# 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.

```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 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 C++ implementation of this approach :

## C++

```// C++ program to find number of subset containing
// consecutive numbers
#include<bits/stdc++.h>
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;
}
```

## Python

```# 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.
```

Output:

```3
```

Time Complexity : O(nlogn)

This article is contributed by Anuj Chauhan(anuj0503). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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