# 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 <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;` `}` |

## 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.` |

## PHP

`<?php` `// PHP program to find number` `// of subset containing` `// consecutive numbers` `// Returns count of subsets` `// with consecutive numbers` `function` `numofsubset( ` `$arr` `, ` `$n` `)` `{` ` ` ` ` `// Sort the array so that` ` ` `// elements which are` ` ` `// consecutive in nature` ` ` `// became consecutive` ` ` `// in the array.` ` ` `sort(` `$arr` `);` ` ` `// Initialize result` ` ` `$count` `= 1;` ` ` `for` `(` `$i` `= 0; ` `$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` `;` `}` ` ` `// Driver Code` ` ` `$arr` `= ` `array` `(100, 56, 5, 6, 102, 58, 101,` ` ` `57, 7, 103, 59 );` ` ` `$n` `= sizeof(` `$arr` `);` ` ` `echo` `numofsubset(` `$arr` `, ` `$n` `);` ` ` `// This code is contributed by Anuj_67` `?>` |

## Javascript

`<script>` `// javascript program to find number of subset` `// containing consecutive numbers` ` ` `// Returns count of subsets with consecutive numbers` ` ` `function` `numofsubset(arr , n) {` ` ` `// Sort the array so that elements` ` ` `// which are consecutive in nature` ` ` `// became consecutive in the array.` ` ` `arr.sort((a,b)=>a-b);` ` ` `// Initialize result` ` ` `var` `count = 1;` ` ` `for` `(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` ` ` ` ` `var` `arr = [ 100, 56, 5, 6, 102, 58, 101, 57, 7, 103, 59 ];` ` ` `var` `n = arr.length;` ` ` `document.write(numofsubset(arr, n));` `// This code contributed by Rajput-Ji` `</script>` |

**Output**

3

**Time Complexity : O(nlogn)**

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