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Check if Array is Consecutive Integers

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


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)


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:



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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)