Skip to content
Related Articles
Open in App
Not now

Related Articles

k-th missing element in sorted array

Improve Article
Save Article
  • Difficulty Level : Medium
  • Last Updated : 16 Oct, 2022
Improve Article
Save Article

Given an increasing sequence a[], we need to find the K-th missing contiguous element in the increasing sequence which is not present in the sequence. If no k-th missing element is there output -1. 

Examples : 

Input : a[] = {2, 3, 5, 9, 10};   
        k = 1;
Output : 1
Explanation: Missing Element in the increasing 
sequence are {1,4, 6, 7, 8}. So k-th missing element
is 1

Input : a[] = {2, 3, 5, 9, 10, 11, 12};       
        k = 4;
Output : 7
Explanation: missing element in the increasing 
sequence are {1, 4, 6, 7, 8}  so k-th missing 
element is 7
Recommended Practice

Approach 1: Start iterating over the array elements, and for every element check if the next element is consecutive or not, if not, then take the difference between these two, and check if the difference is greater than or equal to given k, then calculate ans = a[i] + count, else iterate for next element.

Implementation:

C++




#include <bits/stdc++.h>
using namespace std;
 
// Function to find k-th
// missing element
int missingK(int a[], int k, int n)
{
    int difference = 0, ans = 0, count = k;
    bool flag = 0;
 
    // case when first number is not 1
    if (a[0] != 1) {
        difference = a[0] - 1;
        if (difference >= count)
            return count;
        count -= difference;
    }
 
    // iterating over the array
    for (int i = 0; i < n - 1; i++) {
        difference = 0;
 
        // check if i-th and
        // (i + 1)-th element
        // are not consecutive
        if ((a[i] + 1) != a[i + 1]) {
 
            // save their difference
            difference += (a[i + 1] - a[i]) - 1;
 
            // check for difference
            // and given k
            if (difference >= count) {
                ans = a[i] + count;
                flag = 1;
                break;
            }
            else
                count -= difference;
        }
    }
 
    // if found
    if (flag)
        return ans;
    else
        return -1;
}
 
// Driver code
int main()
{
    // Input array
    int a[] = { 1, 5, 11, 19 };
 
    // k-th missing element
    // to be found in the array
    int k = 11;
    int n = sizeof(a) / sizeof(a[0]);
 
    // calling function to
    // find missing element
    int missing = missingK(a, k, n);
 
    cout << missing << endl;
 
    return 0;
}

Java




// Java program to check for
// even or odd
import java.io.*;
import java.util.*;
 
public class GFG {
 
    // Function to find k-th
    // missing element
    static int missingK(int[] a, int k, int n)
    {
        int difference = 0, ans = 0, count = k;
        boolean flag = false;
 
        // case when first number is not 1
        if (a[0] != 1) {
            difference = a[0] - 1;
            if (difference >= count)
                return count;
            count -= difference;
        }
 
        // iterating over the array
        for (int i = 0; i < n - 1; i++) {
            difference = 0;
 
            // check if i-th and
            // (i + 1)-th element
            // are not consecutive
            if ((a[i] + 1) != a[i + 1]) {
 
                // save their difference
                difference += (a[i + 1] - a[i]) - 1;
 
                // check for difference
                // and given k
                if (difference >= count) {
                    ans = a[i] + count;
                    flag = true;
                    break;
                }
                else
                    count -= difference;
            }
        }
 
        // if found
        if (flag)
            return ans;
        else
            return -1;
    }
 
    // Driver code
    public static void main(String args[])
    {
 
        // Input array
        int[] a = { 1, 5, 11, 19 };
 
        // k-th missing element
        // to be found in the array
        int k = 11;
        int n = a.length;
 
        // calling function to
        // find missing element
        int missing = missingK(a, k, n);
 
        System.out.print(missing);
    }
}
 
// This code is contributed by
// Manish Shaw (manishshaw1)

Python3




# Function to find k-th
# missing element
 
 
def missingK(a, k, n):
 
    difference = 0
    ans = 0
    count = k
    flag = 0
 
    # case when first number is not 1
    if a[0] != 1:
        difference = a[0]-1
        if difference >= count:
            return count
        count -= difference
 
    # iterating over the array
    for i in range(0, n-1):
        difference = 0
 
        # check if i-th and
        # (i + 1)-th element
        # are not consecutive
        if ((a[i] + 1) != a[i + 1]):
 
            # save their difference
            difference += (a[i + 1] - a[i]) - 1
 
            # check for difference
            # and given k
            if (difference >= count):
                ans = a[i] + count
                flag = 1
                break
            else:
                count -= difference
 
    # if found
    if(flag):
        return ans
    else:
        return -1
 
 
# Driver code
# Input array
a = [1, 5, 11, 19]
 
# k-th missing element
# to be found in the array
k = 11
n = len(a)
 
# calling function to
# find missing element
missing = missingK(a, k, n)
 
print(missing)
 
# This code is contributed by
# Manish Shaw (manishshaw1)

C#




// C# program to check for
// even or odd
using System;
using System.Collections.Generic;
 
class GFG {
 
    // Function to find k-th
    // missing element
    static int missingK(int[] a, int k, int n)
    {
        int difference = 0, ans = 0, count = k;
        bool flag = false;
 
        // case when first number is not 1
        if (a[0] != 1) {
            difference = a[0] - 1;
            if (difference >= count)
                return count;
            count -= difference;
        }
 
        // iterating over the array
        for (int i = 0; i < n - 1; i++) {
            difference = 0;
 
            // check if i-th and
            // (i + 1)-th element
            // are not consecutive
            if ((a[i] + 1) != a[i + 1]) {
 
                // save their difference
                difference += (a[i + 1] - a[i]) - 1;
 
                // check for difference
                // and given k
                if (difference >= count) {
                    ans = a[i] + count;
                    flag = true;
                    break;
                }
                else
                    count -= difference;
            }
        }
 
        // if found
        if (flag)
            return ans;
        else
            return -1;
    }
 
    // Driver code
    public static void Main()
    {
 
        // Input array
        int[] a = { 1, 5, 11, 19 };
 
        // k-th missing element
        // to be found in the array
        int k = 11;
        int n = a.Length;
 
        // calling function to
        // find missing element
        int missing = missingK(a, k, n);
 
        Console.Write(missing);
    }
}
 
// This code is contributed by
// Manish Shaw (manishshaw1)

PHP




<?php
// Function to find k-th
// missing element
function missingK(&$a, $k, $n)
{
    $difference = 0;
    $ans = 0;
    $count = $k;
    $flag = 0;
     
    // iterating over the array
    for($i = 0 ; $i < $n - 1; $i++)
    {
        $difference = 0;
         
        // check if i-th and
        // (i + 1)-th element
        // are not consecutive
        if (($a[$i] + 1) != $a[$i + 1])
        {
             
            // save their difference
            $difference += ($a[$i + 1] -
                            $a[$i]) - 1;
             
            // check for difference
            // and given k
            if ($difference >= $count)
                {
                    $ans = $a[$i] + $count;
                    $flag = 1;
                    break;
                }
            else
                $count -= $difference;
        }
    }
     
    // if found
    if($flag)
        return $ans;
    else
        return -1;
}
 
// Driver Code
 
// Input array
$a = array(1, 5, 11, 19);
 
// k-th missing element
// to be found in the array
$k = 11;
$n = count($a);
 
// calling function to
// find missing element
$missing = missingK($a, $k, $n);
 
echo $missing;
 
// This code is contributed by Manish Shaw
// (manishshaw1)
?>

Javascript




<script>
     
// Javascript program to check for
// even or odd
 
// Function to find k-th
// missing element
function missingK(a, k, n)
{
    let difference = 0, ans = 0, count = k;
    let flag = false;
     
    //case when first number is not 1
    if (a[0] != 1){
      difference = a[0]-1;
      if (difference >= count){
          return count;
          }
      count -= difference;
      }
      
    // iterating over the array
    for(let i = 0 ; i < n - 1; i++)
    {
        difference = 0;
          
        // Check if i-th and
        // (i + 1)-th element
        // are not consecutive
        if ((a[i] + 1) != a[i + 1])
        {
              
            // Save their difference
            difference += (a[i + 1] - a[i]) - 1;
              
            // Check for difference
            // and given k
            if (difference >= count)
            {
                ans = a[i] + count;
                flag = true;
                break;
            }
            else
                count -= difference;
        }
    }
     
    // If found
    if (flag)
        return ans;
    else
        return -1;
}
 
// Driver code
 
// Input array
let a = [ 1, 5, 11, 19 ];
 
// k-th missing element
// to be found in the array
let k = 11;
let n = a.length;
 
// Calling function to
// find missing element
let missing = missingK(a, k, n);
 
document.write(missing);
 
// This code is contributed by suresh07
   
</script>

Output

14

Time Complexity: O(n), where n is the number of elements in the array. 
Auxiliary Space: O(1)

Approach 2: 

Apply a binary search. Since the array is sorted we can find at any given index how many numbers are missing as arr[index] – (index+1). We would leverage this knowledge and apply binary search to narrow down our hunt to find that index from which getting the missing number is easier.

Implementation:

C++




// CPP program for above approach
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
 
// Function to find
// kth missing number
int missingK(vector<int>& arr, int k)
{
    int n = arr.size();
    int l = 0, u = n - 1, mid;
 
    while (l <= u) {
        mid = (l + u) / 2;
 
        int numbers_less_than_mid = arr[mid] - (mid + 1);
 
        // If the total missing number
        // count is equal to k we can iterate
        // backwards for the first missing number
        // and that will be the answer.
        if (numbers_less_than_mid == k) {
 
            // To further optimize we check
            // if the previous element's
            // missing number count is equal
            // to k. Eg: arr = [4,5,6,7,8]
            // If you observe in the example array,
            // the total count of missing numbers for all
            // the indices are same, and we are
            // aiming to narrow down the
            // search window and achieve O(logn)
            // time complexity which
            // otherwise would've been O(n).
            if (mid > 0 && (arr[mid - 1] - (mid)) == k) {
                u = mid - 1;
                continue;
            }
            // Else we return arr[mid] - 1.
            return arr[mid] - 1;
        }
 
        // Here we appropriately
        // narrow down the search window.
        if (numbers_less_than_mid < k) {
            l = mid + 1;
        }
        else if (k < numbers_less_than_mid) {
            u = mid - 1;
        }
    }
 
    // In case the upper limit is -ve
    // it means the missing number set
    // is 1,2,..,k and hence we directly return k.
    if (u < 0)
        return k;
 
    // Else we find the residual count
    // of numbers which we'd then add to
    // arr[u] and get the missing kth number.
    int less = arr[u] - (u + 1);
    k -= less;
 
    // Return arr[u] + k
    return arr[u] + k;
}
 
// Driver Code
int main()
{
    vector<int> arr = { 2, 3, 4, 7, 11 };
    int k = 5;
 
    // Function Call
    cout << "Missing kth number = " << missingK(arr, k)
         << endl;
    return 0;
}

Java




// Java program for above approach
public class GFG {
 
    // Function to find
    // kth missing number
    static int missingK(int[] arr, int k)
    {
        int n = arr.length;
        int l = 0, u = n - 1, mid;
        while (l <= u) {
            mid = (l + u) / 2;
            int numbers_less_than_mid
                = arr[mid] - (mid + 1);
 
            // If the total missing number
            // count is equal to k we can iterate
            // backwards for the first missing number
            // and that will be the answer.
            if (numbers_less_than_mid == k) {
 
                // To further optimize we check
                // if the previous element's
                // missing number count is equal
                // to k. Eg: arr = [4,5,6,7,8]
                // If you observe in the example array,
                // the total count of missing numbers for
                // all the indices are same, and we are
                // aiming to narrow down the
                // search window and achieve O(logn)
                // time complexity which
                // otherwise would've been O(n).
                if (mid > 0
                    && (arr[mid - 1] - (mid)) == k) {
                    u = mid - 1;
                    continue;
                }
 
                // Else we return arr[mid] - 1.
                return arr[mid] - 1;
            }
 
            // Here we appropriately
            // narrow down the search window.
            if (numbers_less_than_mid < k) {
                l = mid + 1;
            }
            else if (k < numbers_less_than_mid) {
                u = mid - 1;
            }
        }
 
        // In case the upper limit is -ve
        // it means the missing number set
        // is 1,2,..,k and hence we directly return k.
        if (u < 0)
            return k;
 
        // Else we find the residual count
        // of numbers which we'd then add to
        // arr[u] and get the missing kth number.
        int less = arr[u] - (u + 1);
        k -= less;
 
        // Return arr[u] + k
        return arr[u] + k;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 2, 3, 4, 7, 11 };
        int k = 5;
 
        // Function Call
        System.out.println("Missing kth number = "
                           + missingK(arr, k));
    }
}
 
// This code is contributed by divyesh072019.

Python3




# Python3 program for above approach
 
# Function to find
# kth missing number
 
 
def missingK(arr, k):
    n = len(arr)
    l = 0
    u = n - 1
    mid = 0
    while(l <= u):
        mid = (l + u)//2
        numbers_less_than_mid = arr[mid] - (mid + 1)
 
        # If the total missing number
        # count is equal to k we can iterate
        # backwards for the first missing number
        # and that will be the answer.
        if(numbers_less_than_mid == k):
 
            # To further optimize we check
            # if the previous element's
            # missing number count is equal
            # to k. Eg: arr = [4,5,6,7,8]
            # If you observe in the example array,
            # the total count of missing numbers for all
            # the indices are same, and we are
            # aiming to narrow down the
            # search window and achieve O(logn)
            # time complexity which
            # otherwise would've been O(n).
            if(mid > 0 and (arr[mid - 1] - (mid)) == k):
                u = mid - 1
                continue
 
            # Else we return arr[mid] - 1.
            return arr[mid]-1
 
        # Here we appropriately
        # narrow down the search window.
        if(numbers_less_than_mid < k):
            l = mid + 1
        elif(k < numbers_less_than_mid):
            u = mid - 1
 
    # In case the upper limit is -ve
    # it means the missing number set
    # is 1,2,..,k and hence we directly return k.
    if(u < 0):
        return k
 
    # Else we find the residual count
    # of numbers which we'd then add to
    # arr[u] and get the missing kth number.
    less = arr[u] - (u + 1)
    k -= less
 
    # Return arr[u] + k
    return arr[u] + k
 
 
# Driver Code
if __name__ == '__main__':
 
    arr = [2, 3, 4, 7, 11]
    k = 5
 
    # Function Call
    print("Missing kth number = " + str(missingK(arr, k)))
 
# This code is contributed by rutvik_56.

C#




// C# program for above approach
using System;
class GFG {
 
    // Function to find
    // kth missing number
    static int missingK(int[] arr, int k)
    {
        int n = arr.Length;
        int l = 0, u = n - 1, mid;
 
        while (l <= u) {
            mid = (l + u) / 2;
 
            int numbers_less_than_mid
                = arr[mid] - (mid + 1);
 
            // If the total missing number
            // count is equal to k we can iterate
            // backwards for the first missing number
            // and that will be the answer.
            if (numbers_less_than_mid == k) {
 
                // To further optimize we check
                // if the previous element's
                // missing number count is equal
                // to k. Eg: arr = [4,5,6,7,8]
                // If you observe in the example array,
                // the total count of missing numbers for
                // all the indices are same, and we are
                // aiming to narrow down the
                // search window and achieve O(logn)
                // time complexity which
                // otherwise would've been O(n).
                if (mid > 0
                    && (arr[mid - 1] - (mid)) == k) {
                    u = mid - 1;
                    continue;
                }
 
                // Else we return arr[mid] - 1.
                return arr[mid] - 1;
            }
 
            // Here we appropriately
            // narrow down the search window.
            if (numbers_less_than_mid < k) {
                l = mid + 1;
            }
            else if (k < numbers_less_than_mid) {
                u = mid - 1;
            }
        }
 
        // In case the upper limit is -ve
        // it means the missing number set
        // is 1,2,..,k and hence we directly return k.
        if (u < 0)
            return k;
 
        // Else we find the residual count
        // of numbers which we'd then add to
        // arr[u] and get the missing kth number.
        int less = arr[u] - (u + 1);
        k -= less;
 
        // Return arr[u] + k
        return arr[u] + k;
    }
 
    // Driver code
    static void Main()
    {
        int[] arr = { 2, 3, 4, 7, 11 };
        int k = 5;
 
        // Function Call
        Console.WriteLine("Missing kth number = "
                          + missingK(arr, k));
    }
}
 
// This code is contributed by divyeshrabadiya07.

Javascript




<script>
// JavaScript program for above approach
 
  // Function to find
  // kth missing number
  function missingK(arr, k)
  {
    var n = arr.length;
    var l = 0, u = n - 1, mid;   
    while(l <= u)
    {
      mid = (l + u)/2;       
      var numbers_less_than_mid = arr[mid] -
        (mid + 1);
 
      // If the total missing number
      // count is equal to k we can iterate
      // backwards for the first missing number
      // and that will be the answer.
      if(numbers_less_than_mid == k)
      {
 
        // To further optimize we check
        // if the previous element's
        // missing number count is equal
        // to k. Eg: arr = [4,5,6,7,8]
        // If you observe in the example array,
        // the total count of missing numbers for all
        // the indices are same, and we are
        // aiming to narrow down the
        // search window and achieve O(logn)
        // time complexity which
        // otherwise would've been O(n).
        if(mid > 0 && (arr[mid - 1] - (mid)) == k)
        {
          u = mid - 1;
          continue;
        }
 
        // Else we return arr[mid] - 1.
        return arr[mid] - 1;
      }
 
      // Here we appropriately
      // narrow down the search window.
      if(numbers_less_than_mid < k)
      {
        l = mid + 1;
      }
      else if(k < numbers_less_than_mid)
      {
        u = mid - 1;
      }
    }
 
    // In case the upper limit is -ve
    // it means the missing number set
    // is 1,2,..,k and hence we directly return k.
    if(u < 0)
      return k;
 
    // Else we find the residual count
    // of numbers which we'd then add to
    // arr[u] and get the missing kth number.
    var less = arr[u] - (u + 1);
    k -= less;
 
    // Return arr[u] + k
    return arr[u] + k;
  }
 
  // Driver code
    var arr = [2,3,4,7,11];
    var k = 5;
 
    // Function Call
    document.write("Missing kth number = "+ missingK(arr, k));
   
 
// This code is contributed by shivanisinghss2110
 
</script>

Output

Missing kth number = 9

Time Complexity: O(logn), where n is the number of elements in the array.
Auxiliary Space: O(1)

Approach 3 (Using Map): 

We will traverse the array and mark each of the elements as visited in the map and we will also keep track of the min and max element present so that we know the lower and upper bound for the given particular input. Then we start a loop from lower to upper bound and maintain a count variable. As soon we found an element that is not present in the map we increment the count and until the count becomes equal to k.

Implementation:

C++14




#include <iostream>
#include <unordered_map>
using namespace std;
 
int solve(int arr[], int k, int n)
{
    unordered_map<int, int> umap;
    int mins = 99999;
    int maxs = -99999;
    for (int i = 0; i < n; i++) {
        umap[arr[i]]
            = 1; // mark each element of array in map
        if (mins > arr[i])
            mins = arr[i]; // keeping track of minimum
                           // element
        if (maxs < arr[i])
            maxs = arr[i]; // keeping track of maximum
                           // element i.e. upper bound
    }
    int counts = 0;
    // iterate from lower to upper bound
    for (int i = mins; i <= maxs; i++) {
        if (umap[i] == 0)
            counts++;
        if (counts == k)
            return i;
    }
    return -1;
}
int main()
{
 
    int arr[] = { 2, 3, 5, 9, 10, 11, 12 };
    int k = 4;
    cout << solve(arr, k, 7); //(array , k , size of array)
    return 0;
}

Java




// Java approach
 
// Importing HashMap class
import java.util.HashMap;
class GFG {
 
    public static int solve(int arr[], int k, int n)
    {
        HashMap<Integer, Integer> umap
            = new HashMap<Integer, Integer>();
        int mins = 99999;
        int maxs = -99999;
        for (int i = 0; i < n; i++) {
            umap.put(
                arr[i],
                1); // mark each element of array in map
            if (mins > arr[i])
                mins = arr[i]; // keeping track of minimum
                               // element
            if (maxs < arr[i])
                maxs = arr[i]; // keeping track of maximum
                               // element i.e. upper bound
        }
        int counts = 0;
 
        // iterate from lower to upper bound
        for (int i = mins; i <= maxs; i++) {
            if (umap.get(i) == null)
                counts++;
            if (counts == k)
                return i;
        }
        return -1;
    }
    public static void main(String[] args)
    {
        int arr[] = { 2, 3, 5, 9, 10, 11, 12 };
        int k = 4;
        System.out.println(
            solve(arr, k, 7)); //(array , k , size of array)
    }
}
 
// This code is contributed by Shubham Singh

Python3




# Python approach
def solve(arr,  k,  n):
    umap = {}
    mins = 99999
    maxs = -99999
 
    for i in range(n):
        umap[arr[i]] = 1  # mark each element of array in map
        if(mins > arr[i]):
            mins = arr[i]  # keeping track of minimum element
        if(maxs < arr[i]):
            maxs = arr[i]  # keeping track of maximum element i.e. upper bound
    counts = 0
 
    # iterate from lower to upper bound
    for i in range(mins, maxs+1):
        if(i not in umap):
            counts += 1
        if(counts == k):
            return i
    return -1
 
 
arr = [2, 3, 5, 9, 10, 11, 12]
k = 4
print(solve(arr, k, 7))  # (array , k , size of array)
 
# This code is contributed
# by Shubham Singh

C#




// C# program for above approach
using System;
using System.Collections.Generic;
class GFG {
 
    // Function to find
    // kth missing number
    static int solve(int[] arr, int k, int n)
    {
 
        Dictionary<int, int> umap
            = new Dictionary<int, int>();
        int mins = 99999;
        int maxs = -99999;
        for (int i = 0; i < n; i++) {
            umap.Add(
                arr[i],
                1); // mark each element of array in map
            if (mins > arr[i])
                mins = arr[i]; // keeping track of minimum
                               // element
            if (maxs < arr[i])
                maxs = arr[i]; // keeping track of maximum
                               // element i.e. upper bound
        }
        int counts = 0;
        // iterate from lower to upper bound
        for (int i = mins; i <= maxs; i++) {
            if (!umap.ContainsKey(i))
                counts++;
            if (counts == k)
                return i;
        }
        return -1;
    }
 
    // Driver code
    static void Main()
    {
        int[] arr = { 2, 3, 5, 9, 10, 11, 12 };
        int k = 4;
        int n = arr.Length;
 
        // Function Call
        Console.WriteLine("Missing kth number = "
                          + solve(arr, k, n));
        //(array , k , size of array));
    }
}
 
// This code is contributed by Aarti_Rathi

Javascript




<script>
 
// JavaScript program
function solve(arr ,k ,n){
    let umap = new Map()
    let mins = 99999
    let maxs = -99999
     
    for(let i = 0; i < n; i++){
        umap.set(arr[i] , 1) //mark each element of array in map
        if(mins > arr[i])
            mins = arr[i]  //keeping track of minimum element
        if(maxs < arr[i])
            maxs = arr[i] //keeping track of maximum element i.e. upper bound
    }
    let counts = 0
     
    // iterate from lower to upper bound
    for(let i = mins; i < maxs + 1; i++){
        if(!umap.has(i))
            counts += 1
        if(counts == k)
            return i
    }
    return -1
}
        
// driver code   
let arr = [2, 3, 5, 9, 10, 11, 12]
let k = 4
document.write(solve(arr , k , 7),"</br>"//(array , k , size of array)
 
// This code is contributed by shinjanpatra
 
</script>

Output

8

Time Complexity: O(n+m), where n is the number of elements in the array and m is the difference between the largest and smallest element of the array.
Auxiliary Space: O(n)


My Personal Notes arrow_drop_up
Related Articles

Start Your Coding Journey Now!