Find smallest range containing elements from k lists

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.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 initilized 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 <bits/stdc++.h> 
  
using namespace std; 
  
#define N 5 
  
// array for storing the current index of list i 
int ptr[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. 
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 mainting 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[0]); 
  
    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 mainting 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 mainting 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[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 = int.MaxValue; 
  
        while (true) { 
            // for mainting 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 

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 initilized 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 upadate 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 <bits/stdc++.h> 
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[0]; } 
  
    // to replace root with new node x and heapify() new root 
    void replaceMin(MinHeapNode x) 
    { 
        harr[0] = 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][0]; 
  
        // 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[0]); 
  
    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[0]; 
        } 
  
        // to replace root with new node x 
        // and heapify() new root 
        public 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 = 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 

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)

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article and 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

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python

C#

C++

Java

C#

Recommended Posts: