# 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[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.

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[0]);     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



Output

2

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

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 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[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 Program int main() {     int k = 1;     int arr[] = { 1 };     int n = sizeof(arr) / sizeof(arr[0]);     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[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

 # 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 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[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

 \$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 ?>

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 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[0]);     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

## Python3

 #Python3 code to demonstrate working of above approach def find_kth_smallest_number(arr, k):     unique_nums = set(arr)  # Create a set to store unique elements from the array           num = 1  # Current natural number           while k > 0:         if num in unique_nums:  # If num is already present in the set             num += 1         else:             k -= 1             num += 1           return num - 1  # Subtract 1 to get the k-th smallest number   # Driver code if __name__ == "__main__":     arr = [1, 3]     k = 4           kth_smallest = find_kth_smallest_number(arr, k)     print("K-th smallest number:", kth_smallest)

## C#

 using System; using System.Collections.Generic;   class GFG {     static int FindKthSmallestNumber(int[] arr, int n,                                      int k)     {         HashSet set = new HashSet();           // Add elements from the array to the set         for (int i = 0; i < n; i++) {             set.Add(arr[i]);         }           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     }       static void Main()     {         int[] arr = { 1, 3 };         int n = arr.Length;         int k = 4;           int kthSmallest = FindKthSmallestNumber(arr, n, k);         Console.WriteLine("K-th smallest number: "                           + kthSmallest);     } }

## Javascript

 function findKthSmallestNumber(arr, k) {     const set = new Set(); // Create a new set to store unique elements           // Add elements from the array to the set     for (let i = 0; i < arr.length; i++) {         set.add(arr[i]);     }           let num = 1; // Current natural number           while (k > 0) {         if (set.has(num)) { // Check if the set contains the current number             num++;         } else {             k--;             num++;         }     }           return num - 1; // Subtract 1 to get the k-th smallest number }   // Driver Code const arr = [1, 3]; const k = 4; const kthSmallest = findKthSmallestNumber(arr, k); console.log("K-th smallest number:", kthSmallest);

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[0]);     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

## Python3

 def find_kth_smallest_number(arr, k):     max_num = max(arr)           count = [0] * (max_num + 1)  # Corrected size     for num in arr:         count[num] += 1               num = 1  # Current natural number           while k > 0:         if num < len(count) and count[num] > 0:             count[num] -= 1         else:             k -= 1         num += 1               return num - 1  # Subtract 1 to get the k-th smallest number   def main():     arr = [1, 3]     k = 4           kth_smallest = find_kth_smallest_number(arr, k)     print("K-th smallest number:", kth_smallest)   if __name__ == "__main__":     main()

## C#

 using System;   public class KthSmallestNumber {     public static void Main(string[] args)     {         int[] arr = { 1, 3 };         int k = 4;         int kthSmallest = FindKthSmallestNumber(arr, k);         Console.WriteLine("K-th smallest number: "                           + kthSmallest);     }       public static int FindKthSmallestNumber(int[] arr,                                             int k)     {         int max = int.MinValue;         foreach(int numValue in                     arr) // Change variable name here         {             max = Math.Max(max, numValue);         }           int[] count             = new int[max                       + 1]; // Updated size of count array           foreach(int numValue in                     arr) // Change variable name here         {             count[numValue]++;         }           int currentNum = 1; // Current natural number           while (k > 0) {             if (currentNum < count.Length                 && count[currentNum]                        > 0) // Added boundary check             {                 count[currentNum]--;             }             else {                 k--;             }             currentNum++;         }           return currentNum - 1; // Subtract 1 to get the k-th                                // smallest number     } }

## Javascript

 function findKthSmallestNumber(arr, k) {     let max = Number.MIN_VALUE;       // Find the maximum value in the array     for (let numValue of arr) {         max = Math.max(max, numValue);     }       // Create an array to store the count of each number     const count = new Array(max + 1).fill(0);       // Count the occurrences of each number in the array     for (let numValue of arr) {         count[numValue]++;     }       let currentNum = 1; // Start with the smallest natural number       while (k > 0) {         if (currentNum < count.length && count[currentNum] > 0) {             // If the currentNum is within the array bounds and has occurrences left             count[currentNum]--;         } else {             // Otherwise, decrement k             k--;         }         currentNum++;     }       // Subtract 1 to get the k-th smallest number     return currentNum - 1; }   const arr = [1, 3]; const k = 4; const kthSmallest = findKthSmallestNumber(arr, k); console.log("K-th smallest number: " + kthSmallest);

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
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