30Day LeetCoding Challenge: Counting Elements
Ahmad Ra'fat ・3 min read
The problem
Approach 1:
If you read the problem carefully you will find the solution in the problem statement itself.
It says you have an integer array
and we need to count the element when the element + 1
is also in that array considering if you have duplicates count at as separate so we don't care about duplicates.
Let's actually do whatever the problem says in the following example.
 Example:
arr = [1,2,3]
Start with
element = 1
iselement + 1 = 1 + 1 = 2
in the array or not, yes it is there so increate our answer with oneanswer = 1
Second element is
element = 2
,element + 1 = 2 + 1 = 3
in the array yes it is there soanswer = 2
Third and our last
element = 3
,element + 1 = 3 + 1 = 4
is in the array no it is not so our final answer stays as it is and it will beanswer = 2
.
Try it out with this example arr = [1,1,3,3,5,5,7,7]
Remember we don't care about duplicates.
 Pseudocode:
1. define counts = 0
2. loop for each element in arr
3. if element + 1 is in arr
4. counts += 1
5. return counts
Complexity:
Time Complexity: we do only one path over the whole array so it will be O() where
n
is the size of the array, and inside this loop, we do another check whichelement is in arr
this will take also O(n) so overall the time complexity will be O(n^{2}).Space Complexity: We only use extra space to store our answer which is constant so our overall space is O(1).
Approach 2:
Can we improve the above solution yes we can improve it at least on the time complexity part?
Let's think about the problem again:
We need to know if each
element + 1
is in the array or not and if it is there increase our answer by1
.What if one element is existing more than one time for example we have three twos and we have a three in the array so the condition of
elem + 1
will be valid.Hmm, so each time I will reach a
2
I will add1
to my answer so what if I have a map telling me that 2 already exist 3 times like this{2: 3}
.Now, easily I can say if
2 + 1 = 3
exist just add3
to my answer.
So to be able to have this map which will contain every element and it is occurrence times. Why map (hashmap) because it is search time complexity is O(1) this will improve our search right!

Example:
arr = [1,2,1,3]
We need to define our hashmap so it will be:
counts = {1: 2, 2: 1, 3: 1}
Then go over the
counts
hashmap, starts with1
is1 + 1 = 2
in ourcounts
yes it is there so ouranswer += counts[1] += 2 = 2
.Then with the
2
is2 + 1 = 3
in ourcounts
, yes soanswer += counts[2] += 1 = 3
With the
3
and3 + 1 = 4
not in thecounts
so we stop here and our final answer isanswer = 3
.
 Pseudocode:
1. define counts = Count(arr)
2. define ans = 0
3. for elem in counts
4. if elem + 1 in counts:
5. ans += counts[elem]
6. return ans
Complexity:
Time Complexity: We do two loops here the first one to build our
counts
map from the original array and it will take O(n) wheren
is the length of the array, then we go through thecounts
map which will take O(n) * O(1) where O(1) iselem+1 in counts
so totally the loop will be O(n) and the overall will be O(n) + O(n) = O(n).Space Complexity: we use extra space to store the counts of the elements in hashmap so it will be O(n) as we store all elements in the array with the space to store our answer which is O(1) so the overall will be O(n).