Número mínimo de secuencias consecutivas que se pueden formar en una array

Dada una array de enteros. La tarea es encontrar el número mínimo de secuencias consecutivas que se pueden formar usando los elementos de la array. 
Ejemplos: 
 

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}

Acercarse: 

• Ordenar la array.
• Repita la array y verifique si el elemento actual es solo 1 más pequeño que el siguiente elemento. 
   
• Si es así, incremente el conteo en 1.
• Devuelve el recuento final de secuencias consecutivas.

A continuación se muestra la implementación del enfoque anterior: 
 

// 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  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 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# 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 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
?>

<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>

5

Complejidad de tiempo: O (n log n), donde n es el tamaño de la array.

Espacio Auxiliar: O(1)