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Counting Elements

Easy

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Topics:

ArraysHash Table

Given an integer array arr and an integer k, return the number of good pairs in arr.

A pair of indices (i,j) is called good if arr[i] == arr[j] and i*j is divisible by k.

Example 1:

Input: arr = [1,2,3,1,1,3], k = 2Output: 4Explanation: There are 4 good pairs (0,3), (0,4), (3,4), and (2,5) 0-indexed.

Example 2:

Input: arr = [1,2,3,4], k = 1Output: [0]Explanation: There are no good pairs.

Constraints:

1 ≤ arr.length ≤ 100
1 ≤ arr[i], k ≤ 100

Try coding it in LeetCode

Counting Elements

Easy

Asked by:

22 views

Topics:

ArraysHash Table

Given an integer array arr and an integer k, return the number of good pairs in arr.

A pair of indices (i,j) is called good if arr[i] == arr[j] and i*j is divisible by k.

Example 1:

Input: arr = [1,2,3,1,1,3], k = 2Output: 4Explanation: There are 4 good pairs (0,3), (0,4), (3,4), and (2,5) 0-indexed.

Example 2:

Input: arr = [1,2,3,4], k = 1Output: [0]Explanation: There are no good pairs.

Constraints:

1 ≤ arr.length ≤ 100
1 ≤ arr[i], k ≤ 100

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the input array `arr` and the range of integer values it can contain?
Are duplicate elements allowed in the input array `arr`, and if so, how should they be counted? For example, if `arr = [1, 1, 2]`, should the count be 1 (for the first '1' because '2' exists) or 2 (for both '1's)?
Can the array contain negative numbers or zero?
What should be returned if the input array `arr` is empty?
Does the problem imply that `x + 1` must be *strictly greater* than `x`? (This is implicitly true for integers, but good to confirm no edge cases like extremely large numbers are involved.)

Brute Force Solution

Approach

The brute force way to count elements is to consider every single possible combination of elements and check if each combination satisfies the problem's condition. We'll carefully examine each scenario to see if it meets all the requirements.

Here's how the algorithm would work step-by-step:

Take the first element and see if it can be part of a valid count.
If it can, then consider the next element and see if it can be combined with the first element to form a valid count.
Continue this process, trying to add each subsequent element to the current group and checking if the entire group is valid.
If adding an element makes the group invalid, then you know that combination won't work.
You essentially explore all potential groups of elements, one element at a time, building up possibilities.
Any group that perfectly matches the problem's criteria is a potential solution.
Keep track of all the valid groups you discover.
Finally, review all the valid groups you found and select the one that best fits the overall goal.

Code Implementation

def count_elements_brute_force(input_list):
    valid_groups_count = 0
    n = len(input_list)

    # Iterate through all possible starting elements for a group
    for starting_element_index in range(n):
        # Initialize a group with the current starting element
        current_group = [input_list[starting_element_index]]
        potential_valid_group = True

        # Consider adding subsequent elements to the current group
        for next_element_index in range(starting_element_index + 1, n):
            # Check if adding this element would invalidate the group based on problem conditions
            if len(current_group) > 0 and input_list[next_element_index] == current_group[-1]:
                # This condition would break if adding duplicates breaks the criteria
                potential_valid_group = False
                break
            else:
                current_group.append(input_list[next_element_index])

        # After exploring combinations with the starting element, check if the group meets criteria
        if potential_valid_group:
            # This check assumes a specific condition, e.g., all elements are unique or follow a pattern
            # For a generic 'counting elements' problem, this part needs a specific condition.
            # Assuming the condition is that all elements in the formed group are distinct for demonstration
            if len(current_group) == len(set(current_group)):
                valid_groups_count += 1

    # The brute force approach needs to consider all subsets, not just contiguous ones
    # This implementation is a simplification based on the provided steps and needs a concrete problem condition.
    # For a general 'count elements' problem, one might need to generate all subsets and check each.
    # This simplified version counts groups formed sequentially where each element has a unique successor.
    return valid_groups_count

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores every single possible combination of elements. For an input array of size n, there are 2^n possible subsets (combinations). The algorithm iterates through each of these subsets, and for each subset, it performs operations to check its validity. Therefore, the total number of operations grows exponentially with the input size n, leading to a time complexity of O(2^n).

Space Complexity

–

Optimal Solution

Approach

The problem asks us to count how many numbers in a given collection have a specific relationship with other numbers also present in that collection. Instead of checking every number individually against all others, we can use a clever shortcut.

Here's how the algorithm would work step-by-step:

First, let's make a quick reference of all the unique numbers we have available.
Now, for each number we're considering from the original collection, we'll check if a number that's exactly one greater than it also exists in our quick reference.
If that number one greater does exist, we count the current number as satisfying the condition.
We do this for every number in the original collection.
Finally, we add up all the counts to get our answer.

Code Implementation

def count_elements(collection_of_numbers):
    # Using a set provides efficient O(1) average time complexity for lookups.
    available_numbers = set(collection_of_numbers)

    count_of_matching_elements = 0

    # Iterate through each number in the original collection to check the condition.
    for number_to_check in collection_of_numbers:
        # This check ensures that we only count elements where x + 1 is also present.
        if (number_to_check + 1) in available_numbers:
            count_of_matching_elements += 1

    # The final count represents elements that have a successor within the collection.
    return count_of_matching_elements

Big(O) Analysis

Time Complexity

O(n) – First, creating a quick reference of all unique numbers, typically using a hash set, takes O(n) time, where n is the number of elements in the input collection. Then, iterating through each of the n numbers in the original collection and performing a lookup (checking if a number one greater exists) in the hash set takes O(1) on average for each number. Therefore, the total time complexity is dominated by the initial creation of the hash set and the subsequent single pass through the array, resulting in O(n) time. The total operations are roughly n for set creation plus n lookups, approximating n + n which simplifies to O(n).

Space Complexity

O(k) – The solution uses a set (or hash map) to store the unique numbers from the input collection. In the worst case, if all N numbers are distinct, this auxiliary data structure will store up to k unique numbers, where k is the number of unique elements in the input. The space complexity is therefore dominated by the size of this set, which is O(k). If all elements are unique, k can be equal to N, leading to O(N) space. However, if there are many duplicates, k will be significantly smaller than N.

Edge Cases

Case	How to Handle
Input array is null	A null input should ideally throw an exception or return 0 based on problem constraints.
Input array is empty	An empty array should result in a count of 0.
Input array has only one element	An array with a single element cannot satisfy the condition, so the count should be 0.
Input array contains only identical elements	If all elements are the same, and x+1 is not present, the count will correctly be 0.
Input array contains negative numbers	The presence of negative numbers does not affect the logic as we are checking for x+1, regardless of sign.
Input array contains zero	If 0 is present, we will count it if 1 is also present in the array.
Input array contains duplicate values of x where x+1 also exists	A frequency map or set-based approach will correctly count each instance of x if x+1 is present.
Very large input array size	An efficient solution using a hash set or frequency map will have a time complexity of O(n) and space complexity of O(n), which scales well.

Input array is null

How to Handle:

A null input should ideally throw an exception or return 0 based on problem constraints.

Input array is empty

How to Handle:

An empty array should result in a count of 0.

Input array has only one element

How to Handle:

An array with a single element cannot satisfy the condition, so the count should be 0.

Input array contains only identical elements

How to Handle:

If all elements are the same, and x+1 is not present, the count will correctly be 0.

Input array contains negative numbers

How to Handle:

The presence of negative numbers does not affect the logic as we are checking for x+1, regardless of sign.

Input array contains zero

How to Handle:

If 0 is present, we will count it if 1 is also present in the array.

Input array contains duplicate values of x where x+1 also exists

How to Handle:

A frequency map or set-based approach will correctly count each instance of x if x+1 is present.

Very large input array size

How to Handle:

An efficient solution using a hash set or frequency map will have a time complexity of O(n) and space complexity of O(n), which scales well.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the input array `arr` and the range of integer values it can contain?
Are duplicate elements allowed in the input array `arr`, and if so, how should they be counted? For example, if `arr = [1, 1, 2]`, should the count be 1 (for the first '1' because '2' exists) or 2 (for both '1's)?
Can the array contain negative numbers or zero?
What should be returned if the input array `arr` is empty?
Does the problem imply that `x + 1` must be *strictly greater* than `x`? (This is implicitly true for integers, but good to confirm no edge cases like extremely large numbers are involved.)

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Take the first element and see if it can be part of a valid count.
If it can, then consider the next element and see if it can be combined with the first element to form a valid count.
Continue this process, trying to add each subsequent element to the current group and checking if the entire group is valid.
If adding an element makes the group invalid, then you know that combination won't work.
You essentially explore all potential groups of elements, one element at a time, building up possibilities.
Any group that perfectly matches the problem's criteria is a potential solution.
Keep track of all the valid groups you discover.
Finally, review all the valid groups you found and select the one that best fits the overall goal.

Code Implementation

def count_elements_brute_force(input_list):
    valid_groups_count = 0
    n = len(input_list)

    # Iterate through all possible starting elements for a group
    for starting_element_index in range(n):
        # Initialize a group with the current starting element
        current_group = [input_list[starting_element_index]]
        potential_valid_group = True

        # Consider adding subsequent elements to the current group
        for next_element_index in range(starting_element_index + 1, n):
            # Check if adding this element would invalidate the group based on problem conditions
            if len(current_group) > 0 and input_list[next_element_index] == current_group[-1]:
                # This condition would break if adding duplicates breaks the criteria
                potential_valid_group = False
                break
            else:
                current_group.append(input_list[next_element_index])

        # After exploring combinations with the starting element, check if the group meets criteria
        if potential_valid_group:
            # This check assumes a specific condition, e.g., all elements are unique or follow a pattern
            # For a generic 'counting elements' problem, this part needs a specific condition.
            # Assuming the condition is that all elements in the formed group are distinct for demonstration
            if len(current_group) == len(set(current_group)):
                valid_groups_count += 1

    # The brute force approach needs to consider all subsets, not just contiguous ones
    # This implementation is a simplification based on the provided steps and needs a concrete problem condition.
    # For a general 'count elements' problem, one might need to generate all subsets and check each.
    # This simplified version counts groups formed sequentially where each element has a unique successor.
    return valid_groups_count

Big(O) Analysis

Time Complexity

Space Complexity

–

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, let's make a quick reference of all the unique numbers we have available.
Now, for each number we're considering from the original collection, we'll check if a number that's exactly one greater than it also exists in our quick reference.
If that number one greater does exist, we count the current number as satisfying the condition.
We do this for every number in the original collection.
Finally, we add up all the counts to get our answer.

Code Implementation

def count_elements(collection_of_numbers):
    # Using a set provides efficient O(1) average time complexity for lookups.
    available_numbers = set(collection_of_numbers)

    count_of_matching_elements = 0

    # Iterate through each number in the original collection to check the condition.
    for number_to_check in collection_of_numbers:
        # This check ensures that we only count elements where x + 1 is also present.
        if (number_to_check + 1) in available_numbers:
            count_of_matching_elements += 1

    # The final count represents elements that have a successor within the collection.
    return count_of_matching_elements

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Input array is null	A null input should ideally throw an exception or return 0 based on problem constraints.
Input array is empty	An empty array should result in a count of 0.
Input array has only one element	An array with a single element cannot satisfy the condition, so the count should be 0.
Input array contains only identical elements	If all elements are the same, and x+1 is not present, the count will correctly be 0.
Input array contains negative numbers	The presence of negative numbers does not affect the logic as we are checking for x+1, regardless of sign.
Input array contains zero	If 0 is present, we will count it if 1 is also present in the array.
Input array contains duplicate values of x where x+1 also exists	A frequency map or set-based approach will correctly count each instance of x if x+1 is present.
Very large input array size	An efficient solution using a hash set or frequency map will have a time complexity of O(n) and space complexity of O(n), which scales well.

Input array is null

How to Handle:

A null input should ideally throw an exception or return 0 based on problem constraints.

Input array is empty

How to Handle:

An empty array should result in a count of 0.

Input array has only one element

How to Handle:

An array with a single element cannot satisfy the condition, so the count should be 0.

Input array contains only identical elements

How to Handle:

If all elements are the same, and x+1 is not present, the count will correctly be 0.

Input array contains negative numbers

How to Handle:

The presence of negative numbers does not affect the logic as we are checking for x+1, regardless of sign.

Input array contains zero

How to Handle:

If 0 is present, we will count it if 1 is also present in the array.

Input array contains duplicate values of x where x+1 also exists

How to Handle:

A frequency map or set-based approach will correctly count each instance of x if x+1 is present.

Very large input array size

How to Handle:

An efficient solution using a hash set or frequency map will have a time complexity of O(n) and space complexity of O(n), which scales well.