Open In App

Find missing elements of a range

Last Updated : 14 Aug, 2023
Improve
Improve
Like Article
Like
Save
Share
Report

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:  

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 55

Naive Approach:

The naive approach for the problem can be to use two nested loops: one to traverse numbers from low to high and other one to traverse entire array to find out whether the element of the outer  loop exists in the array or not. If it doesn’t exist we will print it else continue to next iteration.

Algorithm:

  1.    Traverse numbers from low to high using a for loop.
  2.    For each number i in the range, initialize a boolean variable found to false.
  3.    Traverse the array arr to check if i is present in the array.
  4.    If i is found in arr, set found to true and break out of the loop.
  5.    If i is not found in arr, print i.
  6.    Repeat steps 2-5 for all numbers in the range [low, high].

C++




// C++ code for the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find and print missing
// elements in the given range
void findMissing(int arr[], int n, int low, int high) {
    // Loop through the range of numbers from low to high
    for (int i = low; i <= high; i++) {
        bool found = false;
       
        // Loop through the array to check if i exists in it
        for (int j = 0; j < n; j++) {
            if (arr[j] == i) {
                found = true;
                break;
            }
        }
       
        // If i is not found in the array, print it
        if (!found) {
            cout << i << " ";
        }
    }
}
 
// Driver's code
int main() {
      // Input array
    int arr[] = { 1, 3, 5, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int low = 1, high = 10;
   
      // Function call
    findMissing(arr, n, low, high);
    return 0;
}


Java




// Java code for the approach
 
import java.util.*;
 
public class GFG {
    // Function to find and print missing
    // elements in the given range
    static void findMissing(int[] arr, int n, int low,
                            int high)
    {
        // Loop through the range of numbers from low to
        // high
        for (int i = low; i <= high; i++) {
            boolean found = false;
            // Loop through the array to check if i exists
            // in it
            for (int j = 0; j < n; j++) {
                if (arr[j] == i) {
                    found = true;
                    break;
                }
            }
 
            // If i is not found in the array, print it
            if (!found) {
                System.out.print(i + " ");
            }
        }
    }
 
    // Driver's code
    public static void main(String[] args)
    {
        // Input array
        int[] arr = { 1, 3, 5, 4 };
        int n = arr.length;
        int low = 1, high = 10;
 
        // Function call
        findMissing(arr, n, low, high);
    }
}


Python3




# Function to find and print missing
# elements in the given range
def findMissing(arr, n, low, high):
    # Loop through the range of numbers from low to high
    for i in range(low, high+1):
        found = False
     
        # Loop through the array to check if i exists in it
        for j in range(n):
            if arr[j] == i:
                found = True
                break
     
        # If i is not found in the array, print it
        if not found:
            print(i, end=' ')
 
# Driver's code
if __name__ == '__main__':
    # Input array
    arr = [1, 3, 5, 4]
    n = len(arr)
    low = 1
    high = 10
 
    # Function call
    findMissing(arr, n, low, high)


C#




using System;
 
class GFG {
    // Function to find and print missing
    // elements in the given range
    static void FindMissing(int[] arr, int n, int low,
                            int high)
    {
        // Loop through the range of numbers from low to
        // high
        for (int i = low; i <= high; i++) {
            bool found = false;
 
            // Loop through the array to check if i exists
            // in it
            for (int j = 0; j < n; j++) {
                if (arr[j] == i) {
                    found = true;
                    break;
                }
            }
 
            // If i is not found in the array, print it
            if (!found) {
                Console.Write(i + " ");
            }
        }
    }
 
    // Driver's code
    static void Main()
    {
        // Input array
        int[] arr = { 1, 3, 5, 4 };
        int n = arr.Length;
        int low = 1, high = 10;
 
        // Function call
        FindMissing(arr, n, low, high);
    }
}
 
// This code is contributed by rambabuguphka


Javascript




// Function to find and print missing
// elements in the given range
function findMissing(arr, low, high) {
    const missing = [];
   
    // Loop through the range of numbers from low to high
    for (let i = low; i <= high; i++) {
        let found = false;
       
        // Loop through the array to check if i exists in it
        for (let j = 0; j < arr.length; j++) {
            if (arr[j] === i) {
                found = true;
                break;
            }
        }
       
        // If i is not found in the array, add it to the missing array
        if (!found) {
            missing.push(i);
        }
    }
   
    // Print the missing elements
    for (let i = 0; i < missing.length; i++) {
        console.log(missing[i] + " ");
    }
}
 
// Driver's code
const arr = [1, 3, 5, 4];
const low = 1, high = 10;
 
// Function call
findMissing(arr, low, high);
 
// THIS CODE IS CONTRIBUTED BY CHANDAN AGARWAL


Output

2 6 7 8 9 10 




Time Complexity: O( n * (high-low+1) ) as two nested for loops are executing with outer one from low to high and inner one from 1 to n where n is size of the input array.
Space Complexity: O(1) as no extra space has been taken.

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.

Implementation:

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 match 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 that 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 match 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 that 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


Python3




# 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 match 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 that 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 match 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 that
        // 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


Javascript




<script>
        // JavaScript code for the above approach
 
    // Print all elements of range [low, high] that
    // are not present in arr[0..n-1]
    function printMissing(ar, low, high)
    {
        ar.sort();
        // Do binary search for 'low' in sorted
        // array and find index of first element
        // which either equal to or greater than
        // low.
        let index = ceilindex(ar, low, 0, ar.length - 1);
        let 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 match with current element
            // print it
            if (ar[index] != x) {
                document.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 that are greater than the
        // last element of sorted array.
        while (x <= high) {
            document.write(x + " ");
            x++;
        }
    }
  
    // Utility function to find ceil index of given element
    function ceilindex(ar, val, low, high)
    {
  
        if (val < ar[0])
            return 0;
        if (val > ar[ar.length - 1])
            return ar.length;
  
        let mid = Math.floor((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
         let arr = [ 1, 3, 5, 4 ];
        let low = 1, high = 10;
        printMissing(arr, low, high);
         
         
    </script>


Output

2 6 7 8 9 10 




Time Complexity: O(n log n + k) where k is the number of missing elements
Auxiliary Space: O(n) or O(1) depending on the type of the array.

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.

Implementation:

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
 
// Print all elements of range
    // [low, high] that are not present
    // in arr[0..n-1]
    function printMissing(
        arr, low, high)
    {
        // Create boolean array of
        // size high-low+1, each index i
        // representing whether (i+low)th
        // element found or not.
        let points_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)
                points_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 (points_of_range[x] == false)
                document.write((low + x) + " ");
        }
    }
 
// Driver program
         let arr = [ 1, 3, 5, 4 ];
        let low = 1, high = 10;
        printMissing(arr, low, high);
       
      // This code is contributed by code_hunt.
</script>


Output

2 6 7 8 9 10 




Time Complexity: O(n + (high-low+1))
Auxiliary Space: O(n)

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 through 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 through 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 through 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


Javascript




<script>
 
// A hashing based Java program to find missing
// elements from an array
 
 
// Print all elements of range [low, high] that
// are not present in arr[0..n-1]
function printMissing(ar, low, high) {
  let hs = new Set();
 
  // Insert all elements of arr[] in set
  for (let i = 0; i < ar.length; i++)
    hs.add(ar[i]);
 
  // Traverse through the range an print all
  // missing elements
  for (let i = low; i <= high; i++) {
    if (!hs.has(i)) {
      document.write(i + " ");
    }
  }
}
 
// Driver program to test above function
let arr = [1, 3, 5, 4];
let low = 1, high = 10;
printMissing(arr, low, high);
 
// This code is contributed by gfgking
 
</script>


Output

2 6 7 8 9 10 




Time Complexity: O(n + (high-low+1))
Auxiliary Space: O(n)

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 n Log n 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.



Previous Article
Next Article

Similar Reads

Count of missing elements from 1 to maximum in index range [L, R]
Given an array arr[] of length N, and an array queries[] of size Q, the task is to find the number of missing elements from 1 to maximum in the range of indices [L, R] where L is queries[i][0] and R is queries[i][1]. Examples: Input: arr[] = {8, 6, 7, 7, 7}, queries = { {0, 3}, {1, 2}, {0, 2}, {1, 3}, {2, 3} }Output: 5 5 5 5 6Explanation: Maximum e
7 min read
Print missing elements that lie in range 0 - 99
Given an array of integers print the missing elements that lie in range 0-99. If there are more than one missing, collate them, otherwise just print the number.Note that the input array may not be sorted and may contain numbers outside the range [0-99], but only this range is to be considered for printing missing elements. Examples : Input: {88, 10
8 min read
Find the one missing number in range
Given an array of size n. It is also given that range of numbers is from smallestNumber to smallestNumber + n where smallestNumber is the smallest number in array. The array contains number in this range but one number is missing so the task is to find this missing number. Examples: Input: arr[] = {13, 12, 11, 15}Output: 14 Input: arr[] = {33, 36,
9 min read
Find K missing numbers from given Array in range [1, M] such that total average is X
Given an array arr[] of integers of size N where each element can be in the range [1, M], and two integers X and K. The task is to find K possible numbers in range [1, M] such that the average of all the (N + K) numbers is equal to X. If there are multiple valid answers, any one is acceptable. Examples: Input: arr[] = {3, 2, 4, 3}, M = 6, K = 2, X
6 min read
Find the missing number in range [1, N*M+1] represented as Matrix of size N*M
Given an N x M matrix mat[][] where all elements are natural numbers starting from 1 and are continuous except 1 element, find that element. Examples: Input: mat[][] = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 11, 12, 13}}N = 3, M = 4Output: 10Explanation: Missing number is 10 at row no 3. Input: mat[][] = {{1, 2, 3}, {5, 6, 7}, {8, 9, 10}} N = 3, M = 3Outp
15+ min read
Find the missing number in a sorted array of limited range
Given a sorted array of size n and given that there are numbers from 1 to n+1 with one missing, the missing number is to be found. It may be assumed that array has distinct elements. Examples: Input : 1 3 4 5 6Output : 2 Input : 1 2 3 4 5 7 8 9 10Output : 6 Recommended: Please try your approach on {IDE} first, before moving on to the solution.We tr
10 min read
Find the missing elements from 1 to M in given N ranges
Given N segments as ranges [L, R] where ranges are non-intersecting and non-overlapping. The task is to find all number between 1 to M that doesn't belong to any of the given ranges. Examples: Input : N = 2, M = 6 Ranges: [1, 2] [4, 5] Output : 3, 6 Explanation: Only 3 and 6 are missing from the above ranges. Input : N = 1, M = 5 Ranges: [2, 4] Out
8 min read
Find the missing elements from 1 to M in given N ranges | Set-2
Given an integer m and n ranges (e.g. [a, b]) which are intersecting and overlapping. The task is to find all the number within the range that doesn’t belong to any of the given ranges. Examples: Input: m = 6, ranges = {{1, 2}, {4, 5}} Output: 3 6 As only 3 and 6 are missing from the given ranges. Input: m = 5, ranges = {{2, 4}} Output: 1 5 Approac
14 min read
Find missing elements from an Array with duplicates
Given an array arr[] of size N having integers in the range [1, N] with some of the elements missing. The task is to find the missing elements. Note: There can be duplicates in the array. Examples: Input: arr[] = {1, 3, 3, 3, 5}, N = 5Output: 2 4Explanation: The numbers missing from the list are 2 and 4All other elements in the range [1, 5] are pre
12 min read
Find four missing numbers in an array containing elements from 1 to N
Given an array of unique integers where each integer of the given array lies in the range [1, N]. The size of array is (N-4). No Single element is repeated. Hence four numbers from 1 to N are missing in the array. Find the 4 missing numbers in sorted order. Examples: Input : arr[] = {2, 5, 6, 3, 9}Output : 1 4 7 8Input : arr[] = {1, 7, 3, 13, 5, 10
10 min read
Article Tags :
Practice Tags :
three90RightbarBannerImg