**Objective: **Given a array of unsorted numbers, check if all the numbers in the array are consecutive numbers.

**Examples:**

int[] arrA = {21,24,22,26,23,25}; - True (All the integers are consecutive from 21 to 26)int[] arrB = {11,10,12,14,13}; - True (All the integers are consecutive from 10 to 14)int[] arrC = {11,10,14,13}; - False (Integers are not consecutive, 12 is missing)

**Approach:**

**Naive Approach: **Sorting . Time Complexity – O(nlogn).

**Better Approach: Time Complexity – O(n).**

- Find the Maximum and minimum elements in array (Say the array is arrA)
- Check if array length = max-min+1
- Subtract the min from every element of the array.
- Check if array doesn’t have duplicates

for Step 4

**a) If array contains negative elements**

- Create an aux array and put 0 at every position
- Navigate the main array and update the aux array as aux[arrA[i]]=1
- During step 2 if u find any index position already filled with 1, we have duplicates, return false
- This operation performs in O(n) time and O(n) space

**b) If array does not contains negative elements – Time Complexity : O(n), Space Complexity : O(1)**

- Navigate the array.
- Update the array as for ith index :- arrA[arrA[i]] = arrA[arrA[i]]*-1 (if it already not negative).
- If is already negative, we have duplicates, return false.

**Complete Code:**

**Output**:

true true false

We can also sort the array and thn check for consecutive numbers by adding one to previous element – complexity O(nlogn)

we can find the smallest number say smlNo; and we know the size of the array say S, using formula SumOfSNumbers with start element as smlNo, SumOfSNumbers = S/2(2*smlNO + (S-1)*1);

Now sum all the array numbers in tempSum .

Match tempSum with SumOfSNumbers , if both are equal return true otherwise false.

TimeComplexity – O(n)