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

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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}

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 <bits/stdc++.h>
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

<?php
// PHP program find the minimum number 
// of consecutive sequences in an array
 
function countSequences($arr, $n)
{
    $count = 1;
 
    sort($arr);
 
    for ($i = 0; $i < $n - 1; $i++)
        if ($arr[$i] + 1 != $arr[$i + 1])
            $count++;
 
    return $count;
}
 
// Driver Code
$arr = array( 1, 7, 3, 5, 10 );
$n = count($arr);
 
// function call to print required answer
echo countSequences($arr, $n);
 
// This code is contributed by inder_verma
?>

Javascript

<script>
 
    // Javascript program find the 
    // minimum number of consecutive 
    // sequences in an array
     
    function countSequences(arr, n)
    {
        let count = 1;
 
        arr.sort(function(a, b){return a - b});
 
        for (let i = 0; i < n - 1; i++)
            if (arr[i] + 1 != arr[i + 1])
                count++;
 
        return count;
    }
     
    let arr = [ 1, 7, 3, 5, 10 ];
    let n = arr.length;
     
    // function call to print required answer
    document.write(countSequences(arr, n));
     
</script>

Output:

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

Auxiliary Space: O(1)

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