Given an unsorted sequence a[], the task is to find the K-th missing contiguous element in the increasing sequence of the array elements i.e. consider the array in sorted order and find the kth missing number. If no k-th missing element is there output -1.
Note: Only elements exists in the range of minimum and maximum element to be considered.
Examples:
Input: arr[] = {2, 4, 10, 7}, k = 5
Output: 9
Missing elements in the given array: 3, 5, 6, 8, 9
5th missing is 9.
Input: arr[] = {1, 3, 4}, k = 5
Output: -1
Method-1: Sort the array and use the approach used in the k-th missing element in a sorted array.
Method-2:
- Insert all the elements in an unordered_set.
- Find the minimum and maximum element of the array.
- Traverse the elements from minimum to maximum.
- Check if current element is present in the set or not.
- If not then check if this is kth missing by counting the missing elements.
- Return the current element if this is current missing.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;
// Function to find the sum// of minimum of all subarraysint findKth(int arr[], int n, int k)
{ unordered_set<int> missing;
int count = 0;
// Insert all the elements in a set
for (int i = 0; i < n; i++)
missing.insert(arr[i]);
// Find the maximum and minimum element
int maxm = *max_element(arr, arr + n);
int minm = *min_element(arr, arr + n);
// Traverse from the minimum to maximum element
for (int i = minm + 1; i < maxm; i++) {
// Check if "i" is missing
if (missing.find(i) == missing.end())
count++;
// Check if it is kth missing
if (count == k)
return i;
}
// If no kth element is missing
return -1;
}// Driver codeint main()
{ int arr[] = { 2, 10, 9, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 5;
cout << findKth(arr, n, k);
return 0;
} |
Java
// Java implementation of the above approachimport java.util.*;
class GFG
{ // Function to find the sum
// of minimum of all subarrays
static int findKth(int arr[], int n, int k)
{
HashSet<Integer> missing = new HashSet<>();
int count = 0;
// Insert all the elements in a set
for (int i = 0; i < n; i++)
{
missing.add(arr[i]);
}
// Find the maximum and minimum element
int maxm = Arrays.stream(arr).max().getAsInt();
int minm = Arrays.stream(arr).min().getAsInt();
// Traverse from the minimum to maximum element
for (int i = minm+1; i < maxm; i++)
{
// Check if "i" is missing
if (!missing.contains(i))
{
count++;
}
// Check if it is kth missing
if (count == k)
{
return i;
}
}
// If no kth element is missing
return -1;
}
// Driver code
public static void main(String[] args)
{
int arr[] = {2, 10, 9, 4};
int n = arr.length;
int k = 5;
System.out.println(findKth(arr, n, k));
}
}/* This code contributed by PrinciRaj1992 */ |
Python
# Python3 implementation of the above approach# Function to find the sum# of minimum of all subarraysdef findKth( arr, n, k):
missing = dict()
count = 0
# Insert all the elements in a set
for i in range(n):
missing[arr[i]] = 1
# Find the maximum and minimum element
maxm = max(arr)
minm = min(arr)
# Traverse from the minimum to maximum element
for i in range(minm + 1, maxm):
# Check if "i" is missing
if (i not in missing.keys()):
count += 1
# Check if it is kth missing
if (count == k):
return i
# If no kth element is missing
return -1
# Driver codearr = [2, 10, 9, 4 ]
n = len(arr)
k = 5
print(findKth(arr, n, k))
# This code is contributed by Mohit Kumar |
C#
// C# implementation of the above approach using System;
using System.Linq;
using System.Collections.Generic;
class GFG
{ // Function to find the sum
// of minimum of all subarrays
static int findKth(int []arr, int n, int k)
{
HashSet<int> missing = new HashSet<int>();
int count = 0;
// Insert all the elements in a set
for (int i = 0; i < n; i++)
{
missing.Add(arr[i]);
}
// Find the maximum and minimum element
int maxm = arr.Max();
int minm = arr.Min();
// Traverse from the minimum to maximum element
for (int i = minm + 1; i < maxm; i++)
{
// Check if "i" is missing
if (!missing.Contains(i))
{
count++;
}
// Check if it is kth missing
if (count == k)
{
return i;
}
}
// If no kth element is missing
return -1;
}
// Driver code
public static void Main(String[] args)
{
int []arr = {2, 10, 9, 4};
int n = arr.Length;
int k = 5;
Console.WriteLine(findKth(arr, n, k));
}
} // This code has been contributed by 29AjayKumar |
PHP
<?php// PHP implementation of the above approach // Function to find the sum // of minimum of all subarrays function findKth($arr, $n, $k)
{ $missing = array();
$count = 0;
// Insert all the elements in a set
for ($i = 0; $i < $n; $i++)
array_push($missing, $arr[$i]);
$missing = array_unique($missing);
// Find the maximum and minimum element
$maxm = max($arr);
$minm = min($arr);
// Traverse from the minimum to
// maximum element
for ($i = $minm + 1; $i < $maxm; $i++)
{
// Check if "i" is missing
if (!in_array($i, $missing, false))
$count += 1;
// Check if it is kth missing
if ($count == $k)
return $i;
}
// If no kth element is missing
return -1;
}// Driver code $arr = array(2, 10, 9, 4);
$n = sizeof($arr);
$k = 5;
echo findKth($arr, $n, $k);
// This code is contributed by Ryuga?> |
Javascript
<script>// javascript implementation of the above approach// Function to find the sum// of minimum of all subarraysfunction findKth(arr, n, k)
{ var missing = new Set();
var count = 0;
// Insert all the elements in a set
for (var i = 0; i < n; i++)
missing.add(arr[i]);
// Find the maximum and minimum element
var maxm = arr.reduce((a,b)=>Math.max(a,b));
var minm = arr.reduce((a,b)=>Math.min(a,b));
// Traverse from the minimum to maximum element
for (var i = minm + 1; i < maxm; i++) {
// Check if "i" is missing
if (!missing.has(i))
count++;
// Check if it is kth missing
if (count == k)
return i;
}
// If no kth element is missing
return -1;
}// Driver codevar arr = [ 2, 10, 9, 4 ];
var n = arr.length;
var k = 5;
document.write( findKth(arr, n, k));// This code is contributed by noob2000.</script> |
Output:
8
Time complexity: O(n) where n is size of input array
Auxiliary Space: O(n)