Solution Approach

The solution implements the counting approach using Python's Counter class from the collections module.

Step 1: Count frequencies

cnt = Counter(arr)

This creates a dictionary-like object where keys are the unique numbers from arr and values are their frequencies. For example, if arr = [1, 1, 2, 3], then cnt = {1: 2, 2: 1, 3: 1}.

Step 2: Iterate and sum valid counts

return sum(v for x, v in cnt.items() if cnt[x + 1])

This line does several things:

• cnt.items() iterates through all (number, frequency) pairs in the counter
• For each pair (x, v) where x is the number and v is its frequency
• if cnt[x + 1] checks if x + 1 exists in the counter (in Python, this evaluates to True if cnt[x + 1] > 0, and False if x + 1 is not in cnt or has count 0)
• If the condition is true, v (the frequency of x) is included in the sum
• The sum() function adds up all the valid frequencies

Why this works:

• When x + 1 exists in the array, all occurrences of x meet our criteria
• By summing the frequencies (v) of all such valid numbers x, we get the total count of elements that have their successor in the array

Time Complexity: O(n) where n is the length of the array - we iterate through the array once to build the counter, then iterate through the unique elements once.

Space Complexity: O(n) for storing the counter, which in the worst case contains all unique elements from the array.

Example Walkthrough

Let's walk through the solution with arr = [1, 2, 1, 3, 5, 6, 5].

Step 1: Build the frequency counter

First, we create a Counter from the array:

cnt = Counter([1, 2, 1, 3, 5, 6, 5])
cnt = {1: 2, 2: 1, 3: 1, 5: 2, 6: 1}

Step 2: Check each unique number

Now we iterate through each (number, frequency) pair and check if number + 1 exists:

• x = 1, v = 2: Check if 1 + 1 = 2 exists in cnt

  • cnt[2] = 1 (exists), so we count all 2 occurrences of 1
  • Add 2 to our sum

• x = 2, v = 1: Check if 2 + 1 = 3 exists in cnt

  • cnt[3] = 1 (exists), so we count the 1 occurrence of 2
  • Add 1 to our sum

• x = 3, v = 1: Check if 3 + 1 = 4 exists in cnt

  • cnt[4] doesn't exist, so we don't count 3
  • Add 0 to our sum

• x = 5, v = 2: Check if 5 + 1 = 6 exists in cnt

  • cnt[6] = 1 (exists), so we count all 2 occurrences of 5
  • Add 2 to our sum

• x = 6, v = 1: Check if 6 + 1 = 7 exists in cnt

  • cnt[7] doesn't exist, so we don't count 6
  • Add 0 to our sum

Step 3: Calculate final result

Total sum = 2 + 1 + 0 + 2 + 0 = 5

Therefore, 5 elements in the array have their successor (x + 1) also present in the array.

Time and Space Complexity

The time complexity is O(n), where n is the length of the array arr. This is because:

• Creating the Counter object requires iterating through all n elements of the array once, which takes O(n) time
• The Counter object will have at most n key-value pairs (in the worst case where all elements are unique)
• Iterating through the Counter items and checking if cnt[x + 1] exists takes O(1) time per item due to hash table lookup, resulting in O(n) total time for the iteration

The space complexity is O(n), where n is the length of the array arr. This is because:

• The Counter object stores at most n key-value pairs, requiring O(n) space in the worst case when all elements in the array are distinct
• The sum operation uses constant extra space beyond the Counter object

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Counting Unique Elements Instead of All Occurrences

The Mistake: A common error is counting each unique element only once, rather than counting all occurrences when x + 1 exists.

Incorrect Implementation:

def countElements(self, arr: List[int]) -> int:
    unique_elements = set(arr)
    count = 0
    for x in unique_elements:
        if x + 1 in unique_elements:
            count += 1  # Wrong! Only counts once per unique element
    return count

Why It's Wrong: If arr = [1, 1, 2], this would return 1 instead of 2, because it only counts the unique element 1 once, ignoring the duplicate.

Correct Approach: Count the frequency of each element and sum up all occurrences when the condition is met:

def countElements(self, arr: List[int]) -> int:
    frequency_map = Counter(arr)
    count = 0
    for element, frequency in frequency_map.items():
        if element + 1 in frequency_map:
            count += frequency  # Correct! Adds all occurrences
    return count

Pitfall 2: Checking Array Values Directly Without Efficient Lookup

The Mistake: Using repeated linear searches through the array to check if x + 1 exists.

Inefficient Implementation:

def countElements(self, arr: List[int]) -> int:
    count = 0
    for x in arr:
        if (x + 1) in arr:  # O(n) search for each element!
            count += 1
    return count

Why It's Problematic: This creates O(n²) time complexity because in arr performs a linear search for each element. For large arrays, this becomes extremely slow.

Better Solution: Use a set or Counter for O(1) lookup:

def countElements(self, arr: List[int]) -> int:
    element_set = set(arr)
    count = 0
    for x in arr:
        if (x + 1) in element_set:  # O(1) lookup
            count += 1
    return count

Pitfall 3: Misunderstanding the Problem Requirements

The Mistake: Counting elements where x - 1 exists instead of x + 1, or counting both x and x + 1 when the condition is met.

Wrong Interpretation Example:

def countElements(self, arr: List[int]) -> int:
    frequency_map = Counter(arr)
    count = 0
    for element, frequency in frequency_map.items():
        if element + 1 in frequency_map:
            # Wrong! Counting both x and x+1
            count += frequency + frequency_map[element + 1]
    return count

Correct Understanding: Only count occurrences of x when x + 1 exists. The element x + 1 itself is not counted unless x + 2 also exists in the array.