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Given K sorted lists of integers of size N each, find the smallest range that includes at least one element from each of the K lists. If more than one smallest range is found, print any one of them.

Examples: 

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: 

The idea is to 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 minimized then either the minimum value has to be increased or the maximum value has to be decreased. To decrease the maximum value we have to move our pointer of current maximum to the left and since we are currently at 0 the index of every list so we can’t move our pointer to left, hence we can’t decrease the current max. So, the only possible option to get a better range is to increase the current minimum. 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.

Follow the given steps to solve the problem:

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

Below is the Implementation of the above approach:

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;
 
// 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(vector<vector<int> >& arr, 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's code
int main()
{
    vector<vector<int> > arr = { { 4, 7, 9, 12, 15 },
                                 { 0, 8, 10, 14, 20 },
                                 { 6, 12, 16, 30, 50 } };
 
    int K = arr.size();
    int N = arr[0].size();
 
    // Function call
    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 = # 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
if __name__ == '__main__':
    arr = [
        [4, 7, 9, 12, 15],
        [0, 8, 10, 14, 20],
        [6, 12, 16, 30, 50]
    ]
 
    K = len(arr)
 
    # Function call
    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 maintaining 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's 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);
 
        // Function call
        findSmallestRange(arr, N, K);
    }
}
 
// This code has been contributed by 29AjayKumar

                    

Javascript

// Javascript program to finds out smallest range that includes
// elements from each of the given sorted lists.
let N = 5;
 
// array for storing the current index of list i
let ptr=new Array(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.
function findSmallestRange(arr,n,k)
{
    let i, minval, maxval, minrange, minel = 0, maxel = 0, flag, minind;
   
        // initializing to 0 index;
        for (i = 0; i <= k; i++) {
            ptr[i] = 0;
        }
   
        minrange = Number.MAX_VALUE;
   
        while (true) {
            // for maintaining the index of list containing the minimum element
            minind = -1;
            minval = Number.MAX_VALUE;
            maxval = Number.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;
            }
        }
        document.write("The smallest range is ["+minel+", "+maxel+"]<br>");
}
 
// Driver program to test above function
let arr = [
    [4, 7, 9, 12, 15],
    [0, 8, 10, 14, 20],
    [6, 12, 16, 30, 50]
    ]
let k = arr.length;
findSmallestRange(arr, N, k);
 
 
// This code is contributed by unknown2108

                    

Output
The smallest range is [6, 8]








Time complexity: O(N * K2)
Space complexity: O(K)

The smallest range containing elements from k lists using Min-Heap:

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. 

Follow the given steps to solve the problem:

  • create a 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.
  • Initially put the first element from each list and store the maximum value in max.
  • Repeat the following steps until at least one list exhausts : 
    • To find the minimum value or min, use the top or root of the Min heap which is the minimum element.
    • Now update the minrange if the current (max-min) is less than minrange.
    • remove the top or root element from the priority queue, insert the next element from the list containing the min element, and update the max with the new element inserted.

Below is the Implementation of the above approach:

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's code
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]);
 
    // Function call
    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's 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;
 
        // Function call
        findSmallestRange(arr, K);
    }
}
 
// This code is contributed by nobody_cares

                    

Python

import sys
 
# A min heap node
class MinHeapNode:
    def __init__(self, element, i, j):
        self.element = element
        self.i = i
        self.j = j
  
  
# A class for Min Heap
class MinHeap:
    def __init__(self, a, size):
        self.heap_size = size
        self.harr = # store address of array
        i = (self.heap_size - 1) // 2
        while i >= 0:
            self.MinHeapify(i)
            i -= 1
  
    def MinHeapify(self, i):
        l = self.left(i)
        r = self.right(i)
        smallest = i
  
        if l < self.heap_size and self.harr[l].element < self.harr[i].element:
            smallest = l
  
        if r < self.heap_size and self.harr[r].element < self.harr[smallest].element:
            smallest = r
  
        if smallest != i:
            self.swap(i, smallest)
            self.MinHeapify(smallest)
  
    def left(self, i):
        return 2 * i + 1
  
    def right(self, i):
        return 2 * i + 2
  
    def getMin(self):
        return self.harr[0]
  
    def replaceMin(self, x):
        self.harr[0] = x
        self.MinHeapify(0)
  
    def swap(self, i, j):
        self.harr[i], self.harr[j] = self.harr[j], self.harr[i]
  
  
def findSmallestRange(arr, K):
  
    range_val = sys.maxsize
    min_val = sys.maxsize
    max_val = -sys.maxsize
    start, end = 0, 0
  
    harr = [MinHeapNode(0, i, 1) for i in range(K)]
    for i in range(K):
        harr[i].element = arr[i][0]
        if harr[i].element > max_val:
            max_val = harr[i].element
  
    hp = MinHeap(harr, K)
  
    while True:
        root = hp.getMin()
  
        min_val = root.element
  
        if range_val > max_val - min_val + 1:
            range_val = max_val - min_val + 1
            start = min_val
            end = max_val
  
        if root.j < N:
            root.element = arr[root.i][root.j]
            root.j += 1
  
            if root.element > max_val:
                max_val = root.element
        else:
            break
  
        hp.replaceMin(root)
  
    print("The smallest range is [{} {}]".format(start, end))
 
  
# Driver's code
if __name__ == "__main__":
    arr = [[4, 7, 9, 12, 15], [0, 8, 10, 14, 20], [6, 12, 16, 30, 50]]
    K = len(arr)
    N = len(arr[0])
    findSmallestRange(arr, K)

                    

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

                    

Javascript

// Javascript program to find out smallest
// range that includes elements from
// each of the given sorted lists.
class Node
{
    constructor(a, b, c)
    {
        this.ele = a;
        this.i = b;
        this.j = c;
    }
}
 
// A class for Min Heap
class MinHeap
{
     
    // Array of elements in heap
    harr;
     
    // Size of min heap
    size;
     
    // Constructor: creates a min heap
    // of given size
    constructor(arr,size)
    {
        this.harr = arr;
        this.size = size;
        let i = Math.floor((size - 1) / 2);
         
        while (i >= 0)
        {
            this.MinHeapify(i);
            i--;
        }
    }
     
    // To get index of left child
    // of node at index i
    left(i)
    {
        return 2 * i + 1;
    }
     
    // To get index of right child
    // of node at index i
    right(i)
    {
        return 2 * i + 2;
    }
     
    // To heapify a subtree with
    // root at given index
    MinHeapify(i)
    {
        let l = this.left(i);
        let r = this.right(i);
        let small = i;
         
        if (l < this.size &&
                this.harr[l].ele <
                this.harr[i].ele)
            small = l;
        if (r < this.size &&
                this.harr[r].ele <
                this.harr[small].ele)
            small = r;
        if (small != i)
        {
            this.swap(small, i);
            this.MinHeapify(small);
        }
    }
      
    swap(i, j)
    {
         
        let temp = this.harr[i];
        this.harr[i] = this.harr[j];
        this.harr[j] = temp;
    }
     
    // To get the root
    getMin()
    {
        return this.harr[0];
    }
     
    // To replace root with new node x
    // and heapify() new root
    replaceMin(x)
    {
        this.harr[0] = x;
        this.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.
function findSmallestRange(arr, k)
{
    let range = Number.MAX_VALUE;
    let min = Number.MAX_VALUE;
    let max = Number.MIN_VALUE;
    let start = -1, end = -1;
    let n = arr[0].length;
 
    // Create a min heap with k heap nodes.
    // Every heap node has first element of an list
    let arr1 = new Array(k);
    for(let i = 0; i < k; i++)
    {
        let node = new Node(arr[i][0], i, 1);
        arr1[i] = node;
 
        // Store max element
        max = Math.max(max, node.ele);
    }
 
    // Create the heap
    let 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
        let 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);
    }
    document.write("The smallest range is [" +
                   start + " " + end + "]");
}
 
// Driver Code
let arr = [ [ 4, 7, 9, 12, 15 ],
            [ 0, 8, 10, 14, 20 ],
            [ 6, 12, 16, 30, 50 ] ];
let k = arr.length;
 
findSmallestRange(arr, k);
 
// This code is contributed by rag2127

                    

Output
The smallest range is [6 8]








Time complexity: O(N * K * log K)
Auxiliary Space: O(K)

ANOTHER APPROACH USING PRIORITY QUEUE AND BUILT-IN METHODS

Intuition:

  1. We declare a Priority Queue<Node> ,Node class consists of current row,current column and current data.
  2. We first push every details of first column in the queue.
  3. while pushing elements in the queue , we always keep a track of the maximum element achieved so far.
  4. While poping the smallest element , we update the min value and compare with range ,ie max-min and if the smaller than previous then we update start to min and end to max.
  5. Atlast we print start and end.

Implementation

C++

// c++ program to find the smallest range that includes
// elements from each of the given sorted lists.
#include <bits/stdc++.h>
using namespace std;
 
// Node structure
class Node {
public:
    int data;
    int row;
    int nextCol;
    Node(int data, int row, int nextCol)
    {
        this->data = data;
        this->row = row;
        this->nextCol = nextCol;
    }
};
 
// Custom comparator for min heap
struct compare {
    bool operator()(Node a, Node b)
    {
        return a.data > b.data;
    }
};
 
// function to get the smallest range
void findSmallestRange(vector<vector<int> >& arr, int n,
                       int k)
{
    // initialize a min heap
    priority_queue<Node, vector<Node>, compare> pq;
 
    int maxi = 0; // Maximum element found so far
    int range = INT_MAX; // Smallest range found so far
 
    // push first element of each list and also found max
    // element
    for (int i = 0; i < k; i++) {
        pq.push(Node(arr[i][0], i, 0));
        maxi = max(maxi, arr[i][0]);
    }
 
    int start = -1; // Start of the smallest range
    int end = -1; // End of the smallest range
 
    while (!pq.empty()) {
        Node n1 = pq.top(); // Get the minimum element from
                            // the priority queue
        pq.pop();
        int min = n1.data; // Current minimum element
 
        // is smaller range found
        if (range > maxi - min) {
            range = maxi - min; // Update the smallest range
            start = min; // Update the start of the smallest
                         // range
            end = maxi; // Update the end of the smallest
                        // range
        }
 
        int nextRow = n1.row; // Next row index
        int nextColumn = n1.nextCol; // Next column index
        if (n1.nextCol + 1 < n) {
            pq.push(Node(
                arr[nextRow][nextColumn + 1], nextRow,
                nextColumn + 1)); // Push the next element
                                  // from the same list into
                                  // the priority queue
            maxi = max(
                maxi,
                arr[nextRow]
                   [nextColumn
                    + 1]); // Update the maximum element
        }
        else
            break; // Break if reached the end of any list
    }
    cout << start << " " << end << endl;
}
 
// Driver Code
int main()
{
    vector<vector<int> > arr = { { 1, 3, 5, 7, 9 },
                                 { 0, 2, 4, 6, 8 },
                                 { 2, 3, 5, 7, 11 } };
 
    int k = 3; // Number of sorted lists
 
    findSmallestRange(arr, k, 3); // Find the smallest range
 
    return 0;
}
// This code is contributed by Tapesh(tapeshdua420)

                    

Java

// Java program to finds out smallest range that includes
// elements from each of the given sorted lists.
 
import java.io.*;
import java.util.*;
 
class Node {
    int data;
    int row;
    int nextCol;
    Node(int data, int row, int nextCol)
    {
        this.data = data;
        this.row = row;
        this.nextCol = nextCol;
    }
}
class GFG {
    static void findSmallestRange(int[][] arr, int n, int k)
    {
        PriorityQueue<Node> pq = new PriorityQueue<Node>(
            (a, b) -> (a.data - b.data));
 
        int max = 0;
        int range = Integer.MAX_VALUE;
        for (int i = 0; i < k; i++) {
            pq.add(new Node(arr[i][0], i, 0));
            max = Math.max(max, arr[i][0]);
        }
        int res[] = new int[2];
        int start = -1;
        int end = -1;
        while (!pq.isEmpty()) {
            Node n1 = pq.poll();
            int min = n1.data;
            if (range > max - min) {
                range = max - min;
                start = min;
                end = max;
            }
            int nextRow = n1.row;
            int nextColumn = n1.nextCol;
            if (n1.nextCol + 1 < n) {
                pq.add(
                    new Node(arr[nextRow][nextColumn + 1],
                             nextRow, nextColumn + 1));
                max = Math.max(
                    max, arr[nextRow][nextColumn + 1]);
            }
            else
                break;
        }
        System.out.println(start + " " + end);
    }
    public static void main(String[] args)
    {
        int arr[][]
            = { { 1, 3, 5, 7, 9 },
                { 0, 2, 4, 6, 8 },
                { 2, 3, 5, 7, 11 } };
 
        int k
            = arr.length;
 
        findSmallestRange(arr, k, 3);
    }
}
// This code is contributed by Raunak Singh

                    

Python3

import heapq
 
class Node:
    def __init__(self, data, row, next_col):
        self.data = data
        self.row = row
        self.next_col = next_col
 
    def __lt__(self, other):
        return self.data < other.data
 
def find_smallest_range(arr, n, k):
    min_heap = []        # Create a min-heap to store elements from different arrays in sorted order
    max_value = 0        # Initialize the maximum value found so far in the min_heap
    smallest_range = float("inf"# Initialize the smallest range found so far as infinity
    start = -1           # Initialize the start of the smallest range
    end = -1             # Initialize the end of the smallest range
 
    # Add the first element from each array to the min_heap
    for i in range(k):
        heapq.heappush(min_heap, Node(arr[i][0], i, 0))
        max_value = max(max_value, arr[i][0])  # Update the maximum value found so far
 
    # Process the min_heap until it is empty
    while min_heap:
        node = heapq.heappop(min_heap)  # Get the minimum element from the min_heap
        min_value = node.data          # Get the value of the minimum element
 
        # Check if the range formed by the current element and the maximum element in min_heap is smaller
        # than the smallest range found so far.
        if smallest_range > max_value - min_value:
            smallest_range = max_value - min_value
            start = min_value
            end = max_value
 
        row = node.row          # Get the row of the current element
        next_col = node.next_col  # Get the next column index of the current element
 
        # If the current row has more elements, add the next element to the min_heap and update max_value
        if next_col + 1 < n:
            heapq.heappush(min_heap, Node(arr[row][next_col + 1], row, next_col + 1))
            max_value = max(max_value, arr[row][next_col + 1])
        else:
            break   # If the current row has no more elements, exit the loop
 
    print(start, end)  # Print the smallest range found
 
if __name__ == "__main__":
    arr = [[1, 3, 5, 7, 9], [0, 2, 4, 6, 8], [2, 3, 5, 7, 11]]
    k = 3
    find_smallest_range(arr, k, 3)

                    

C#

// C# program to find the smallest range that includes
// elements from each of the given sorted lists.
using System;
using System.Collections.Generic;
 
public class Node : IComparable<Node>
{
    public int Data { get; private set; }
    public int Row { get; private set; }
    public int NextCol { get; private set; }
 
    public Node(int data, int row, int nextCol)
    {
        Data = data;
        Row = row;
        NextCol = nextCol;
    }
 
    public int CompareTo(Node other)
    {
        return Data.CompareTo(other.Data);
    }
}
 
public class Program
{
    public static void FindSmallestRange(List<List<int>> arr, int k, int n)
    {
        var minHeap = new MinHeap<Node>();  // Create a min-heap to store elements from different arrays in sorted order
        int maxValue = 0;                  // Initialize the maximum value found so far in the min_heap
        int smallestRange = int.MaxValue;  // Initialize the smallest range found so far as infinity
        int start = -1;                    // Initialize the start of the smallest range
        int end = -1;                      // Initialize the end of the smallest range
 
        // Add the first element from each array to the min_heap
        for (int i = 0; i < k; i++)
        {
            var node = new Node(arr[i][0], i, 0);
            minHeap.Add(node);
            maxValue = Math.Max(maxValue, arr[i][0]);  // Update the maximum value found so far
        }
 
        // Process the min_heap until it is empty
        while (minHeap.Count > 0)
        {
            var node = minHeap.ExtractMin();  // Get the minimum element from the min_heap
            int minValue = node.Data;        // Get the value of the minimum element
 
            // Check if the range formed by the current element and the maximum element in min_heap is smaller
            // than the smallest range found so far.
            if (smallestRange > maxValue - minValue)
            {
                smallestRange = maxValue - minValue;
                start = minValue;
                end = maxValue;
            }
 
            int row = node.Row;               // Get the row of the current element
            int nextCol = node.NextCol;       // Get the next column index of the current element
 
            // If the current row has more elements, add the next element to the min_heap and update maxValue
            if (nextCol + 1 < n)
            {
                var nextNode = new Node(arr[row][nextCol + 1], row, nextCol + 1);
                minHeap.Add(nextNode);
                maxValue = Math.Max(maxValue, arr[row][nextCol + 1]);
            }
            else
            {
                break;   // If the current row has no more elements, exit the loop
            }
        }
 
        Console.WriteLine($"{start} {end}");  // Print the smallest range found
    }
//Driver Code
    public static void Main(string[] args)
    {
        List<List<int>> arr = new List<List<int>>()
        {
            new List<int>(){1, 3, 5, 7, 9},
            new List<int>(){0, 2, 4, 6, 8},
            new List<int>(){2, 3, 5, 7, 11}
        };
        int k = 3;
        int n = 5;
        FindSmallestRange(arr, k, n);
    }
}
 
public class MinHeap<T> where T : IComparable<T>
{
    private List<T> heap;
 
    public MinHeap()
    {
        heap = new List<T>();
    }
 
    public int Count { get { return heap.Count; } }
 
    public void Add(T item)
    {
        heap.Add(item);
        int currentIndex = heap.Count - 1;
 
        while (currentIndex > 0)
        {
            int parentIndex = (currentIndex - 1) / 2;
 
            if (heap[currentIndex].CompareTo(heap[parentIndex]) >= 0)
                break;
 
            Swap(currentIndex, parentIndex);
            currentIndex = parentIndex;
        }
    }
 
    public T ExtractMin()
    {
        int lastIndex = heap.Count - 1;
        T min = heap[0];
        heap[0] = heap[lastIndex];
        heap.RemoveAt(lastIndex);
 
        Heapify(0);
 
        return min;
    }
 
    private void Heapify(int index)
    {
        int left = index * 2 + 1;
        int right = index * 2 + 2;
        int smallest = index;
 
        if (left < heap.Count && heap[left].CompareTo(heap[smallest]) < 0)
            smallest = left;
 
        if (right < heap.Count && heap[right].CompareTo(heap[smallest]) < 0)
            smallest = right;
 
        if (smallest != index)
        {
            Swap(index, smallest);
            Heapify(smallest);
        }
    }
 
    private void Swap(int i, int j)
    {
        T temp = heap[i];
        heap[i] = heap[j];
        heap[j] = temp;
    }
}

                    

Javascript

// Node class
class Node {
    constructor(data, row, nextCol) {
        this.data = data;
        this.row = row;
        this.nextCol = nextCol;
    }
}
 
// Custom comparator for min heap
class MinHeap {
    constructor() {
        this.heap = [];
    }
 
    push(node) {
        this.heap.push(node);
        this.heapifyUp();
    }
 
    pop() {
        if (this.heap.length === 0) return null;
        const root = this.heap[0];
        const lastNode = this.heap.pop();
        if (this.heap.length > 0) {
            this.heap[0] = lastNode;
            this.heapifyDown();
        }
        return root;
    }
 
    heapifyUp() {
        let currentIndex = this.heap.length - 1;
        while (currentIndex > 0) {
            const parentIndex = Math.floor((currentIndex - 1) / 2);
            if (this.heap[currentIndex].data < this.heap[parentIndex].data) {
                [this.heap[currentIndex], this.heap[parentIndex]] =
                [this.heap[parentIndex], this.heap[currentIndex]];
                currentIndex = parentIndex;
            } else {
                break;
            }
        }
    }
 
    heapifyDown() {
        let currentIndex = 0;
        while (true) {
            let leftChildIndex = 2 * currentIndex + 1;
            let rightChildIndex = 2 * currentIndex + 2;
            let swapIndex = currentIndex;
 
            if (leftChildIndex < this.heap.length &&
                this.heap[leftChildIndex].data < this.heap[swapIndex].data) {
                swapIndex = leftChildIndex;
            }
 
            if (rightChildIndex < this.heap.length &&
                this.heap[rightChildIndex].data < this.heap[swapIndex].data) {
                swapIndex = rightChildIndex;
            }
 
            if (swapIndex !== currentIndex) {
                [this.heap[currentIndex], this.heap[swapIndex]] =
                [this.heap[swapIndex], this.heap[currentIndex]];
                currentIndex = swapIndex;
            } else {
                break;
            }
        }
    }
}
 
// Function to find the smallest range
function findSmallestRange(arr, k, n) {
    const pq = new MinHeap();
    let maxi = 0;
    let range = Number.MAX_SAFE_INTEGER;
    for (let i = 0; i < k; i++) {
        pq.push(new Node(arr[i][0], i, 0));
        maxi = Math.max(maxi, arr[i][0]);
    }
     
    let start = -1;
    let end = -1;
 
    while (pq.heap.length > 0) {
        const n1 = pq.pop();
        const min = n1.data;
 
        if (range > maxi - min) {
            range = maxi - min;
            start = min;
            end = maxi;
        }
 
        const nextRow = n1.row;
        const nextColumn = n1.nextCol;
 
        if (n1.nextCol + 1 < n) {
            pq.push(new Node(arr[nextRow][nextColumn + 1], nextRow, nextColumn + 1));
            maxi = Math.max(maxi, arr[nextRow][nextColumn + 1]);
        } else {
            break;
        }
    }
 
    console.log(start, end);
}
 
// Driver Code
const arr = [
    [1, 3, 5, 7, 9],
    [0, 2, 4, 6, 8],
    [2, 3, 5, 7, 11]
];
const k = 3;
const n = 5;
 
findSmallestRange(arr, k, n);

                    

Output
1 2








Time complexity: O(N * K * log K)
Auxiliary Space: O(K) since at worst case only k elements are in the priority Queue.



Last Updated : 19 Sep, 2023
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