Replace root with next element of list``        ``hp.replaceMin(root);``    ``}` `    ``cout << ``"The smallest range is "``         ``<< ``"["``         ``<< start << ``" "` `<< end << ``"]"` `<< endl;``    ``;``}` `// Driver program to test above functions``int` `main()``{``    ``int` `arr[][N] = {``        ``{ 4, 7, 9, 12, 15 },``        ``{ 0, 8, 10, 14, 20 },``        ``{ 6, 12, 16, 30, 50 }``    ``};``    ``int` `k = ``sizeof``(arr) / ``sizeof``(arr);` `    ``findSmallestRange(arr, k);` `    ``return` `0;``}`

## Java

 `// Java program to find out smallest``// range that includes elements from``// each of the given sorted lists.``class` `GFG {` `    ``// A min heap node``    ``static` `class` `Node {``        ``// The element to be stored``        ``int` `ele;` `        ``// index of the list from which``        ``// the element is taken``        ``int` `i;` `        ``// index of the next element``        ``// to be picked from list``        ``int` `j;` `        ``Node(``int` `a, ``int` `b, ``int` `c)``        ``{``            ``this``.ele = a;``            ``this``.i = b;``            ``this``.j = c;``        ``}``    ``}` `    ``// A class for Min Heap``    ``static` `class` `MinHeap {``        ``Node[] harr; ``// array of elements in heap``        ``int` `size; ``// size of min heap` `        ``// Constructor: creates a min heap``        ``// of given size``        ``MinHeap(Node[] arr, ``int` `size)``        ``{``            ``this``.harr = arr;``            ``this``.size = size;``            ``int` `i = (size - ``1``) / ``2``;``            ``while` `(i >= ``0``) {``                ``MinHeapify(i);``                ``i--;``            ``}``        ``}` `        ``// 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 heapify a subtree with``        ``// root at given index``        ``void` `MinHeapify(``int` `i)``        ``{``            ``int` `l = left(i);``            ``int` `r = right(i);``            ``int` `small = i;``            ``if` `(l < size && harr[l].ele < harr[i].ele)``                ``small = l;``            ``if` `(r < size && harr[r].ele < harr[small].ele)``                ``small = r;``            ``if` `(small != i) {``                ``swap(small, i);``                ``MinHeapify(small);``            ``}``        ``}` `        ``void` `swap(``int` `i, ``int` `j)``        ``{``            ``Node temp = harr[i];``            ``harr[i] = harr[j];``            ``harr[j] = temp;``        ``}` `        ``// to get the root``        ``Node getMin()``        ``{``            ``return` `harr[``0``];``        ``}` `        ``// to replace root with new node x``        ``// and heapify() new root``        ``void` `replaceMin(Node x)``        ``{``            ``harr[``0``] = x;``            ``MinHeapify(``0``);``        ``}``    ``}` `    ``// This function takes an k sorted lists``    ``// in the form of 2D array as an argument.``    ``// It finds out smallest range that includes``    ``// elements from each of the k lists.``    ``static` `void` `findSmallestRange(``int``[][] arr, ``int` `k)``    ``{``        ``int` `range = Integer.MAX_VALUE;``        ``int` `min = Integer.MAX_VALUE;``        ``int` `max = Integer.MIN_VALUE;``        ``int` `start = -``1``, end = -``1``;` `        ``int` `n = arr[``0``].length;` `        ``// Create a min heap with k heap nodes.``        ``// Every heap node has first element of an list``        ``Node[] arr1 = ``new` `Node[k];``        ``for` `(``int` `i = ``0``; i < k; i++) {``            ``Node node = ``new` `Node(arr[i][``0``], i, ``1``);``            ``arr1[i] = node;` `            ``// store max element``            ``max = Math.max(max, node.ele);``        ``}` `        ``// Create the heap``        ``MinHeap mh = ``new` `MinHeap(arr1, k);` `        ``// Now one by one get the minimum element``        ``// from min heap and replace it with``        ``// next element of its list``        ``while` `(``true``) {``            ``// Get the minimum element and``            ``// store it in output``            ``Node root = mh.getMin();` `            ``// update min``            ``min = root.ele;` `            ``// update range``            ``if` `(range > max - min + ``1``) {``                ``range = max - min + ``1``;``                ``start = min;``                ``end = max;``            ``}` `            ``// Find the next element that will``            ``// replace current root of heap.``            ``// The next element belongs to same``            ``// list as the current root.``            ``if` `(root.j < n) {``                ``root.ele = arr[root.i][root.j];``                ``root.j++;` `                ``// update max element``                ``if` `(root.ele > max)``                    ``max = root.ele;``            ``}``            ``// break if we have reached``            ``// end of any list``            ``else``                ``break``;` `            ``// Replace root with next element of list``            ``mh.replaceMin(root);``        ``}``        ``System.out.print(``"The smallest range is ["` `+ start + ``" "` `+ end + ``"]"``);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[][] = { { ``4``, ``7``, ``9``, ``12``, ``15` `},``                        ``{ ``0``, ``8``, ``10``, ``14``, ``20` `},``                        ``{ ``6``, ``12``, ``16``, ``30``, ``50` `} };` `        ``int` `k = arr.length;` `        ``findSmallestRange(arr, k);``    ``}``}` `// This code is contributed by nobody_cares`

## C#

 `// C# program to find out smallest``// range that includes elements from``// each of the given sorted lists.``using` `System;``using` `System.Collections.Generic;` `class` `GFG {` `    ``// A min heap node``    ``public` `class` `Node {``        ``// The element to be stored``        ``public` `int` `ele;` `        ``// index of the list from which``        ``// the element is taken``        ``public` `int` `i;` `        ``// index of the next element``        ``// to be picked from list``        ``public` `int` `j;` `        ``public` `Node(``int` `a, ``int` `b, ``int` `c)``        ``{``            ``this``.ele = a;``            ``this``.i = b;``            ``this``.j = c;``        ``}``    ``}` `    ``// A class for Min Heap``    ``public` `class` `MinHeap {``        ``// array of elements in heap``        ``public` `Node[] harr;` `        ``// size of min heap``        ``public` `int` `size;` `        ``// Constructor: creates a min heap``        ``// of given size``        ``public` `MinHeap(Node[] arr, ``int` `size)``        ``{``            ``this``.harr = arr;``            ``this``.size = size;``            ``int` `i = (size - 1) / 2;``            ``while` `(i >= 0) {``                ``MinHeapify(i);``                ``i--;``            ``}``        ``}` `        ``// 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 heapify a subtree with``        ``// root at given index``        ``void` `MinHeapify(``int` `i)``        ``{``            ``int` `l = left(i);``            ``int` `r = right(i);``            ``int` `small = i;``            ``if` `(l < size && harr[l].ele < harr[i].ele)``                ``small = l;``            ``if` `(r < size && harr[r].ele < harr[small].ele)``                ``small = r;``            ``if` `(small != i) {``                ``swap(small, i);``                ``MinHeapify(small);``            ``}``        ``}` `        ``void` `swap(``int` `i, ``int` `j)``        ``{``            ``Node temp = harr[i];``            ``harr[i] = harr[j];``            ``harr[j] = temp;``        ``}` `        ``// to get the root``        ``public` `Node getMin()``        ``{``            ``return` `harr;``        ``}` `        ``// to replace root with new node x``        ``// and heapify() new root``        ``public` `void` `replaceMin(Node x)``        ``{``            ``harr = x;``            ``MinHeapify(0);``        ``}``    ``}` `    ``// This function takes an k sorted lists``    ``// in the form of 2D array as an argument.``    ``// It finds out smallest range that includes``    ``// elements from each of the k lists.``    ``static` `void` `findSmallestRange(``int``[, ] arr, ``int` `k)``    ``{``        ``int` `range = ``int``.MaxValue;``        ``int` `min = ``int``.MaxValue;``        ``int` `max = ``int``.MinValue;``        ``int` `start = -1, end = -1;` `        ``int` `n = arr.GetLength(0);` `        ``// Create a min heap with k heap nodes.``        ``// Every heap node has first element of an list``        ``Node[] arr1 = ``new` `Node[k];``        ``for` `(``int` `i = 0; i < k; i++) {``            ``Node node = ``new` `Node(arr[i, 0], i, 1);``            ``arr1[i] = node;` `            ``// store max element``            ``max = Math.Max(max, node.ele);``        ``}` `        ``// Create the heap``        ``MinHeap mh = ``new` `MinHeap(arr1, k);` `        ``// Now one by one get the minimum element``        ``// from min heap and replace it with``        ``// next element of its list``        ``while` `(``true``) {``            ``// Get the minimum element and``            ``// store it in output``            ``Node root = mh.getMin();` `            ``// update min``            ``min = root.ele;` `            ``// update range``            ``if` `(range > max - min + 1) {``                ``range = max - min + 1;``                ``start = min;``                ``end = max;``            ``}` `            ``// Find the next element that will``            ``// replace current root of heap.``            ``// The next element belongs to same``            ``// list as the current root.``            ``if` `(root.j < n) {``                ``root.ele = arr[root.i, root.j];``                ``root.j++;` `                ``// update max element``                ``if` `(root.ele > max)``                    ``max = root.ele;``            ``}``            ``else``                ``break``; ``// break if we have reached``            ``// end of any list` `            ``// Replace root with next element of list``            ``mh.replaceMin(root);``        ``}``        ``Console.Write(``"The smallest range is ["` `+ start + ``" "` `+ end + ``"]"``);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[, ] arr = { { 4, 7, 9, 12, 15 },``                        ``{ 0, 8, 10, 14, 20 },``                        ``{ 6, 12, 16, 30, 50 } };` `        ``int` `k = arr.GetLength(0);` `        ``findSmallestRange(arr, k);``    ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output:

`The smallest range is [6 8]`
• Complexity Analysis:
• Time complexity : O(n * k *log k).
To find the maximum and minimum in an Min Heap of length k the time required is O(log k), and to traverse all the k arrays of length n (in worst case), the time complexity is O(n*k), then the total time complexity is O(n * k *log k).
• Space complexity: O(k).
The priority queue will store k elements so the space complexity os O(k)

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