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

Given an array arr[] of size ‘n’ and a positive integer k. Consider series of natural numbers and remove arr, arr, arr, …, 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.

Steps to solve the problem:

1. declare an array b of size max.

2. mark complete array as unmarked by 0.

3. iterate through i=0 till n:

* update b[arr[i]] to 1.

4. iterate through j=0 till max:

* check if b[j] is not equal to 1 then decrement k.

* check if k is not equal to 0 then return j.

Below is implementation of this approach:

## C++

 `// C++ program to find the K-th smallest element` `// after removing some integers from natural number.` `#include ` `#define MAX 1000000` `using` `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 Program` `int` `main()` `{` `    ``int` `k = 1;` `    ``int` `arr[] = { 1 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << ksmallest(arr, n, k);` `    ``return` `0;` `}`

## Java

 `// 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`

## Python3

 `# 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 Program` `k ``=` `1``;` `arr ``=` `[ ``1` `];` `n ``=` `len``(arr);` `print``(ksmallest(arr, n, k));`   `# This code contributed by Rajput-Ji`

## C#

 `// 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

 ``

## Javascript

 ``

Output

`2`

Time Complexity: O(MAX)
Auxiliary Space: O(MAX)

Method 2 (Efficient):
First, sort the array arr[]. Observe, there will be arr – 1 numbers between 0 and arr, similarly, arr – arr – 1 numbers between arr and arr 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 n. 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++

 `// C++ program to find the Kth smallest element` `// after removing some integer from first n` `// natural number.` `#include ` `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)` `        ``return` `k;`   `    ``// If k is the first element of array arr[].` `    ``if` `(k == arr)` `        ``return` `arr + 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 == 1)` `        ``k--;`   `    ``// Reducing k by numbers before arr.` `    ``else` `        ``k -= (arr - 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 Program` `int` `main()` `{` `    ``int` `k = 1;` `    ``int` `arr[] = { 1 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << ksmallest(arr, n, k);` `    ``return` `0;` `}`

## Java

 `// 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.` `        ``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

 `# 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.` `    ``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 Code` `k ``=` `1``;` `arr ``=``[ ``1` `];` `n ``=` `len``(arr);` `print``(ksmallest(arr, n, k));`   `# This code is contributed by mits`

## C#

 `// 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)` `            ``return` `k;`   `        ``// If k is the first` `        ``// element of array arr[].` `        ``if` `(k == arr)` `            ``return` `arr + 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 == 1)` `            ``k--;`   `        ``// Reducing k by numbers` `        ``// before arr.` `        ``else` `            ``k -= (arr - 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

 ` ``\$arr``[``\$n` `- 1])` `        ``return` `\$k` `+ ``\$n``;`   `    ``// If first element ` `    ``// of array is 1.` `    ``if` `(``\$arr`` == 1)` `        ``\$k``--;`   `    ``// Reducing k by numbers` `    ``// before arr.` `    ``else` `        ``\$k` `-= (``\$arr`` - 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` `?>`

## Javascript

 ``

Output

`2`

Time Complexity: O(nlog(n))
Auxiliary Space: O(1)

Set Approach:

• Create a set to store the elements of the array.
• Iterate through the array and add each element to the set.
• Start changing num to 1, which represents the current generation number.
If k is greater than 0:
If the set contains num, continue by incrementing num.
If num is not available, reduce k by 1 .
Increase num by1.
• Returns the value of num as the k-th smallest number.

## C++

 `#include ` `#include ` `using` `namespace` `std;`   `int` `findKthSmallestNumber(``int` `arr[], ``int` `n, ``int` `k) {` `    ``unordered_set<``int``> set;` `    `  `    ``// Add elements from the array to the set` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``set.insert(arr[i]);` `    ``}` `    `  `    ``int` `num = 1; ``// Current natural number` `    `  `    ``while` `(k > 0) {` `        ``if` `(set.count(num)) {` `            ``num++;` `        ``} ``else` `{` `            ``k--;` `            ``num++;` `        ``}` `    ``}` `    `  `    ``return` `num - 1; ``// Subtract 1 to get the k-th smallest number` `}`   `int` `main() {` `    ``int` `arr[] = {1, 3};` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``int` `k = 4;` `    `  `    ``int` `kthSmallest = findKthSmallestNumber(arr, n, k);` `    ``cout << ``"K-th smallest number: "` `<< kthSmallest << endl;` `    `  `    ``return` `0;` `}`

## Java

 `/*package whatever //do not write package name here */`   `import` `java.util.HashSet;` `import` `java.util.Set;`   `public` `class` `KthSmallestNumber {` `    `  `    ``public` `static` `void` `main(String[] args) {` `        ``int``[] arr = {``1``, ``3``};` `        ``int` `k = ``4``;` `        ``int` `kthSmallest = findKthSmallestNumber(arr, k);` `        ``System.out.println(``"K-th smallest number: "` `+ kthSmallest);` `    ``}` `    `  `    ``public` `static` `int` `findKthSmallestNumber(``int``[] arr, ``int` `k) {` `        ``Set set = ``new` `HashSet<>();` `        `  `        ``// Add elements from the array to the set` `        ``for` `(``int` `num : arr) {` `            ``set.add(num);` `        ``}` `        `  `        ``int` `num = ``1``; ``// Current natural number` `        `  `        ``while` `(k > ``0``) {` `            ``if` `(set.contains(num)) {` `                ``num++;` `            ``} ``else` `{` `                ``k--;` `                ``num++;` `            ``}` `        ``}` `        `  `        ``return` `num - ``1``; ``// Subtract 1 to get the k-th smallest number` `    ``}` `}`   `//This code is contributed by Sovi`

Output

`K-th smallest number: 6`

Time Complexity: It takes O(n) time to build a set by iterating through an array, where n is the size of the array.
Iterating through the natural numbers until the kth smallest number is reached takes O(k) time in the worst case.
Thus, the total time complexity is O(n + k).
Space Complexity: O(n), where n is the size of the array, as it can potentially store all the distinct elements of the array.

Counting Approach:

• Create an array of size max+2, where max is the maximum number of elements in the given array.
• Initialize all elements of the count array to 0 .
• Increase the number of each element in the count array by iterating through the given array.
• Start changing num to 1, which represents the current generation number.
If k is greater than 0:
If the number of nums in the count array is greater than 0, decrement the count by 1 and continue.
If the count is 0, decrease k by 1 .
Increase num by
• Returns the num value as the k-th smallest number

## C++

 `#include ` `#include ` `using` `namespace` `std;`   `int` `findKthSmallestNumber(``int` `arr[], ``int` `n, ``int` `k) {` `    ``int` `max_num = *max_element(arr, arr + n);` `    `  `    ``int``* count = ``new` `int``[max_num + 2]();` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``count[arr[i]]++;` `    ``}` `    `  `    ``int` `num = 1; ``// Current natural number` `    `  `    ``while` `(k > 0) {` `        ``if` `(count[num] > 0) {` `            ``count[num]--;` `        ``} ``else` `{` `            ``k--;` `        ``}` `        ``num++;` `    ``}` `    `  `    ``delete``[] count;` `    ``return` `num - 1; ``// Subtract 1 to get the k-th smallest number` `}`   `int` `main() {` `    ``int` `arr[] = {1, 3};` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``int` `k = 4;` `    `  `    ``int` `kthSmallest = findKthSmallestNumber(arr, n, k);` `    ``cout << ``"K-th smallest number: "` `<< kthSmallest << endl;` `    `  `    ``return` `0;` `}`

## Java

 `/*package whatever //do not write package name here */`   `public` `class` `KthSmallestNumber {` `    `  `    ``public` `static` `void` `main(String[] args) {` `        ``int``[] arr = {``1``, ``3``};` `        ``int` `k = ``4``;` `        ``int` `kthSmallest = findKthSmallestNumber(arr, k);` `        ``System.out.println(``"K-th smallest number: "` `+ kthSmallest);` `    ``}` `    `  `    ``public` `static` `int` `findKthSmallestNumber(``int``[] arr, ``int` `k) {` `        ``int` `max = Integer.MIN_VALUE;` `        ``for` `(``int` `num : arr) {` `            ``max = Math.max(max, num);` `        ``}` `        `  `        ``int``[] count = ``new` `int``[max + ``1``]; ``// Updated size of count array` `        `  `        ``for` `(``int` `num : arr) {` `            ``count[num]++;` `        ``}` `        `  `        ``int` `num = ``1``; ``// Current natural number` `        `  `        ``while` `(k > ``0``) {` `            ``if` `(num < count.length && count[num] > ``0``) { ``// Added boundary check` `                ``count[num]--;` `            ``} ``else` `{` `                ``k--;` `            ``}` `            ``num++;` `        ``}` `        `  `        ``return` `num - ``1``; ``// Subtract 1 to get the k-th smallest number` `    ``}` `}`   `//This code is contributed by Sovi`

Output

`K-th smallest number: 6`

Time Complexity: It takes O(n) time to find the largest element in the array, where n is the size of the array.
When you start an accounting system, it takes O(n) time to count every occurrence in the system.
Iterating through the natural numbers until the kth smallest number is reached takes O(k) time in the worst case.
Thus, the total time complexity is O(n + k).

Space Complexity: O(n), where n is the size of the array.

More efficient method : K-th smallest element after removing given integers from natural numbers | Set 2
This 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 review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.