Check if Array is Consecutive Integers

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).

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

for Step 4

a) If array contains negative elements

  1. Create an aux array and put 0 at every position
  2. Navigate the main array and update the aux array as aux[arrA[i]]=1
  3. During step 2 if u find any index position already filled with 1, we have duplicates, return false
  4. 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)

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

Complete Code:


Output:

true
true
false

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If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.
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  • kaushal prajapati

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

  • Nishant Dehariya

    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)

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