# Minimum number of consecutive sequences that can be formed in an array

Given an array of integers. The task is to find the minimum number of consecutive sequences that can be formed using the elements of the array.

Examples:

```Input: arr[] = { -3, -2, -1, 0, 2 }
Output: 2
Consecutive sequences are (-3, -2, -1, 0), (2).

Input: arr[] = { 3, 4, 0, 2, 6, 5, 10 }
Output: 3
Consecutive sequences are (0), {2, 3, 4, 5, 6} and {10}
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Sort the array.
• Iterate the array, and check if current element is just 1 smaller than the next element.
• If it is then increment the count by 1.
• Return the final count of consecutive sequences.

Below is the implementation of above approach :

## C++

 `// C++ program find the minimum number of consecutive  ` `// sequences in an array ` `#include ` `using` `namespace` `std; ` ` `  `int` `countSequences(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `count = 1; ` ` `  `    ``sort(arr, arr + n); ` ` `  `    ``for` `(``int` `i = 0; i < n - 1; i++) ` `        ``if` `(arr[i] + 1 != arr[i + 1]) ` `            ``count++; ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver program ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 7, 3, 5, 10 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``// function call to print required answer ` `    ``cout << countSequences(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// Java  program find the minimum number of consecutive  ` `// sequences in an array ` ` `  `import` `java.util.Arrays;  ` `import` `java.io.*; ` ` `  `class` `GFG { ` `     `  `static` `int` `countSequences(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `count = ``1``; ` ` `  `    ``Arrays.sort(arr); ` ` `  `    ``for` `(``int` `i = ``0``; i < n - ``1``; i++) ` `        ``if` `(arr[i] + ``1` `!= arr[i + ``1``]) ` `            ``count++; ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver program ` `    ``public` `static` `void` `main (String[] args) { ` ` `  `    ``int` `arr[] = { ``1``, ``7``, ``3``, ``5``, ``10` `}; ` `    ``int` `n = arr.length; ` `    ``// function call to print required answer ` `    ``System.out.println( countSequences(arr, n)); ` ` `  `    ``} ` `//This code is contributed by ajit.     ` `} `

## Python3

 `# Python3 program find the minimum number of consecutive  ` `# sequences in an array  ` ` `  `def` `countSequences(arr, n) : ` `    ``count ``=` `1` ` `  `    ``arr.sort() ` ` `  `    ``for` `i ``in` `range``( n ``-``1``) :  ` `        ``if` `(arr[i] ``+` `1` `!``=` `arr[i ``+` `1``]) : ` `            ``count ``+``=` `1` ` `  `    ``return` `count  ` `  `  ` `  `# Driver program  ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``arr ``=` `[ ``1``, ``7``, ``3``, ``5``, ``10` `]  ` `    ``n ``=` `len``(arr) ` ` `  `    ``# function call to print required answer  ` `    ``print``(countSequences(arr, n))  ` ` `  `# This code is contributed by Ryuga `

## C#

 `// C# program find the minimum number of consecutive  ` `// sequences in an array ` ` ``using` `System; ` `class` `GFG { ` `      `  `static` `int` `countSequences(``int` `[]arr, ``int` `n) ` `{ ` `    ``int` `count = 1; ` `  `  `    ``Array.Sort(arr); ` `  `  `    ``for` `(``int` `i = 0; i < n - 1; i++) ` `        ``if` `(arr[i] + 1 != arr[i + 1]) ` `            ``count++; ` `  `  `    ``return` `count; ` `} ` `  `  `// Driver program ` `    ``static` `public` `void` `Main (String []args) { ` `  `  `    ``int` `[]arr = { 1, 7, 3, 5, 10 }; ` `    ``int` `n = arr.Length; ` `    ``// function call to print required answer ` `    ``Console.WriteLine( countSequences(arr, n)); ` `  `  `    ``} ` `} ` `//This code is contributed by Arnab Kundu    `

## PHP

 ` `

Output:

```5
```

Time Complexity : O(n log n), where n is the size of the array.

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