Check if array elements are consecutive

Given an unsorted array of numbers, write a function that returns true if the array consists of consecutive numbers. 
Examples: 
a) If the array is {5, 2, 3, 1, 4}, then the function should return true because the array has consecutive numbers from 1 to 5.
b) If the array is {83, 78, 80, 81, 79, 82}, then the function should return true because the array has consecutive numbers from 78 to 83.
c) If the array is {34, 23, 52, 12, 3}, then the function should return false because the elements are not consecutive.
d) If the array is {7, 6, 5, 5, 3, 4}, then the function should return false because 5 and 5 are not consecutive.

Method 1 (Use Sorting)
1) Sort all the elements. 
2) Do a linear scan of the sorted array. If the difference between the current element and the next element is anything other than 1, then return false. If all differences are 1, then return true.

C++

#include <bits/stdc++.h>
using namespace std;

// Function to Check if array
// elements are consecutive

bool areConsecutive(int arr[], int n)
{
    //Sort the array
    sort(arr,arr+n);
    // checking the adjacent elements
    for(int i=1;i<n;i++)
    {
        if(arr[i]!=arr[i-1]+1)
        {
            return false;
        }
    }
    return true;
}

/* Driver program to test above functions */
int main()
{
    int arr[]= {5, 4, 2, 3, 1, 6};
    int n = sizeof(arr)/sizeof(arr[0]);
    if(areConsecutive(arr, n) == true)
        cout<<" Array elements are consecutive ";
    else
        cout<<" Array elements are not consecutive ";
    return 0;
}

// This code is contributed by Aarti_Rathi

Java

// Java implementation of the approach
import java.util.Arrays;

class AreConsecutive {

/* The function checks if the array elements are consecutive
If elements are consecutive, then returns true, else returns
false */

boolean areConsecutive(int arr[], int n)
{
    //Sort the array
    Arrays.sort(arr);
    // checking the adjacent elements
    for(int i=1;i<n;i++)
    {
        if(arr[i]!=arr[i-1]+1)
        {
            return false;
        }
    }
    return true;
}



    public static void main(String[] args)
    {
        AreConsecutive consecutive = new AreConsecutive();
        int arr[] = {5, 4, 2, 3, 1, 6};
        int n = arr.length;
        if (consecutive.areConsecutive(arr, n) == true)
            System.out.println("Array elements are consecutive");
        else
            System.out.println("Array elements are not consecutive");
    }
}

// This code is contributed by Aarti_Rathi

Python

# The function checks if the array elements
# are consecutive. If elements are consecutive,
# then returns true, else returns false
def areConsecutive(arr, n):

    # Sort the array
    arr.sort()
    # checking the adjacent elements
    for i in range (1,n):
        if(arr[i]!=arr[i-1]+1):
            return False;

    return True;

# Driver Code
arr = [5, 4, 2, 3, 1, 6]
n = len(arr)
if(areConsecutive(arr, n) == True):
    print("Array elements are consecutive ")
else:
    print("Array elements are not consecutive ")

# This code is contributed by Aarti_Rathi

C#

using System;

class GFG {

// Function to Check if array
// elements are consecutive

static bool areConsecutive(int []arr, int n)
{
    //Sort the array
    Array.Sort(arr);
    // checking the adjacent elements
    for(int i=1;i<n;i++)
    {
        if(arr[i]!=arr[i-1]+1)
        {
            return false;
        }
    }
    return true;
}

    /* Driver program to test above
    functions */
    public static void Main()
    {
        int []arr = {5, 4, 2, 3, 1, 6};
        int n = arr.Length;

        if (areConsecutive(arr, n) == true)
            Console.Write("Array elements "
                      + "are consecutive");
        else
            Console.Write("Array elements "
                  + "are not consecutive");
    }
}

// This code is contributed by Aarti_Rathi

Javascript

//JS code to implement the approach

// The function checks if the array elements
// are consecutive. If elements are consecutive,
// then returns true, else returns false
function areConsecutive(arr, n)
{

    // Sort the array
    arr.sort();
    // checking the adjacent elements
    for (var i = 1; i < n; i++)
        if(arr[i]!=arr[i-1]+1)
            return false;

    return true;
}

// Driver Code
var arr = [5, 4, 2, 3, 1, 6];
var n = arr.length;
if(areConsecutive(arr, n) == true)
    console.log("Array elements are consecutive ");
else
    console.log("Array elements are not consecutive ");

// This code is contributed by phasing17

Output

 Array elements are consecutive 

Time Complexity: O(n log n) 
Space Complexity: O(1) 

Method 2 (Use visited array)
The idea is to check for the following two conditions. If the following two conditions are true, then return true. 
1) max – min + 1 = n where max is the maximum element in the array, min is the minimum element in the array and n is the number of elements in the array. 
2) All elements are distinct.
To check if all elements are distinct, we can create a visited[] array of size n. We can map the ith element of input array arr[] to the visited array by using arr[i] – min as the index in visited[]. 

C++

#include<stdio.h>
#include<stdlib.h>

/* Helper functions to get minimum and maximum in an array */
int getMin(int arr[], int n);
int getMax(int arr[], int n);

/* The function checks if the array elements are consecutive
  If elements are consecutive, then returns true, else returns
  false */
bool areConsecutive(int arr[], int n)
{
  if ( n <  1 )
    return false;

  /* 1) Get the minimum element in array */
  int min = getMin(arr, n);

  /* 2) Get the maximum element in array */
  int max = getMax(arr, n);

  /* 3) max - min + 1 is equal to n,  then only check all elements */
  if (max - min  + 1 == n)
  {
      /* Create a temp array to hold visited flag of all elements.
         Note that, calloc is used here so that all values are initialized
         as false */
      bool *visited = (bool *) calloc (n, sizeof(bool));
      int i;
      for (i = 0; i < n; i++)
      {
         /* If we see an element again, then return false */
         if ( visited[arr[i] - min] != false )
           return false;

         /* If visited first time, then mark the element as visited */
         visited[arr[i] - min] = true;
      }

      /* If all elements occur once, then return true */
      return true;
  }

  return false; // if (max - min  + 1 != n)
}

/* UTILITY FUNCTIONS */
int getMin(int arr[], int n)
{
  int min = arr[0];
  for (int i = 1; i < n; i++)
   if (arr[i] < min)
     min = arr[i];
  return min;
}

int getMax(int arr[], int n)
{
  int max = arr[0];
  for (int i = 1; i < n; i++)
   if (arr[i] > max)
     max = arr[i];
  return max;
}

/* Driver program to test above functions */
int main()
{
    int arr[]= {5, 4, 2, 3, 1, 6};
    int n = sizeof(arr)/sizeof(arr[0]);
    if(areConsecutive(arr, n) == true)
        printf(" Array elements are consecutive ");
    else
        printf(" Array elements are not consecutive ");
    getchar();
    return 0;
}

Java

class AreConsecutive
{
    /* The function checks if the array elements are consecutive
       If elements are consecutive, then returns true, else returns
       false */
    boolean areConsecutive(int arr[], int n)
    {
        if (n < 1)
            return false;

        /* 1) Get the minimum element in array */
        int min = getMin(arr, n);

        /* 2) Get the maximum element in array */
        int max = getMax(arr, n);

        /* 3) max - min + 1 is equal to n,  then only check all elements */
        if (max - min + 1 == n)
        {
            /* Create a temp array to hold visited flag of all elements.
               Note that, calloc is used here so that all values are initialized
               as false */
            boolean visited[] = new boolean[n];
            int i;
            for (i = 0; i < n; i++)
            {
                /* If we see an element again, then return false */
                if (visited[arr[i] - min] != false)
                    return false;

                /* If visited first time, then mark the element as visited */
                visited[arr[i] - min] = true;
            }

            /* If all elements occur once, then return true */
            return true;
        }
        return false; // if (max - min  + 1 != n)
    }

    /* UTILITY FUNCTIONS */
    int getMin(int arr[], int n)
    {
        int min = arr[0];
        for (int i = 1; i < n; i++)
        {
            if (arr[i] < min)
                min = arr[i];
        }
        return min;
    }

    int getMax(int arr[], int n)
    {
        int max = arr[0];
        for (int i = 1; i < n; i++)
        {
            if (arr[i] > max)
                max = arr[i];
        }
        return max;
    }

    /* Driver program to test above functions */
    public static void main(String[] args)
    {
        AreConsecutive consecutive = new AreConsecutive();
        int arr[] = {5, 4, 2, 3, 1, 6};
        int n = arr.length;
        if (consecutive.areConsecutive(arr, n) == true)
            System.out.println("Array elements are consecutive");
        else
            System.out.println("Array elements are not consecutive");
    }
}

// This code has been contributed by Mayank Jaiswal

Python3

# Helper functions to get Minimum and
# Maximum in an array

# The function checks if the array elements
# are consecutive. If elements are consecutive,
# then returns true, else returns false
def areConsecutive(arr, n):

    if ( n < 1 ):
        return False

    # 1) Get the Minimum element in array */
    Min = min(arr)

    # 2) Get the Maximum element in array */
    Max = max(arr)

    # 3) Max - Min + 1 is equal to n,
    # then only check all elements */
    if (Max - Min + 1 == n):

        # Create a temp array to hold visited
        # flag of all elements. Note that, calloc
        # is used here so that all values are
        # initialized as false
        visited = [False for i in range(n)]

        for i in range(n):

            # If we see an element again,
            # then return false */
            if (visited[arr[i] - Min] != False):
                return False

            # If visited first time, then mark
            # the element as visited */
            visited[arr[i] - Min] = True

        # If all elements occur once,
        # then return true */
        return True

    return False # if (Max - Min + 1 != n)

# Driver Code
arr = [5, 4, 2, 3, 1, 6]
n = len(arr)
if(areConsecutive(arr, n) == True):
    print("Array elements are consecutive ")
else:
    print("Array elements are not consecutive ")

# This code is contributed by mohit kumar

C#

using System;

class GFG {

    /* The function checks if the array elements
    are consecutive If elements are consecutive,
    then returns true, else returns    false */
    static bool areConsecutive(int []arr, int n)
    {
        if (n < 1)
            return false;

        /* 1) Get the minimum element in array */
        int min = getMin(arr, n);

        /* 2) Get the maximum element in array */
        int max = getMax(arr, n);

        /* 3) max - min + 1 is equal to n, then
        only check all elements */
        if (max - min + 1 == n)
        {

            /* Create a temp array to hold visited
            flag of all elements. Note that, calloc
            is used here so that all values are
            initialized as false */
            bool []visited = new bool[n];
            int i;

            for (i = 0; i < n; i++)
            {

                /* If we see an element again, then
                return false */
                if (visited[arr[i] - min] != false)
                    return false;

                /* If visited first time, then mark
                the element as visited */
                visited[arr[i] - min] = true;
            }

            /* If all elements occur once, then
            return true */
            return true;
        }
        return false; // if (max - min + 1 != n)
    }

    /* UTILITY FUNCTIONS */
    static int getMin(int []arr, int n)
    {
        int min = arr[0];

        for (int i = 1; i < n; i++)
        {
            if (arr[i] < min)
                min = arr[i];
        }

        return min;
    }

    static int getMax(int []arr, int n)
    {
        int max = arr[0];

        for (int i = 1; i < n; i++)
        {
            if (arr[i] > max)
                max = arr[i];
        }

        return max;
    }

    /* Driver program to test above functions */
    public static void Main()
    {
        int []arr = {5, 4, 2, 3, 1, 6};
        int n = arr.Length;

        if (areConsecutive(arr, n) == true)
            Console.Write("Array elements are"
                              + " consecutive");
        else
            Console.Write("Array elements are"
                         + " not consecutive");
    }
}

// This code is contributed by nitin mittal.

PHP

<?php
// PHP Program for above approach

// The function checks if the array elements
// are consecutive. If elements are consecutive,
// then returns true, else returns false
function areConsecutive($arr, $n)
{
    if ( $n < 1 )
        return false;

    // 1) Get the minimum element in array
    $min = getMin($arr, $n);

    // 2) Get the maximum element in array
    $max = getMax($arr, $n);

    // 3) $max - $min + 1 is equal to $n,
    // then only check all elements
    if ($max - $min + 1 == $n)
    {

        // Create a temp array to hold
        // visited flag of all elements.
        $visited = array();
        for ($i = 0; $i < $n; $i++)
        {
            $visited[$i] = false;
        }
        for ($i = 0; $i < $n; $i++)
        {
            // If we see an element again,
            // then return false
            if ( $visited[$arr[$i] - $min] != false )
            return false;

            // If visited first time, then mark
            // the element as visited
            $visited[$arr[$i] - $min] = true;
        }

        // If all elements occur once,
        // then return true
        return true;
    }

    return false; // if ($max - $min + 1 != $n)
}

// UTILITY FUNCTIONS
function getMin($arr, $n)
{
    $min = $arr[0];
    for ($i = 1; $i < $n; $i++)
        if ($arr[$i] < $min)
            $min = $arr[$i];
    return $min;
}

function getMax($arr, $n)
{
    $max = $arr[0];
    for ($i = 1; $i < $n; $i++)
        if ($arr[$i] > $max)
            $max = $arr[$i];
    return $max;
}

// Driver Code
$arr = array(5, 4, 2, 3, 1, 6);
$n = count($arr);
if(areConsecutive($arr, $n) == true)
    echo "Array elements are consecutive ";
else
    echo "Array elements are not consecutive ";

// This code is contributed by rathbhupendra
?>

Javascript

<script>

    /* The function checks if the array elements are consecutive
       If elements are consecutive, then returns true, else returns
       false */
    function areConsecutive(arr,n)
    {
        if (n < 1)
            return false;

        /* 1) Get the minimum element in array */
        let min = getMin(arr, n);

         /* 2) Get the maximum element in array */
        let max = getMax(arr, n);

        /* 3) max - min + 1 is equal to n,  then only check all elements */
        if (max - min + 1 == n)
        {
            /* Create a temp array to hold visited flag of all elements.
               Note that, calloc is used here so that all values are initialized
               as false */
            let visited = new Array(n);
            for(let i=0;i<n;i++)
            {
                visited[i]=false;
            }
            let i;
            for (i = 0; i < n; i++)
            {
                /* If we see an element again, then return false */
                if (visited[arr[i] - min] != false)
                {
                    return false;
                }
                /* If visited first time, then mark the element as visited */
                visited[arr[i] - min] = true;
            }
            /* If all elements occur once, then return true */
            return true;
        }
        return false; // if (max - min  + 1 != n)
    }

    /* UTILITY FUNCTIONS */
    function getMin(arr, n)
    {
        let min = arr[0];
        for (let i = 1; i < n; i++)
        {
            if (arr[i] < min)
                min = arr[i];
        }
        return min;
    }
    function getMax(arr,n)
    {
        let max = arr[0];
        for (let i = 1; i < n; i++)
        {
            if (arr[i] > max)
                max = arr[i];

        }
        return max;
    }

    /* Driver program to test above functions */
    let arr=[5, 4, 2, 3, 1, 6]
    let  n = arr.length;
    if (areConsecutive(arr, n))
    {
        document.write("Array elements are consecutive");
    }
    else
    {
        document.write("Array elements are not consecutive");
    }
    // This code is contributed by avanitrachhadiya2155

</script>

Output

 Array elements are consecutive 

Time Complexity: O(n) 
Auxiliary Space: O(n) 

Method 3 (Mark visited array elements as negative)
This method is O(n) time complexity and O(1) extra space, but it changes the original array, and it works only if all numbers are positive. We can get the original array by adding an extra step though. It is an extension of method 2, and it has the same two steps. 
1) max – min + 1 = n where max is the maximum element in the array, min is the minimum element in the array and n is the number of elements in the array. 
2) All elements are distinct.
In this method, the implementation of step 2 differs from method 2. Instead of creating a new array, we modify the input array arr[] to keep track of visited elements. The idea is to traverse the array and for each index i (where 0 ≤ i < n), make arr[arr[i] – min]] as a negative value. If we see a negative value again then there is repetition. 

C++

#include<stdio.h>
#include<stdlib.h>

/* Helper functions to get minimum and maximum in an array */
int getMin(int arr[], int n);
int getMax(int arr[], int n);

/* The function checks if the array elements are consecutive
  If elements are consecutive, then returns true, else returns
  false */
bool areConsecutive(int arr[], int n)
{

    if ( n <  1 )
        return false;

    /* 1) Get the minimum element in array */
    int min = getMin(arr, n);

    /* 2) Get the maximum element in array */
    int max = getMax(arr, n);

    /* 3) max - min + 1 is equal to n then only check all elements */
    if (max - min  + 1 == n)
    {
        int i;
        for(i = 0; i < n; i++)
        {
            int j;

            if (arr[i] < 0)
                j = -arr[i] - min;
            else
                j = arr[i] - min;

            // if the value at index j is negative then
            // there is repetition
            if (arr[j] > 0)
                arr[j] = -arr[j];
            else
                return false;
        }

        /* If we do not see a negative value then all elements
           are distinct */
        return true;
    }

    return false; // if (max - min  + 1 != n)
}

/* UTILITY FUNCTIONS */
int getMin(int arr[], int n)
{
    int min = arr[0];
    for (int i = 1; i < n; i++)
        if (arr[i] < min)
            min = arr[i];
    return min;
}

int getMax(int arr[], int n)
{
    int max = arr[0];
    for (int i = 1; i < n; i++)
        if (arr[i] > max)
            max = arr[i];
    return max;
}

/* Driver program to test above functions */
int main()
{
    int arr[]= {1, 4, 5, 3, 2, 6};
    int n = sizeof(arr)/sizeof(arr[0]);
    if(areConsecutive(arr, n) == true)
        printf(" Array elements are consecutive ");
    else
        printf(" Array elements are not consecutive ");
    getchar();
    return 0;
}

Java

class AreConsecutive
{
    /* The function checks if the array elements are consecutive
       If elements are consecutive, then returns true, else returns
       false */
    boolean areConsecutive(int arr[], int n)
    {
        if (n < 1)
            return false;

        /* 1) Get the minimum element in array */
        int min = getMin(arr, n);

        /* 2) Get the maximum element in array */
        int max = getMax(arr, n);

        /* 3) max-min+1 is equal to n then only check all elements */
        if (max - min + 1 == n)
        {
            int i;
            for (i = 0; i < n; i++)
            {
                int j;

                if (arr[i] < 0)
                    j = -arr[i] - min;
                else
                    j = arr[i] - min;

                // if the value at index j is negative then
                // there is repetition
                if (arr[j] > 0)
                    arr[j] = -arr[j];
                else
                    return false;
            }

            /* If we do not see a negative value then all elements
               are distinct */
            return true;
        }

        return false; // if (max-min+1 != n)
    }

    /* UTILITY FUNCTIONS */
    int getMin(int arr[], int n)
    {
        int min = arr[0];
        for (int i = 1; i < n; i++)
        {
            if (arr[i] < min)
                min = arr[i];
        }
        return min;
    }

    int getMax(int arr[], int n)
    {
        int max = arr[0];
        for (int i = 1; i < n; i++)
        {
            if (arr[i] > max)
                max = arr[i];
        }
        return max;
    }

    /* Driver program to test above functions */
    public static void main(String[] args)
    {
        AreConsecutive consecutive = new AreConsecutive();
        int arr[] = {5, 4, 2, 3, 1, 6};
        int n = arr.length;
        if (consecutive.areConsecutive(arr, n) == true)
            System.out.println("Array elements are consecutive");
        else
            System.out.println("Array elements are not consecutive");
    }
}

// This code is contributed by Mayank Jaiswal

Python 3

# Helper functions to get minimum and
# maximum in an array

# The function checks if the array
# elements are consecutive. If elements
# are consecutive, then returns true,
# else returns false
def areConsecutive(arr, n):

    if ( n < 1 ):
        return False

    # 1) Get the minimum element in array
    min = getMin(arr, n)

    # 2) Get the maximum element in array
    max = getMax(arr, n)

    # 3) max - min + 1 is equal to n
    # then only check all elements
    if (max - min + 1 == n):

        for i in range(n):

            if (arr[i] < 0):
                j = -arr[i] - min
            else:
                j = arr[i] - min

            # if the value at index j is negative
            # then there is repetition
            if (arr[j] > 0):
                arr[j] = -arr[j]
            else:
                return False

        # If we do not see a negative value
        # then all elements are distinct
        return True

    return False     # if (max - min + 1 != n)

# UTILITY FUNCTIONS
def getMin(arr, n):

    min = arr[0]
    for i in range(1, n):
        if (arr[i] < min):
            min = arr[i]
    return min

def getMax(arr, n):
    max = arr[0]
    for i in range(1, n):
        if (arr[i] > max):
            max = arr[i]
    return max

# Driver Code
if __name__ == "__main__":

    arr = [1, 4, 5, 3, 2, 6]
    n = len(arr)
    if(areConsecutive(arr, n) == True):
        print(" Array elements are consecutive ")
    else:
        print(" Array elements are not consecutive ")

# This code is contributed by ita_c

C#

using System;

class GFG {

    /* The function checks if the array
    elements are consecutive If elements
    are consecutive, then returns true,
    else returns false */
    static bool areConsecutive(int []arr, int n)
    {
        if (n < 1)
            return false;

        /* 1) Get the minimum element in
        array */
        int min = getMin(arr, n);

        /* 2) Get the maximum element in
        array */
        int max = getMax(arr, n);

        /* 3) max-min+1 is equal to n then
        only check all elements */
        if (max - min + 1 == n)
        {
            int i;
            for (i = 0; i < n; i++)
            {
                int j;

                if (arr[i] < 0)
                    j = -arr[i] - min;
                else
                    j = arr[i] - min;

                // if the value at index j
                // is negative then
                // there is repetition
                if (arr[j] > 0)
                    arr[j] = -arr[j];
                else
                    return false;
            }

            /* If we do not see a negative
            value then all elements
            are distinct */
            return true;
        }

        // if (max-min+1 != n)
        return false;
    }

    /* UTILITY FUNCTIONS */
    static int getMin(int []arr, int n)
    {
        int min = arr[0];
        for (int i = 1; i < n; i++)
        {
            if (arr[i] < min)
                min = arr[i];
        }
        return min;
    }

    static int getMax(int []arr, int n)
    {
        int max = arr[0];
        for (int i = 1; i < n; i++)
        {
            if (arr[i] > max)
                max = arr[i];
        }
        return max;
    }

    /* Driver program to test above
    functions */
    public static void Main()
    {
        int []arr = {5, 4, 2, 3, 1, 6};
        int n = arr.Length;

        if (areConsecutive(arr, n) == true)
            Console.Write("Array elements "
                      + "are consecutive");
        else
            Console.Write("Array elements "
                  + "are not consecutive");
    }
}

// This code is contributed by nitin mittal.

PHP

<?php

/* The function checks if the array elements
are consecutive If elements are consecutive,
then returns true, else returns false */
function areConsecutive( $arr, $n)
{

    if ( $n < 1 )
        return false;

    /* 1) Get the minimum element in array */
    $min = getMin($arr, $n);

    /* 2) Get the maximum element in array */
    $max = getMax($arr, $n);

    /* 3) max - min + 1 is equal to n then
          only check all elements */
    if ($max - $min + 1 == $n)
    {
    $i;
        for($i = 0; $i < $n; $i++)
        {
            $j;

            if ($arr[$i] < 0)
                $j = -$arr[$i] - $min;
            else
                $j = $arr[$i] - $min;

            // if the value at index j is
            // negative then there is
            // repetition
            if ($arr[$j] > 0)
                $arr[$j] = -$arr[$j];
            else
                return false;
        }

        /* If we do not see a negative value
        then all elements are distinct */
        return true;
    }

    return false; // if (max - min + 1 != n)
}

/* UTILITY FUNCTIONS */
function getMin( $arr, $n)
{
    $min = $arr[0];
    for ( $i = 1; $i < $n; $i++)
        if ($arr[$i] < $min)
            $min = $arr[$i];
    return $min;
}

function getMax( $arr, $n)
{
    $max = $arr[0];
    for ( $i = 1; $i < $n; $i++)
        if ($arr[$i] > $max)
            $max = $arr[$i];
    return $max;
}

/* Driver program to test above functions */
    $arr= array(1, 4, 5, 3, 2, 6);
    $n = count($arr);
    if(areConsecutive($arr, $n) == true)
        echo " Array elements are consecutive ";
    else
        echo " Array elements are not consecutive ";


// This code is contributed by anuj_67.
?>

Javascript

<script>

    /* The function checks if the array elements are consecutive
       If elements are consecutive, then returns true, else returns
       false */
    function areConsecutive(arr,n)
    {
        if (n < 1)
            return false;

        /* 1) Get the minimum element in array */
        let min = getMin(arr, n);

        /* 2) Get the maximum element in array */
        let max = getMax(arr, n);

        /* 3) max-min+1 is equal to n then only check all elements */
        if (max - min + 1 == n)
        {
            let i;
            for (i = 0; i < n; i++)
            {
                let j;

                if (arr[i] < 0)
                    j = -arr[i] - min;
                else
                    j = arr[i] - min;

                // if the value at index j is negative then
                // there is repetition
                if (arr[j] > 0)
                    arr[j] = -arr[j];
                else
                    return false;
            }

            /* If we do not see a negative value then all elements
               are distinct */
            return true;
        }

        return false; // if (max-min+1 != n)
    }

    /* UTILITY FUNCTIONS */
    function getMin(arr,n)
    {
         let min = arr[0];
        for (let i = 1; i < n; i++)
        {
            if (arr[i] < min)
                min = arr[i];
        }
        return min;
    }

    function getMax(arr,n)
    {
        let max = arr[0];
        for (let i = 1; i < n; i++)
        {
            if (arr[i] > max)
                max = arr[i];
        }
        return max;

    }

    /* Driver program to test above functions */
    let arr=[5, 4, 2, 3, 1, 6];
    let n = arr.length;
    if (areConsecutive(arr, n) == true)
        document.write("Array elements are consecutive");
    else
        document.write("Array elements are not consecutive");



    // This code is contributed by unknown2108
</script>

Output

 Array elements are consecutive 

Note that this method might not work for negative numbers. For example, it returns false for {2, 1, 0, -3, -1, -2}.
Time Complexity: O(n) 
Auxiliary Space: O(1) 

Check if array elements are consecutive in O(n) time and O(1) space (Handles Both Positive and negative numbers)

Method 4 (Using XOR property)

This method is O(n) time complexity and O(1) extra space, does not changes the original array, and it works every time.

1. As elements should be consecutive, let’s find minimum element or maximum element in array.
2. Now if we take xor of two same elements it will result in zero (a^a = 0).
3. Suppose array is {-2, 0, 1, -3, 4, 3, 2, -1}, now if we xor all array elements with minimum element and keep increasing minimum element, the resulting xor will become 0 only if elements are consecutive
   

C

//Code  is contributed by Dhananjay Dhawale @chessnoobdj

#include<stdio.h>
#include<stdlib.h>

/* UTILITY FUNCTIONS */
int getMin(int arr[], int n)
{
    int min = arr[0];
    for (int i = 1; i < n; i++)
        if (arr[i] < min)
            min = arr[i];
    return min;
}
int areConsecutive(int arr[], int n)
{
    int min_ele = getMin(arr, n), num = 0;
    for(int i=0; i<n; i++){
        num ^= min_ele^arr[i];
        min_ele += 1;
    }
    if(num == 0)
        return 1;
    return 0;
}

/* Driver program to test above functions */
int main()
{
    int arr[]= {1, 4, 5, 3, 2, 6};
    int n = sizeof(arr)/sizeof(arr[0]);
    if(areConsecutive(arr, n) == 1)
        printf(" Array elements are consecutive ");
    else
        printf(" Array elements are not consecutive ");
    getchar();
    return 0;
}

C++

//Code  is contributed by Dhananjay Dhawale @chessnoobdj

#include <iostream>
#include <algorithm>
using namespace std;

bool areConsecutive(int arr[], int n)
{
    int min_ele = *min_element(arr, arr+n), num = 0;
    for(int i=0; i<n; i++){
        num ^= min_ele^arr[i];
        min_ele += 1;
    }
    if(num == 0)
        return 1;
    return 0;
}

/* Driver program to test above functions */
int main()
{
    int arr[]= {1, 4, 5, 3, 2, 6};
    int n = sizeof(arr)/sizeof(arr[0]);
    if(areConsecutive(arr, n) == true)
        printf(" Array elements are consecutive ");
    else
        printf(" Array elements are not consecutive ");
    getchar();
    return 0;
}

Java

// Java implementation of the approach
import java.util.Arrays;
import java.util.Collections;

class AreConsecutive {

boolean areConsecutive(int arr[], int n)
{
    int min_ele = Arrays.stream(arr).min().getAsInt();
    int num = 0;
    for(int i=0; i<n; i++){
        num = num ^ min_ele ^ arr[i];
        min_ele += 1;
    }
    if(num == 0)
        return true;
    return false;
}



    public static void main(String[] args)
    {
        AreConsecutive consecutive = new AreConsecutive();
        int arr[] = {5, 4, 2, 3, 1, 6};
        int n = arr.length;
        if (consecutive.areConsecutive(arr, n) == true)
            System.out.println("Array elements are consecutive");
        else
            System.out.println("Array elements are not consecutive");
    }
}

// This code is contributed by Aarti_Rathi

Python3

# Function to Check if array
# elements are consecutive
def areConsecutive(arr, n):
    min_ele = arr.index(min(arr))
    num = 0
    for i in range(0, n):
        num ^= arr[min_ele] ^ arr[i]
        arr[min_ele] += 1
    if num == 0:
        return True
    return False

# Driver program to test above
# functions
if __name__ == "__main__":
    arr = [1, 4, 5, 3, 2, 6]
    n = len(arr)
    if areConsecutive(arr, n) == True:
        print(" Array elements are consecutive ", end=' ')
    else:
        print(" Array elements are not consecutive ", end=' ')

# This code is contributed by Aarti_Rathi

C#

using System;
using System.Linq;

class GFG {

// Function to Check if array
// elements are consecutive

static bool areConsecutive(int []arr, int n)
{
    int min_ele = arr.Min();
    int num = 0;
    for(int i=0; i<n; i++){
        num ^= min_ele^arr[i];
        min_ele += 1;
    }
    if(num == 0)
        return true;
    return false;
}

    /* Driver program to test above
    functions */
    public static void Main()
    {
        int []arr = {5, 4, 2, 3, 1, 6};
        int n = arr.Length;

        if (areConsecutive(arr, n) == true)
            Console.Write("Array elements "
                      + "are consecutive");
        else
            Console.Write("Array elements "
                  + "are not consecutive");
    }
}

// This code is contributed by Aarti_Rathi

Javascript

// Javascript Program
var areConsecutive = function(arr)
{
    var min_ele = Math.min.apply(Math, arr);
    var num = 0;
    for(var i = 0; i < arr.length; i++){
        num = num ^ min_ele ^ arr[i];
        min_ele += 1;
    }
    if(num == 0)
        return 1;
    return 0;
}

/* Driver program to test above functions */
arr = [1, 2, 3, 4, 5, 6];
if(areConsecutive(arr) == 1){
    console.log(" Array elements are consecutive ");}
else
    console.log(" Array elements are not consecutive ");

// This code is contributed by Sajal Aggarwal.

Output

 Array elements are consecutive 

Time Complexity: O(n) 
Auxiliary Space: O(1) 

Please suggest if someone has a better solution which is more efficient in terms of space and time.
This article is contributed by Aarti_Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above