K-th smallest element after removing some integers from natural numbers - GeeksforGeeks

28 Apr, 2021
Given an array arr[] of size ‘n’ and a positive integer k. Consider series of natural numbers and remove arr[0], arr[1], arr[2], ..., arr[p] from it. Now the task is to find k-th smallest number in the remaining set of natural numbers. If no such number exists print “-1”.
Examples :  
Input : arr[] = { 1 } and k = 1.
Output: 2
Natural numbers are {1, 2, 3, 4, .... }
After removing {1}, we get {2, 3, 4, ...}.
Now, K-th smallest element = 2.

Input : arr[] = {1, 3}, k = 4.
Output : 6
First 5 Natural number {1, 2, 3, 4, 5, 6,  .. }
After removing {1, 3}, we get {2, 4, 5, 6, ... }.
Method 1 (Simple): Make an auxiliary array b[] for presence/absence of natural numbers and initialize all with 0. Make all the integer equal to 1 which are present in array arr[] i.e b[arr[i]] = 1. Now, run a loop and decrement k whenever unmarked cell is encountered. When the value of k is 0, we get the answer.
Below is implementation of this approach:  
C++
Java
Python3
C#
PHP
Javascript
// C++ program to find the K-th smallest element// after removing some integers from natural number.#include <bits/stdc++.h>#define MAX 1000000using namespace std; // Return the K-th smallest element.int ksmallest(int arr[], int n, int k){    // Making an array, and mark all number as unmarked.    int b[MAX];    memset(b, 0, sizeof b);     // Marking the number present in the given array.    for (int i = 0; i < n; i++)        b[arr[i]] = 1;     for (int j = 1; j < MAX; j++) {        // If j is unmarked, reduce k by 1.        if (b[j] != 1)            k--;         // If k is 0 return j.        if (!k)            return j;    }} // Driven Programint main(){    int k = 1;    int arr[] = { 1 };    int n = sizeof(arr) / sizeof(arr[0]);    cout << ksmallest(arr, n, k);    return 0;}
// Java program to find the K-th smallest element// after removing some integers from natural number.class GFG {     static final int MAX = 1000000;     // Return the K-th smallest element.    static int ksmallest(int arr[], int n, int k)    {        // Making an array, and mark        // all number as unmarked.        int b[] = new int[MAX];         // Marking the number present        // in the given array.        for (int i = 0; i < n; i++) {            b[arr[i]] = 1;        }         for (int j = 1; j < MAX; j++) {            // If j is unmarked, reduce k by 1.            if (b[j] != 1) {                k--;            }             // If k is 0 return j.            if (k != 1) {                return j;            }        }        return Integer.MAX_VALUE;    }     // Driven code    public static void main(String[] args)    {        int k = 1;        int arr[] = { 1 };        int n = arr.length;        System.out.println(ksmallest(arr, n, k));    }} // This code has been contributed by 29AjayKumar
# Python program to find the K-th smallest element# after removing some integers from natural number.MAX = 1000000  # Return the K-th smallest element.def ksmallest(arr, n, k):         # Making an array, and mark all number as unmarked.    b = [0]*MAX;     # Marking the number present in the given array.    for i in range(n):        b[arr[i]] = 1;     for j in range(1, MAX):        # If j is unmarked, reduce k by 1.        if (b[j] != 1):            k-= 1;         # If k is 0 return j.        if (k is not 1):            return j;             # Driven Programk = 1;arr = [ 1 ];n = len(arr);print(ksmallest(arr, n, k)); # This code contributed by Rajput-Ji
// C# program to find the K-th smallest element// after removing some integers from natural number.using System; class GFG {     static int MAX = 1000000;     // Return the K-th smallest element.    static int ksmallest(int[] arr, int n, int k)    {        // Making an array, and mark        // all number as unmarked.        int[] b = new int[MAX];         // Marking the number present        // in the given array.        for (int i = 0; i < n; i++) {            b[arr[i]] = 1;        }         for (int j = 1; j < MAX; j++) {            // If j is unmarked, reduce k by 1.            if (b[j] != 1) {                k--;            }             // If k is 0 return j.            if (k != 1) {                return j;            }        }        return int.MaxValue;    }     // Driven code    public static void Main()    {        int k = 1;        int[] arr = { 1 };        int n = arr.Length;        Console.WriteLine(ksmallest(arr, n, k));    }} /* This code contributed by PrinciRaj1992 */
<?php// PHP program to find the K-th smallest element// after removing some integers from natural number.$MAX = 10000; // Return the K-th smallest element.function ksmallest($arr, $n, $k){    global $MAX;         // Making an array, and mark all number as unmarked.    $b=array_fill(0, $MAX, 0);     // Marking the number present in the given array.    for ($i = 0; $i < $n; $i++)        $b[$arr[$i]] = 1;     for ($j = 1; $j < $MAX; $j++)    {        // If j is unmarked, reduce k by 1.        if ($b[$j] != 1)            $k--;         // If k is 0 return j.        if ($k == 0)            return $j;    }} // Driver code    $k = 1;    $arr = array( 1 );    $n = count($arr);    echo ksmallest($arr, $n, $k); // This code is contributed by mits?>
<script> // Javascript program to find the K-th// smallest element after removing some// integers from natural number. let MAX = 1000000; // Return the K-th smallest element.function ksmallest(arr, n, k){         // Making an array, and mark    // all number as unmarked.    let b = [];     // Marking the number present    // in the given array.    for(let i = 0; i < n; i++)    {        b[arr[i]] = 1;    }     for(let j = 1; j < MAX; j++)    {                 // If j is unmarked, reduce k by 1.        if (b[j] != 1)        {            k--;        }         // If k is 0 return j.        if (k != 1)        {            return j;        }    }    return Number.MAX_VALUE;}  // Driver code   let k = 1;let arr = [1];let n = arr.length; document.write(ksmallest(arr, n, k));         // This code is contributed by susmitakundugoaldanga                </script>
Output :  
2
Time Complexity : O(n). Method 2 (Efficient): First, sort the array arr[]. Observe, there will be arr[0] – 1 numbers between 0 and arr[0], similarly, arr[1] – arr[0] – 1 numbers between arr[0] and arr[1] and so on. So, if k lies between arr[i] – arr[i+1] – 1, then return K-th smallest element in the range. Else reduce k by arr[i] – arr[i+1] – 1 i.e., k = k – (arr[i] – arr[i+1] – 1).
Algorithm to solve the problem: 
1. Sort the array arr[].
2. For i = 1 to k. Find c = arr[i+1] - arr[i] -1.
  a) if k - c <= 0, return arr[i-1] + k.
  b) else k = k - c.
Below is implementation of this approach: 
C++
Java
Python3
C#
PHP
Javascript
// C++ program to find the Kth smallest element// after removing some integer from first n// natural number.#include <bits/stdc++.h>using namespace std; // Return the K-th smallest element.int ksmallest(int arr[], int n, int k){    sort(arr, arr + n);     // Checking if k lies before 1st element    if (k < arr[0])        return k;     // If k is the first element of array arr[].    if (k == arr[0])        return arr[0] + 1;     // If k is more than last element    if (k > arr[n - 1])        return k + n;     // If first element of array is 1.    if (arr[0] == 1)        k--;     // Reducing k by numbers before arr[0].    else        k -= (arr[0] - 1);     // Finding k'th smallest element after removing    // array elements.    for (int i = 1; i < n; i++) {        // Finding count of element between i-th        // and (i-1)-th element.        int c = arr[i] - arr[i - 1] - 1;        if (k <= c)            return arr[i - 1] + k;        else            k -= c;    }     return arr[n - 1] + k;} // Driven Programint main(){    int k = 1;    int arr[] = { 1 };    int n = sizeof(arr) / sizeof(arr[0]);    cout << ksmallest(arr, n, k);    return 0;}
// Java program to find the// Kth smallest element after// removing some integer from// first n natural number.import java.util.Arrays;import java.io.*; class GFG {     // Return the K-th    // smallest element.    static int ksmallest(int arr[],                         int n, int k)    {        // sort(arr, arr+n);        Arrays.sort(arr);         // Checking if k lies        // before 1st element        if (k < arr[0])            return k;         // If k is the first        // element of array arr[].        if (k == arr[0])            return arr[0] + 1;         // If k is more        // than last element        if (k > arr[n - 1])            return k + n;         // If first element        // of array is 1.        if (arr[0] == 1)            k--;         // Reducing k by numbers        // before arr[0].        else            k -= (arr[0] - 1);         // Finding k'th smallest        // element after removing        // array elements.        for (int i = 1; i < n; i++) {            // Finding count of            // element between i-th            // and (i-1)-th element.            int c = arr[i] - arr[i - 1] - 1;            if (k <= c)                return arr[i - 1] + k;            else                k -= c;        }         return arr[n - 1] + k;    }     // Driven Code    public static void main(String[] args)    {        int k = 1;        int arr[] = { 1 };        int n = arr.length;        System.out.println(ksmallest(arr, n, k));    }} // This code is contributed// by ajit
# Python3 program to find the Kth# smallest element after# removing some integer from# first n natural number. # Return the K-th# smallest element.def ksmallest(arr, n, k):     arr.sort();     # Checking if k lies    # before 1st element    if (k < arr[0]):        return k;     # If k is the first    # element of array arr[].    if (k == arr[0]):        return arr[0] + 1;     # If k is more    # than last element    if (k > arr[n - 1]):        return k + n;     # If first element    # of array is 1.    if (arr[0] == 1):        k-= 1;     # Reducing k by numbers    # before arr[0].    else:        k -= (arr[0] - 1);     # Finding k'th smallest element    # after removing array elements.    for i in range(1, n):        # Finding count of element between        # i-th and (i-1)-th element.        c = arr[i] - arr[i - 1] - 1;        if (k <= c):            return arr[i - 1] + k;        else:            k -= c;     return arr[n - 1] + k; # Driver Codek = 1;arr =[ 1 ];n = len(arr);print(ksmallest(arr, n, k)); # This code is contributed by mits
// C# program to find the// Kth smallest element after// removing some integer from// first n natural number.using System; class GFG {    // Return the K-th    // smallest element.    static int ksmallest(int[] arr,                         int n, int k)    {        // sort(arr, arr+n);        Array.Sort(arr);         // Checking if k lies        // before 1st element        if (k < arr[0])            return k;         // If k is the first        // element of array arr[].        if (k == arr[0])            return arr[0] + 1;         // If k is more        // than last element        if (k > arr[n - 1])            return k + n;         // If first element        // of array is 1.        if (arr[0] == 1)            k--;         // Reducing k by numbers        // before arr[0].        else            k -= (arr[0] - 1);         // Finding k'th smallest        // element after removing        // array elements.        for (int i = 1; i < n; i++) {            // Finding count of            // element between i-th            // and (i-1)-th element.            int c = arr[i] - arr[i - 1] - 1;            if (k <= c)                return arr[i - 1] + k;            else                k -= c;        }         return arr[n - 1] + k;    }     // Driver Code    static public void Main()    {        int k = 1;        int[] arr = { 1 };        int n = arr.Length;        Console.WriteLine(ksmallest(arr, n, k));    }} // This code is contributed// by ajit
<?php// PHP program to find the Kth// smallest element after// removing some integer from// first n natural number. // Return the K-th// smallest element.function ksmallest($arr, $n, $k){    sort($arr);     // Checking if k lies    // before 1st element    if ($k < $arr[0])        return $k;     // If k is the first    // element of array arr[].    if ($k == $arr[0])        return $arr[0] + 1;     // If k is more    // than last element    if ($k > $arr[$n - 1])        return $k + $n;     // If first element    // of array is 1.    if ($arr[0] == 1)        $k--;     // Reducing k by numbers    // before arr[0].    else        $k -= ($arr[0] - 1);     // Finding k'th smallest element    // after removing array elements.    for ($i = 1; $i < $n; $i++)    {        // Finding count of element between        // i-th and (i-1)-th element.        $c = $arr[$i] - $arr[$i - 1] - 1;        if ($k <= $c)            return $arr[$i - 1] + $k;        else            $k -= $c;    }     return $arr[$n - 1] + $k;} // Driver Code$k = 1;$arr = array ( 1 );$n = sizeof($arr);echo ksmallest($arr, $n, $k); // This code is contributed by aj_36?>
<script> // Javascript program to find the// Kth smallest element after// removing some integer from// first n natural number. // Return the K-th// smallest element.function ksmallest(arr, n, k){         // sort(arr, arr+n);    arr.sort(function(a, b){return a - b});     // Checking if k lies    // before 1st element    if (k < arr[0])        return k;     // If k is the first    // element of array arr[].    if (k == arr[0])        return arr[0] + 1;     // If k is more    // than last element    if (k > arr[n - 1])        return k + n;     // If first element    // of array is 1.    if (arr[0] == 1)        k--;     // Reducing k by numbers    // before arr[0].    else        k -= (arr[0] - 1);     // Finding k'th smallest    // element after removing    // array elements.    for(let i = 1; i < n; i++)    {                 // Finding count of        // element between i-th        // and (i-1)-th element.        let c = arr[i] - arr[i - 1] - 1;        if (k <= c)            return arr[i - 1] + k;        else            k -= c;    }    return arr[n - 1] + k;} // Driver codelet k = 1;let arr = [1];let n = arr.length; document.write(ksmallest(arr, n, k)); // This code is contributed by divyesh072019 </script>
Output : 
2
More efficient method : K-th smallest element after removing given integers from natural numbers | Set 2This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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