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 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)

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