# Day- 2 Given an integer array arr, count element x such that x + 1 is also in arr.

### Mridu Bhatnagar ・1 min read

This was one of the problems in LeetCode's ongoing 30-day challenge. Published on 7th April 2020.

Given an integer array arr, count element x such that x + 1 is also in arr.

If there're duplicates in arr, count them separately.

##### Example 1:

```
Input: arr = [1,2,3]
Output: 2
Explanation: 1 and 2 are counted cause 2 and 3 are in arr.
```

##### Example 2:

```
Input: arr = [1,1,3,3,5,5,7,7]
Output: 0
Explanation: No numbers are counted, cause there's no 2, 4, 6, or 8 in arr.
```

##### Example 3:

```
Input: arr = [1,3,2,3,5,0]
Output: 3
Explanation: 0, 1 and 2 are counted cause 1, 2 and 3 are in arr.
```

##### Example 4:

```
Input: arr = [1,1,2,2]
Output: 2
Explanation: Two 1s are counted cause 2 is in arr.
```

##### Solution Approach

- As the original list contains duplicate elements as well. Convert the list to set to get unique elements.
- Iterate over the set and check if element+1 exists in the original list.
- If it exists. Find the count of the element in the original list.
- Keep adding the count to the new list.
- Sum of all the counts. Returns the required output.

##### Code snippet

```
class Solution:
def countElements(self, arr: List[int]) -> int:
unique_elements = set(arr)
L = []
for element in unique_elements:
if element + 1 in arr:
item_count = arr.count(element)
L.append(item_count)
return sum(L)
```

Discussion on more optimized solutions are welcome.

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Classic DEV Post from Aug 1 '19

I suspect using a collections.Counter dict would be beneficial. If you convert the input list to one of those you can then just iterate over the keys and check if dict[dict[key]+1] > 0, and if yes, add dict[key] to a running summation.

The loop would be O(N) and dictionary lookup is typically O(1), which I think would be faster than your algorithm above.

The question is whether the initial conversion of input data is performant, which I don't know! But from a practical point of view, collections.Counter is good to know about!

Hey Phil, thank you for suggesting the collections.Counter approach. I am yet to check out that module. But, yes I did read it somewhere it helps in achieving efficiency.

Also, with these, I am purposefully avoiding using libraries. :)