# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 // A sorting based C++ program to find missing // elements from an array #include 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; }

## Java

 // 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

## C#

 // 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

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++

 // A hashing based C++ program to find missing // elements from an array #include 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 s;    for (int i=0; i

## Java

 // 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 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

## Python3

 # 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

## C#

 // 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 s = new HashSet();     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

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 attributed to GeeksforGeeks.org

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