Counting Elements

The problem is asking to count how many elements x in the given array have a corresponding x + 1 also in the array. Duplicates are counted separately.

Here’s a step-by-step approach to solving this problem:

  1. Create a Set of the Array Elements: We can use a set to keep track of the unique values present in the array. This will help us quickly check whether x + 1 is present for a given x.

  2. Iterate Through the Array and Count Elements: We’ll iterate through the original array and check whether x + 1 is present in the set for each element x. If it is present, we increment our count.

  3. Return the Count: Finally, we return the count of elements that meet the condition.

Here’s the code:

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class Solution:
    def countElements(self, arr: List[int]) -> int:
        # Create a set of unique elements in the array
        unique_elements = set(arr)

        # Initialize a counter for elements x such that x + 1 is in the array
        count = 0

        # Iterate through the array and count elements that meet the condition
        for x in arr:
            if x + 1 in unique_elements:
                count += 1

        return count

Explanation:

  • The set unique_elements helps us quickly determine whether x + 1 is present for any given x in the array.
  • By iterating through the array and checking whether x + 1 is in the set, we can count the elements that meet the condition.
  • The use of the set ensures that the lookup time for checking the presence of x + 1 is (O(1)).

The time complexity of this solution is (O(n)), where (n) is the length of the input array, and the space complexity is also (O(n)).