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Find missing elements of a range

  • Difficulty Level : Easy
  • Last Updated : 30 Jun, 2021

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

Examples:  

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

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

There can be two approaches to solve the problem. 



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

C++




// 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;
}

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

Python




# 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

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

 

Using Arrays: Create a boolean array, where each index will represent whether the (i+low)th element is present in the array or not. Mark all those elements which are in the given range and are present in the array. Once all array items present in the given range have been marked true in the array, we traverse through the boolean array and print all elements whose value is false.

C++14




// An array 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)
{
    // Create boolean array of size
    // high-low+1, each index i representing
    // whether (i+low)th element found or not.
    bool points_of_range[high - low + 1] = { false };
 
    for (int i = 0; i < n; i++) {
        // if ith element of arr is in
        // range low to high then mark
        // corresponding index as true in array
        if (low <= arr[i] && arr[i] <= high)
            points_of_range[arr[i] - low] = true;
    }
 
    // Traverse through the range and
    // print all elements  whose value
    // is false
    for (int x = 0; x <= high - low; x++) {
        if (points_of_range[x] == false)
            cout << low + 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;
}
 
// This code is contributed by Shubh Bansal

Java




// An array based Java program
// to find missing elements from
// an array
 
import java.util.Arrays;
 
public class Print {
    // Print all elements of range
    // [low, high] that are not present
    // in arr[0..n-1]
    static void printMissing(
        int arr[], int low,
        int high)
    {
        // Create boolean array of
        // size high-low+1, each index i
        // representing whether (i+low)th
        // element found or not.
        boolean[] points_of_range = new boolean
            [high - low + 1];
 
        for (int i = 0; i < arr.length; i++) {
            // if ith element of arr is in
            // range low to high then mark
            // corresponding index as true in array
            if (low <= arr[i] && arr[i] <= high)
                points_of_range[arr[i] - low] = true;
        }
 
        // Traverse through the range and print all
        // elements whose value is false
        for (int x = 0; x <= high - low; x++) {
            if (points_of_range[x] == false)
                System.out.print((low + x) + " ");
        }
    }
 
    // 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 Shubh Bansal

Python3




# An array-based Python3 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):
 
    # Create boolean list of size
    # high-low+1, each index i
    # representing whether (i+low)th
    # element found or not.
    points_of_range = [False] * (high-low+1)
     
    for i in range(n) :
        # if ith element of arr is in range
        # low to high then mark corresponding
        # index as true in array
        if ( low <= arr[i] and arr[i] <= high ) :
            points_of_range[arr[i]-low] = True
 
    # Traverse through the range
    # and print all elements  whose value
    # is false
    for x in range(high-low+1) :
        if (points_of_range[x]==False) :
            print(low+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 Shubh Bansal

C#




// An array based C# 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[] arr, int n,
                         int low, int high)
{
     
    // Create boolean array of size
    // high-low+1, each index i representing
    // whether (i+low)th element found or not.
      bool[] points_of_range = new bool[high - low + 1];
       
      for(int i = 0; i < high - low + 1; i++)
          points_of_range[i] = false;
 
    for(int i = 0; i < n; i++)
    {
         
        // If ith element of arr is in
        // range low to high then mark
        // corresponding index as true in array
        if (low <= arr[i] && arr[i] <= high)
            points_of_range[arr[i] - low] = true;
    }
 
    // Traverse through the range and
    // print all elements  whose value
    // is false
    for(int x = 0; x <= high - low; x++)
    {
        if (points_of_range[x] == false)
            Console.Write("{0} ", low + 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 subhammahato348

Javascript




<script>
 
// Javascript program to find missing elements from
// an array
 
// Prlet all elements of range
    // [low, high] that are not present
    // in arr[0..n-1]
    function prletMissing(
        arr, low, high)
    {
        // Create boolean array of
        // size high-low+1, each index i
        // representing whether (i+low)th
        // element found or not.
        let polets_of_range = Array(high - low + 1).fill(0);           
  
        for (let i = 0; i < arr.length; i++) {
            // if ith element of arr is in
            // range low to high then mark
            // corresponding index as true in array
            if (low <= arr[i] && arr[i] <= high)
                polets_of_range[arr[i] - low] = true;
        }
  
        // Traverse through the range and print all
        // elements whose value is false
        for (let x = 0; x <= high - low; x++) {
            if (polets_of_range[x] == false)
                document.write((low + x) + " ");
        }
    }
 
 
// Driver program
 
         let arr = [ 1, 3, 5, 4 ];
        let low = 1, high = 10;
        prletMissing(arr, low, high);
       
</script>

Output:

2 6 7 8 9 10

 

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 <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;
}

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<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

Python3




# A hashing based Python3 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<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

Output:

2 6 7 8 9 10

Which approach is better? 
The time complexity of the first approach is O(nLogn + k) where k is the number of missing elements (Note that k may be more than nLogn if the array is small and the range is big)
The time complexity of the second and third solutions is O(n + (high-low+1)). 

If the given array has almost all elements of the range, i.e., n is close to the value of (high-low+1), then the second and third approaches are definitely better as there is no Log n factor. But if n is much smaller than the range, then the first approach is better as it doesn’t require extra space for hashing. We can also modify the 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 the first method. And if printing this way is allowed, the first approach takes only O(n Log n) time. Out of the Second and Third Solutions, the second solution is better because the worst-case time complexity of the second solution is better than the third.

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

 




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