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Easy

Counter

Easy

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 data type will the counter operate on, and what is the range of possible values?
What should the counter return if it reaches its maximum or minimum value?
Is the counter expected to be thread-safe?
Should the counter support incrementing or decrementing by values other than one?
What is the expected behavior if an invalid operation is attempted (e.g., decrementing below the minimum value)?

Brute Force Solution

Approach

The brute force method for the Counter problem is essentially to try every possible combination. We start by picking a potential configuration and seeing if it is valid. If it is valid and satisfies our objective, we return it, otherwise we try the next possible configuration until we find one that works.

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

Start by considering every possible number you can pick for the first position.
For each of those choices, explore every possible number for the second position.
Continue this pattern for all the positions.
Once you've assigned a number to every position, check if the current configuration is a valid solution.
If it is, then you've found a solution.
If not, move on to the next possible arrangement and repeat the checking process until you find a valid solution or have exhausted all options.

Code Implementation

def counter_brute_force(digits, target_sum):
    possible_combination = [0] * digits

    def backtrack(index):
        # If we've filled all positions, check if the sum is equal to the target
        if index == digits:
            current_sum = sum(possible_combination)
            if current_sum == target_sum:
                return possible_combination[:]
            else:
                return None

        # Try every possible value (0-9) for the current digit
        for current_value in range(10):
            possible_combination[index] = current_value

            # Recursively call backtrack to fill the next digit
            result = backtrack(index + 1)
            if result:
                return result

        return None

    return backtrack(0)

Big(O) Analysis

Time Complexity

O(k^n) – The brute force approach explores every possible combination of numbers for each of the n positions. Assuming each position can take k different values, we have k options for the first position, k options for the second position, and so on. This leads to k * k * ... * k (n times) possible combinations, which is equal to k^n. Therefore, checking all combinations takes O(k^n) time.

Space Complexity

O(N) – The brute force method described uses recursion to explore every possible combination. The depth of the recursion will be determined by the number of positions to be filled, which we can denote as N. At each level of recursion, we store the current state (e.g., the numbers assigned so far). Therefore, the auxiliary space is proportional to the maximum depth of the recursion tree, which is N. This results in an auxiliary space complexity of O(N) due to the call stack.

Optimal Solution

Approach

The goal is to create a data structure that keeps track of how many times each item appears. We want to do this efficiently, so we can quickly update counts and look them up later. We accomplish this by using a concept that allows us to easily access any item's count.

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

Imagine a special notebook where you can write down each unique item you see.
Each time you see a particular item, you'll go to its page in the notebook and increase its count by one.
If you see a new item that isn't in the notebook yet, you'll create a new page for it and set its count to one.
To find out how many times an item appears, just look up its page in the notebook and read its count.
This way, finding the count for any item will be very fast because you can go straight to its page.

Code Implementation

class Counter:
    def __init__(self):
        self.item_counts = {}

    def increment(self, item):
        # Check if the item is already in the counts.
        if item in self.item_counts:

            self.item_counts[item] += 1
        else:
            # If it's not, initialize its count to one.

            self.item_counts[item] = 1

    def get_count(self, item):
        # Returns the number of times the item appears.
        if item in self.item_counts:

            return self.item_counts[item]
        else:
            # If the item is not found, return zero.
            return 0

Big(O) Analysis

Time Complexity

O(1) – The data structure described uses a concept analogous to a hash map or dictionary. Inserting a new item, incrementing its count, and looking up its count all involve accessing or modifying a specific entry in this data structure. Hash map operations (insertion, lookup, and deletion) have an average time complexity of O(1), regardless of the number of unique items already stored, assuming a good hash function and collision handling. Therefore, each of these operations take constant time.

Space Complexity

O(N) – The algorithm described uses a "special notebook" to store each unique item encountered and its corresponding count. This "notebook" can be implemented using a hash map or dictionary. In the worst-case scenario, all N items in the input are unique, so the hash map would need to store N entries. Therefore, the auxiliary space used by the algorithm grows linearly with the number of unique items, leading to a space complexity of O(N), where N represents the number of items processed.

Edge Cases

Case	How to Handle
Null input	Check for null input and return an appropriate error or empty result.
Empty list	Return an empty list or appropriate default value as there's nothing to count.
Very large input list (potential memory issues)	Consider using generators or iterators to avoid loading the entire list into memory at once.
Input list contains non-numeric values	Validate the input to ensure it contains only numbers and raise an exception if not.
Input list contains extreme values (near max/min integer)	Handle potential overflow during counting by using appropriate data types (e.g., long) or checking for overflow conditions.
Input list contains duplicate values and the goal is distinct counts	Use a set or dictionary to ensure each value is counted only once.
Input list is already sorted	The counting method should still work correctly whether the list is sorted or not.
The maximum count exceeds the representable range of the chosen data type	Consider using a larger data type or a specialized library for arbitrarily large numbers.

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 data type will the counter operate on, and what is the range of possible values?
What should the counter return if it reaches its maximum or minimum value?
Is the counter expected to be thread-safe?
Should the counter support incrementing or decrementing by values other than one?
What is the expected behavior if an invalid operation is attempted (e.g., decrementing below the minimum value)?

Brute Force Solution

Approach

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

Start by considering every possible number you can pick for the first position.
For each of those choices, explore every possible number for the second position.
Continue this pattern for all the positions.
Once you've assigned a number to every position, check if the current configuration is a valid solution.
If it is, then you've found a solution.
If not, move on to the next possible arrangement and repeat the checking process until you find a valid solution or have exhausted all options.

Code Implementation

def counter_brute_force(digits, target_sum):
    possible_combination = [0] * digits

    def backtrack(index):
        # If we've filled all positions, check if the sum is equal to the target
        if index == digits:
            current_sum = sum(possible_combination)
            if current_sum == target_sum:
                return possible_combination[:]
            else:
                return None

        # Try every possible value (0-9) for the current digit
        for current_value in range(10):
            possible_combination[index] = current_value

            # Recursively call backtrack to fill the next digit
            result = backtrack(index + 1)
            if result:
                return result

        return None

    return backtrack(0)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Imagine a special notebook where you can write down each unique item you see.
Each time you see a particular item, you'll go to its page in the notebook and increase its count by one.
If you see a new item that isn't in the notebook yet, you'll create a new page for it and set its count to one.
To find out how many times an item appears, just look up its page in the notebook and read its count.
This way, finding the count for any item will be very fast because you can go straight to its page.

Code Implementation

class Counter:
    def __init__(self):
        self.item_counts = {}

    def increment(self, item):
        # Check if the item is already in the counts.
        if item in self.item_counts:

            self.item_counts[item] += 1
        else:
            # If it's not, initialize its count to one.

            self.item_counts[item] = 1

    def get_count(self, item):
        # Returns the number of times the item appears.
        if item in self.item_counts:

            return self.item_counts[item]
        else:
            # If the item is not found, return zero.
            return 0

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null input	Check for null input and return an appropriate error or empty result.
Empty list	Return an empty list or appropriate default value as there's nothing to count.
Very large input list (potential memory issues)	Consider using generators or iterators to avoid loading the entire list into memory at once.
Input list contains non-numeric values	Validate the input to ensure it contains only numbers and raise an exception if not.
Input list contains extreme values (near max/min integer)	Handle potential overflow during counting by using appropriate data types (e.g., long) or checking for overflow conditions.
Input list contains duplicate values and the goal is distinct counts	Use a set or dictionary to ensure each value is counted only once.
Input list is already sorted	The counting method should still work correctly whether the list is sorted or not.
The maximum count exceeds the representable range of the chosen data type	Consider using a larger data type or a specialized library for arbitrarily large numbers.