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 first list, 8 from second
list and 6 from third list.


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

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

Naive approach : The idea is to maintain pointers to every list using array ptr[k].Below are the steps :

Initially the index of every list is 0,therefore initialize every element of ptr[0..k] to 0;
Repeat the following steps until atleast one list exhausts :
- Now find the minimum and maximum value among the current elements of all the list pointed by the ptr[0…k] array.
- Now update the minrange if current (max-min) is less than minrange.
- increment the pointer pointing to current minimum element.

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]

Time complexity : O(n² k)

A Better efficient approach is to use min heap. Below are the steps –

Create a min heap of size k and insert first elements of all k lists into the heap.
Maintain two variables min and max to store minimum and maximum values present in the heap at any point. Note min will always contain value of the root of the heap.
Repeat following steps
- Get minimum element from heap (minimum is always at root) and compute the range.
- Replace heap root with next element of the list from which the min element is extracted. After replacing the root, heapify the tree. Update max if next element is greater. If the list doesn’t have any more elements, break the loop.

Below is C++ implementation of above approach –

// C++ program to finds out smallest range that includes 
// elements from each of the given sorted lists. 
#include<iostream> 
#include<limits.h> 
using namespace std; 
  
#define N 5 
  
// A min heap node 
struct MinHeapNode 
{ 
    int element; // The element to be stored 
    int i; // index of the list from which the element is taken 
    int j; // index of the next element to be picked from list 
}; 
  
// Prototype of a utility function to swap two min heap nodes 
void swap(MinHeapNode *x, MinHeapNode *y); 
  
// A class for Min Heap 
class MinHeap 
{ 
    MinHeapNode *harr; // pointer to array of elements in heap 
    int heap_size; // size of min heap 
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++) 
    { 
        harr[i].element = arr[i][0]; // Store the first element 
        harr[i].i = i; // index of list 
        harr[i].j = 1; // Index of next element to be stored 
                       // from list 
  
        // store max element 
        if (harr[i].element > max) 
            max = harr[i].element; 
    } 
  
    MinHeap hp(harr, k); // Create the heap 
  
    // 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; 
} 

Output :

The smallest range is [6 8]

Time Complexity: The while loop inside findSmallestRange() function can run maximum n*k times. In every iteration of loop, we call heapify which takes O(Logk) time. Therefore, the time complexity is O(nk Logk).

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.

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