30-Day LeetCoding Challenge: Counting Elements

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]

1. Start with element = 1 is element + 1 = 1 + 1 = 2 in the array or not, yes it is there so increate our answer with one answer = 1

2. Second element is element = 2, element + 1 = 2 + 1 = 3 in the array yes it is there so answer = 2

3. 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 be answer = 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 which element is in arr this will take also O(n) so overall the time complexity will be O(n²).

• Space Complexity: We only use extra space to store our answer which is constant so our overall space is O(1).

• Solution in python

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 by 1.

• 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 add 1 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 add 3 to my answer.

So to be able to have this map which will contain every element and it is occurrence times. Why map (hash-map) because it is search time complexity is O(1) this will improve our search right!

• Example: arr = [1,2,1,3]

1. We need to define our hash-map so it will be: counts = {1: 2, 2: 1, 3: 1}

2. Then go over the counts hash-map, starts with 1 is 1 + 1 = 2 in our counts yes it is there so our answer += counts[1] += 2 = 2.

3. Then with the 2 is 2 + 1 = 3 in our counts, yes so answer += counts[2] += 1 = 3

4. With the 3 and 3 + 1 = 4 not in the counts so we stop here and our final answer is answer = 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) where n is the length of the array, then we go through the counts map which will take O(n) * O(1) where O(1) is elem+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 hash-map 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).

• Solution in python