# Count minimum number of subsets (or subsequences) with consecutive numbers

Given an array of distinct positive numbers, the task is to calculate the number of subsets (or subsequences) from the array such that each subset contains consecutive numbers.

Examples:

`Input :  arr[] = {100, 56, 5, 6, 102, 58,                             101, 57, 7, 103, 59}Output : 3{5, 6, 7}, { 56, 57, 58, 59}, {100, 101, 102, 103}are 3 subset in which numbers are consecutive.Input :  arr[] = {10, 100, 105}Output : 3{10}, {100} and {105} are 3 subset in which numbers are consecutive. `

The idea is to sort the array and traverse the sorted array to count the number of such subsets. To count the number of such subsets, we need to count the consecutive numbers such that the difference between them is not equal to one.

Following is the algorithm for the finding number of subset containing consecutive numbers:

`1. Sort the array arr[ ] and count = 1.2. Traverse the sorted array and for each element arr[i].   If arr[i] + 1 != arr[i+1],        then increment the count by one.3. Return the count. `

Below is the implementation of this approach :

## C++

 `// C++ program to find number of subset containing``// consecutive numbers``#include ``using` `namespace` `std;` `// Returns count of subsets with consecutive numbers``int` `numofsubset(``int` `arr[], ``int` `n)``{``    ``// Sort the array so that elements which are``    ``// consecutive in nature became consecutive``    ``// in the array.``    ``sort(arr, arr + n);` `    ``int` `count = 1; ``// Initialize result``    ``for` `(``int` `i = 0; i < n - 1; i++) {``        ``// Check if there is beginning of another``        ``// subset of consecutive number``        ``if` `(arr[i] + 1 != arr[i + 1])``            ``count++;``    ``}` `    ``return` `count;``}` `// Driven Program``int` `main()``{``    ``int` `arr[] = { 100, 56, 5, 6, 102, 58, 101,``                  ``57, 7, 103, 59 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << numofsubset(arr, n) << endl;``    ``return` `0;``}`

## Java

 `// Java program to find number of subset``// containing consecutive numbers``import` `java.util.*;``class` `GFG {` `    ``// Returns count of subsets with consecutive numbers``    ``static` `int` `numofsubset(``int` `arr[], ``int` `n)``    ``{``        ``// Sort the array so that elements``        ``// which are consecutive in nature``        ``// became consecutive in the array.``        ``Arrays.sort(arr);` `        ``// Initialize result``        ``int` `count = ``1``;``        ``for` `(``int` `i = ``0``; i < n - ``1``; i++) {``            ``// Check if there is beginning``            ``// of another subset of``            ``// consecutive number``            ``if` `(arr[i] + ``1` `!= arr[i + ``1``])``                ``count++;``        ``}` `        ``return` `count;``    ``}` `    ``// Driven Program``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``100``, ``56``, ``5``, ``6``, ``102``, ``58``, ``101``,``                      ``57``, ``7``, ``103``, ``59` `};``        ``int` `n = arr.length;``        ``System.out.println(numofsubset(arr, n));``    ``}``}` `// This code is contributed by prerna saini.`

## Python3

 `# Python program to find number of subset containing ``# consecutive numbers``def` `numofsubset(arr, n):` `  ``# Sort the array so that elements which are consecutive``  ``# in nature became consecutive in the array.``  ``x ``=` `sorted``(arr)`` ` `  ``count ``=` `1`` ` `  ``for` `i ``in` `range``(``0``, n``-``1``):` `    ``# Check if there is beginning of another subset of ``    ``# consecutive number``    ``if` `(x[i] ``+` `1` `!``=` `x[i ``+` `1``]):``      ``count ``=` `count ``+` `1`` ` `  ``return` `count` `# Driven Program``arr ``=` `[ ``100``, ``56``, ``5``, ``6``, ``102``, ``58``, ``101``, ``57``, ``7``, ``103``, ``59` `]``n ``=` `len``(arr)``print` `(numofsubset(arr, n))` `# This code is contributed by Afzal Ansari.`

## C#

 `// C# program to find number of subset``// containing consecutive numbers``using` `System;` `class` `GFG {` `    ``// Returns count of subsets with ``    ``// consecutive numbers``    ``static` `int` `numofsubset(``int``[] arr, ``int` `n)``    ``{``        ``// Sort the array so that elements``        ``// which are consecutive in nature``        ``// became consecutive in the array.``        ``Array.Sort(arr);` `        ``// Initialize result``        ``int` `count = 1;``        ``for` `(``int` `i = 0; i < n - 1; i++) {``            ` `            ``// Check if there is beginning``            ``// of another subset of``            ``// consecutive number``            ``if` `(arr[i] + 1 != arr[i + 1])``                ``count++;``        ``}` `        ``return` `count;``    ``}` `    ``// Driven Program``    ``public` `static` `void` `Main()``    ``{``        ``int``[] arr = { 100, 56, 5, 6, 102, 58, 101,``                                 ``57, 7, 103, 59 };``        ``int` `n = arr.Length;``        ``Console.WriteLine(numofsubset(arr, n));``    ``}``}` `// This code is contributed by vt_m.`

## Javascript

 ``

## PHP

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Output
```3

```

Time Complexity : O(nlogn)

Auxiliary Space :O(1) , since no extra space required.

### Using Hashing:

Follow the steps below to implement the approach:

1. Create an unordered set to keep track of the presence of each element in the array.
2. Iterate through the array and add each element to the set.
3. Iterate through the array again and for each element, check if its previous element is present in the subset. If not, start a new subset and add all consecutive elements to it.
4. Increment the count variable for each new subset.
5. Return the count variable which represents the minimum number of subsets required.

Below is the implementation of the above approach:

## C++

 `// C++ code to implement hashing approach``#include ``using` `namespace` `std;` `int` `countSubsets(``int` `arr[], ``int` `n) {``    ``unordered_set<``int``> s;``    ``int` `count = 0;``    ``// iterate through the array and add each element to the set``    ``for``(``int` `i = 0; i < n; i++) {``        ``s.insert(arr[i]);``    ``}``    ``// iterate through the array again and for each element, check if it is the starting element of a subset``    ``for``(``int` `i = 0; i < n; i++) {``        ``if``(s.find(arr[i]-1) == s.end()) {``            ``int` `j = arr[i];``            ``// count the number of consecutive elements and add them to the set``            ``while``(s.find(j) != s.end()) {``                ``j++;``            ``}``            ``count++;``        ``}``    ``}``    ``return` `count;``}``// driver code``int` `main() {``    ``int` `arr[] = {100, 56, 5, 6, 102, 58, 101, 57, 7, 103};``    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);``    ``int` `subsets = countSubsets(arr, n);``    ``cout  << subsets << endl;``    ``return` `0;``}``// This code is contributed by Veerendra_Singh_Rajpoot`

## Java

 `// Java code to implement hashing approach``import` `java.util.HashSet;` `public` `class` `GFG {``    ``public` `static` `int` `countSubsets(``int``[] arr, ``int` `n) {``        ``HashSet set = ``new` `HashSet<>();``        ``int` `count = ``0``;``        ``// Iterate through the array and add each element to the set``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``set.add(arr[i]);``        ``}``        ``// Iterate through the array again and for each element, ``       ``// check if it is the starting element of a subset``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``if` `(!set.contains(arr[i] - ``1``)) {``                ``int` `j = arr[i];``                ``// Count the number of consecutive elements ``               ``// and add them to the set``                ``while` `(set.contains(j)) {``                    ``j++;``                ``}``                ``count++;``            ``}``        ``}``        ``return` `count;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args) {``        ``int``[] arr = {``100``, ``56``, ``5``, ``6``, ``102``, ``58``, ``101``, ``57``, ``7``, ``103``};``        ``int` `n = arr.length;``        ``int` `subsets = countSubsets(arr, n);``        ``System.out.println(subsets);``    ``}``}`

## Python3

 `def` `count_subsets(arr, n):``    ``s ``=` `set``()``    ``count ``=` `0``    ` `    ``# Iterate through the array and add each element to the set``    ``for` `i ``in` `range``(n):``        ``s.add(arr[i])``    ` `    ``# Iterate through the array again and for each element, check if it is the starting element of a subset``    ``for` `i ``in` `range``(n):``        ``if` `(arr[i] ``-` `1``) ``not` `in` `s:``            ``j ``=` `arr[i]``            ``# Count the number of consecutive elements and add them to the set``            ``while` `j ``in` `s:``                ``j ``+``=` `1``            ``count ``+``=` `1``            ` `    ``return` `count` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ``arr ``=` `[``100``, ``56``, ``5``, ``6``, ``102``, ``58``, ``101``, ``57``, ``7``, ``103``]``    ``n ``=` `len``(arr)``    ``subsets ``=` `count_subsets(arr, n)``    ``print``(subsets)``# This code is contributed by akshitaguprzj3`

## C#

 `using` `System;``using` `System.Collections.Generic;` `class` `SubsetsCount``{``    ``static` `int` `CountSubsets(``int``[] arr, ``int` `n)``    ``{``        ``// Use HashSet for efficient lookup``        ``HashSet<``int``> ``set` `= ``new` `HashSet<``int``>();``        ``int` `count = 0;` `        ``// Iterate through the array and add each element to the set``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``set``.Add(arr[i]);``        ``}` `        ``// Iterate through the array again and for each element, check if it is the starting``        ``// element of a subset``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``if` `(!``set``.Contains(arr[i] - 1))``            ``{``                ``int` `j = arr[i];` `                ``// Count the number of consecutive elements and add them to the set``                ``while` `(``set``.Contains(j))``                ``{``                    ``j++;``                ``}` `                ``count++;``            ``}``        ``}` `        ``return` `count;``    ``}` `    ``// Driver code``    ``static` `void` `Main()``    ``{``        ``int``[] arr = { 100, 56, 5, 6, 102, 58, 101, 57, 7, 103 };``        ``int` `n = arr.Length;` `        ``// Function call``        ``int` `subsets = CountSubsets(arr, n);``        ``Console.WriteLine(subsets);``    ``}``}`

## Javascript

 `// Javascript code to implement hashing approach``function` `countSubsets(arr) {``    ``let s = ``new` `Set();``    ``let count = 0;``    ``// iterate through the array and add each element to the set``    ``for``(let i = 0; i < arr.length; i++) {``        ``s.add(arr[i]);``    ``}``    ``// iterate through the array again and for each element, ``    ``// check if it is the starting element of a subset``    ``for``(let i = 0; i < arr.length; i++) {``        ``if``(!s.has(arr[i]-1)) {``            ``let j = arr[i];``            ``// count the number of consecutive elements and ``            ``// add them to the set``            ``while``(s.has(j)) {``                ``j++;``            ``}``            ``count++;``        ``}``    ``}``    ``return` `count;``}``// driver code``let arr = [100, 56, 5, 6, 102, 58, 101, 57, 7, 103];``let subsets = countSubsets(arr);``console.log(subsets);`

Output
```3

```

Time Complexity : O(n) , where n is the number of element in the array

Auxiliary Space :O(n) , As we need to store all the element in the Unordered set.

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