Find missing elements of a range

Given an array arr[0..n-1] of distinct elements and a range [low, high], find all numbers that are in range, but not in array. The missing elements should be printed in sorted order.

Examples:

Input: arr[] = {10, 12, 11, 15}, 
       low = 10, hight = 15
Output: 13, 14

Input: arr[] = {1, 14, 11, 51, 15}, 
       low = 50, hight = 55
Output: 50, 52, 53, 54

There can be following two approaches to solve the problem.

  1. Use Sorting : Sort the array, then do binary search for ‘low’. Once location of low is find, start traversing array from that location and keep printing all missing numbers.

    C++

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // A sorting based C++ program to find missing
    // elements from an array
    #include<bits/stdc++.h>
    using namespace std;
      
    // Print all elements of range [low, high] that
    // are not present in arr[0..n-1]
    void printMissing(int arr[], int n, int low,
                                       int high)
    {
       // Sort the array
       sort(arr, arr+n);
      
       // Do binary search for 'low' in sorted
       // array and find index of first element
       // which either equal to or greater than
       // low.
       int *ptr = lower_bound(arr, arr+n, low);
       int index = ptr - arr;
      
       // Start from the found index and linearly
       // search every range element x after this
       // index in arr[]
       int i = index, x = low;
       while (i < n && x<=high)
       {
           // If x doesn't math with current element
           // print it
           if (arr[i] != x)
              cout << x << " ";
      
           // If x matches, move to next element in arr[]
           else
              i++;
      
           // Move to next element in range [low, high]
           x++;
       }
      
       // Print range elements thar are greater than the
       // last element of sorted array.
       while (x <= high)
         cout << x++ << " ";
    }
      
    // Driver program
    int main()
    {
       int arr[] = {1, 3, 5, 4};
       int n = sizeof(arr)/sizeof(arr[0]);
       int low = 1, high = 10;
       printMissing(arr, n, low, high);
       return 0;
    }

    chevron_right

    
    

    Java

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // A sorting based Java program to find missing
    // elements from an array
      
    import java.util.Arrays;
      
    public class PrintMissing 
    {
        // Print all elements of range [low, high] that
        // are not present in arr[0..n-1]
        static void printMissing(int ar[], int low, int high) 
        {
            Arrays.sort(ar);
            // Do binary search for 'low' in sorted
            // array and find index of first element
            // which either equal to or greater than
            // low.
            int index = ceilindex(ar, low, 0, ar.length - 1);
            int x = low;
      
            // Start from the found index and linearly
            // search every range element x after this
            // index in arr[]
            while (index < ar.length && x <= high) 
            {
                // If x doesn't math with current element
                // print it
                if (ar[index] != x) 
                {
                    System.out.print(x + " ");
                }
                  
                // If x matches, move to next element in arr[]
                else
                    index++;
                // Move to next element in range [low, high]
                x++;
            }
              
            // Print range elements thar are greater than the
            // last element of sorted array.
            while (x <= high) 
            {
                System.out.print(x + " ");
                x++;
            }
      
        }
          
        // Utility function to find ceil index of given element
        static int ceilindex(int ar[], int val, int low, int high) 
        {
      
            if (val < ar[0])
                return 0;
            if (val > ar[ar.length - 1])
                return ar.length;
      
            int mid = (low + high) / 2;
            if (ar[mid] == val)
                return mid;
            if (ar[mid] < val) 
            {
                if (mid + 1 < high && ar[mid + 1] >= val)
                    return mid + 1;
                return ceilindex(ar, val, mid + 1, high);
            
            else 
            {
                if (mid - 1 >= low && ar[mid - 1] < val)
                    return mid;
                return ceilindex(ar, val, low, mid - 1);
            }
      
        }
      
        // Driver program to test above function
        public static void main(String[] args) 
        {
            int arr[] = { 1, 3, 5, 4 };
            int low = 1, high = 10;
            printMissing(arr, low, high);
        }
    }
      
    // This code is contributed by Rishabh Mahrsee

    chevron_right

    
    

    Python

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    # Python library for binary search 
    from bisect import bisect_left 
      
    # A sorting based C++ program to find missing 
    # elements from an array 
      
    # Print all elements of range [low, high] that 
    # are not present in arr[0..n-1] 
      
    def printMissing(arr,n,low,high):
          
        # Sort the array
        arr.sort()
          
        # Do binary search for 'low' in sorted 
        # array and find index of first element 
        # which either equal to or greater than 
        # low. 
        ptr = bisect_left(arr,low)
        index = ptr
          
        # Start from the found index and linearly 
        # search every range element x after this 
        # index in arr[] 
        i = index
        x = low
        while (i < n and x <=high):
        # If x doesn't math with current element 
        # print it 
            if(arr[i] != x):
                print(x,end=" ")
      
        # If x matches, move to next element in arr[] 
            else:
                i = i + 1
        # Move to next element in range [low, high] 
            x = x + 1
      
        # Print range elements thar are greater than the 
        # last element of sorted array. 
        while (x <= high): 
            print(x,end=" ")
            x = x+ 1
      
      
    # Driver code 
      
    arr = [1, 3, 5, 4
    n = len(arr)
    low = 1
    high = 10
    printMissing(arr, n, low, high); 
      
    # This code is contributed by YatinGupta

    chevron_right

    
    

    C#

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // A sorting based Java program to 
    // find missing elements from an array 
    using System;
      
    class GFG
    {
      
    // Print all elements of range 
    // [low, high] that are not 
    // present in arr[0..n-1] 
    static void printMissing(int []ar,
                             int low, int high) 
        Array.Sort(ar); 
          
        // Do binary search for 'low' in sorted 
        // array and find index of first element 
        // which either equal to or greater than 
        // low. 
        int index = ceilindex(ar, low, 0, 
                              ar.Length - 1); 
        int x = low; 
      
        // Start from the found index and linearly 
        // search every range element x after this 
        // index in arr[] 
        while (index < ar.Length && x <= high) 
        
            // If x doesn't math with current 
            // element print it 
            if (ar[index] != x) 
            
                    Console.Write(x + " "); 
            
              
            // If x matches, move to next 
            // element in arr[] 
            else
                index++; 
                  
            // Move to next element in 
            // range [low, high] 
            x++; 
        
          
        // Print range elements thar
        // are greater than the 
        // last element of sorted array. 
        while (x <= high) 
        
            Console.Write(x + " "); 
            x++; 
        
      
      
    // Utility function to find 
    // ceil index of given element 
    static int ceilindex(int []ar, int val, 
                         int low, int high) 
        if (val < ar[0]) 
            return 0; 
        if (val > ar[ar.Length - 1]) 
            return ar.Length; 
      
        int mid = (low + high) / 2; 
        if (ar[mid] == val) 
            return mid; 
        if (ar[mid] < val) 
        
            if (mid + 1 < high && ar[mid + 1] >= val) 
                return mid + 1; 
            return ceilindex(ar, val, mid + 1, high); 
        
        else
        
            if (mid - 1 >= low && ar[mid - 1] < val) 
                return mid; 
            return ceilindex(ar, val, low, mid - 1); 
        
      
      
    // Driver Code
    static public void Main ()
    {
        int []arr = { 1, 3, 5, 4 }; 
        int low = 1, high = 10; 
        printMissing(arr, low, high); 
      
    // This code is contributed 
    // by Sach_Code

    chevron_right

    
    


    Output:

    2 6 7 8 9 10 
  2. Use Hashing : Create a hash table and insert all array items into the hash table. Once all items are in hash table, traverse through the range and print all missing elements.

    C++

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // A hashing based C++ program to find missing
    // elements from an array
    #include<bits/stdc++.h>
    using namespace std;
      
    // Print all elements of range [low, high] that
    // are not present in arr[0..n-1]
    void printMissing(int arr[], int n, int low,
                                       int high)
    {
       // Insert all elements of arr[] in set
       unordered_set<int> s;
       for (int i=0; i<n; i++)
          s.insert(arr[i]);
      
       // Traverse throught the range an print all
       // missing elements
       for (int x=low; x<=high; x++)
          if (s.find(x) == s.end())
             cout << x << " ";
    }
      
    // Driver program
    int main()
    {
       int arr[] = {1, 3, 5, 4};
       int n = sizeof(arr)/sizeof(arr[0]);
       int low = 1, high = 10;
       printMissing(arr, n, low, high);
       return 0;
    }

    chevron_right

    
    

    Java

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // A hashing based Java program to find missing
    // elements from an array
      
    import java.util.Arrays;
    import java.util.HashSet;
      
    public class Print 
    {
        // Print all elements of range [low, high] that
        // are not present in arr[0..n-1]
        static void printMissing(int ar[], int low, int high) 
        {
            HashSet<Integer> hs = new HashSet<>();
              
            // Insert all elements of arr[] in set
            for (int i = 0; i < ar.length; i++)
                hs.add(ar[i]);
              
            // Traverse throught the range an print all
            // missing elements
            for (int i = low; i <= high; i++) 
            {
                if (!hs.contains(i)) 
                {
                    System.out.print(i + " ");
                }
            }
        }
      
        // Driver program to test above function
        public static void main(String[] args) 
        {
            int arr[] = { 1, 3, 5, 4 };
            int low = 1, high = 10;
            printMissing(arr, low, high);
        }
    }
      
    // This code is contributed by Rishabh Mahrsee

    chevron_right

    
    

    Python3

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    # A hashing based Python 3 program to 
    # find missing elements from an array 
      
    # Print all elements of range 
    # [low, high] that are not
    # present in arr[0..n-1] 
    def printMissing(arr, n, low, high):
      
        # Insert all elements of 
        # arr[] in set 
        s = set(arr)
      
        # Traverse through the range 
        # and print all missing elements 
        for x in range(low, high + 1):
            if x not in s:
                print(x, end = ' ')
      
    # Driver Code 
    arr = [1, 3, 5, 4]
    n = len(arr)
    low, high = 1, 10
    printMissing(arr, n, low, high)
      
    # This code is contributed 
    # by SamyuktaSHegde 

    chevron_right

    
    

    C#

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // A hashing based C# program to 
    // find missing elements from an array
    using System;
    using System.Collections.Generic; 
      
    class GFG
    {
      
    // Print all elements of range 
    // [low, high] that are not 
    // present in arr[0..n-1]
    static void printMissing(int []arr, int n, 
                             int low, int high)
    {
        // Insert all elements of arr[] in set
        HashSet<int> s = new HashSet<int>();
        for (int i = 0; i < n; i++)
        {
            s.Add(arr[i]);
        }        
          
        // Traverse throught the range 
        // an print all missing elements
        for (int x = low; x <= high; x++)
            if (!s.Contains(x))
                Console.Write(x + " ");
    }
      
    // Driver Code
    public static void Main()
    {
        int []arr = {1, 3, 5, 4};
        int n = arr.Length;
        int low = 1, high = 10;
        printMissing(arr, n, low, high);
    }
    }
      
    // This code is contributed by ihritik

    chevron_right

    
    


    Output:

    2 6 7 8 9 10 

Which approach is better?
Time complexity of first approach is O(nLogn + k) where k is number of missing elements (Note that k may be more than nLogn if array is small and range is big)

Time complexity of second solution is O(n + (high-low+1)).

If the given array has almost elements of range i.e., n is close to value of (high-low+1), then second approach is definitely better as there is no Log n factor. But if n is much smaller than range, then first approach is better as it doesn’t require extra space for hashing. We can also modify first approach to print adjacent missing elements as range to save time. For example if 50, 51, 52, 53, 54, 59 are missing, we can print them as 50-54, 59 in first method. And if printing this way is allowed, the first approach takes only O(n Log n) time.

This article is contributed by Piyush Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



My Personal Notes arrow_drop_up