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.

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 <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 wether the (i+low)th element is present in 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 given range have been marked true in the array we traverse through the boolean array and print all elements whose value is false.

C++

// 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 
    // wether (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 wether (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

# n array-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): 
  
    # Create boolean list of size  
    # high-low+1, each index i  
    # representing wether (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 

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 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<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?
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 and third 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 and third approaches are 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.

Out of Second and Third Solution the second solution is better because worst case time complexity of second solution is better than 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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My Personal Notes

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

C++

Java

Python

C#

C++

Java

Python3

C++

Java

Python3

C#

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