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Maximum Difference Between Even and Odd Frequency I

Easy

Google

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Topics:

Strings

You are given a string s consisting of lowercase English letters.

Your task is to find the maximum difference diff = freq(a1) - freq(a2) between the frequency of characters a1 and a2 in the string such that:

a1 has an odd frequency in the string.
a2 has an even frequency in the string.

Return this maximum difference.

Example 1:

Input: s = "aaaaabbc" Output: 3

Explanation:

The character 'a' has an odd frequency of 5, and 'b' has an even frequency of 2.
The maximum difference is 5 - 2 = 3.

Example 2:

Input: s = "abcabcab" Output: 1

Explanation:

The character 'a' has an odd frequency of 3, and 'c' has an even frequency of 2.
The maximum difference is 3 - 2 = 1.

Constraints:

3 <= s.length <= 100
s consists only of lowercase English letters.
s contains at least one character with an odd frequency and one with an even frequency.

Maximum Difference Between Even and Odd Frequency I

Easy

Google

3 views

Topics:

Strings

You are given a string s consisting of lowercase English letters.

Your task is to find the maximum difference diff = freq(a1) - freq(a2) between the frequency of characters a1 and a2 in the string such that:

a1 has an odd frequency in the string.
a2 has an even frequency in the string.

Return this maximum difference.

Example 1:

Input: s = "aaaaabbc" Output: 3

Explanation:

The character 'a' has an odd frequency of 5, and 'b' has an even frequency of 2.
The maximum difference is 5 - 2 = 3.

Example 2:

Input: s = "abcabcab" Output: 1

Explanation:

The character 'a' has an odd frequency of 3, and 'c' has an even frequency of 2.
The maximum difference is 3 - 2 = 1.

Constraints:

3 <= s.length <= 100
s consists only of lowercase English letters.
s contains at least one character with an odd frequency and one with an even frequency.

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 is the expected return value if the array is empty or if there are no even or odd frequency elements?
Are the array elements guaranteed to be integers, or could they be other data types like floating-point numbers?
How large can the input array be, and what is the range of possible values for the elements within the array?
If multiple pairs of elements yield the same maximum difference, is any one of them acceptable as a valid solution?
Can you provide an example input and its corresponding expected output to illustrate the problem more clearly?

Brute Force Solution

Approach

The brute force method for this problem means we'll check every possible group of numbers in our set. We'll calculate the even and odd frequency for each group to find the biggest difference between them.

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

Consider all possible starting points in the set of numbers.
For each starting point, create a group of numbers by adding one number at a time.
For each group of numbers, count how many times each distinct number appears. This gives us the frequencies.
Separate the numbers that appeared an even number of times from the numbers that appeared an odd number of times.
Add up all the even frequencies. This is the total even frequency.
Add up all the odd frequencies. This is the total odd frequency.
Subtract the total odd frequency from the total even frequency. This is the difference for that group.
Compare this difference to the largest difference we've found so far. If it's bigger, remember it.
Repeat steps 2-8 for every starting point.
After checking every possible group, the largest difference we remembered is our answer.

Code Implementation

def maximum_difference_between_even_and_odd_frequency_brute_force(numbers):
    maximum_difference = float('-inf')

    for start_index in range(len(numbers)):
        for end_index in range(start_index, len(numbers)):
            sub_array = numbers[start_index : end_index + 1]
            
            frequency_map = {}
            for number in sub_array:
                frequency_map[number] = frequency_map.get(number, 0) + 1
            
            even_frequency_sum = 0
            odd_frequency_sum = 0

            #Iterate through the map to aggregate even and odd frequencies
            for number, frequency in frequency_map.items():
                if frequency % 2 == 0:
                    even_frequency_sum += frequency
                else:
                    odd_frequency_sum += frequency

            difference = even_frequency_sum - odd_frequency_sum

            # Update maximum difference if necessary.
            if difference > maximum_difference:
                maximum_difference = difference

    return maximum_difference

Big(O) Analysis

Time Complexity

O(n^3) – The algorithm iterates through all possible subarrays. The outer loop iterates 'n' times to select the starting index of a subarray. The inner loop iterates up to 'n' times to expand the subarray. Inside the inner loop, calculating the frequencies of each element in the subarray takes O(n) in the worst case (where each element is distinct). Therefore, the overall time complexity is O(n * n * n) which simplifies to O(n^3).

Space Complexity

O(N) – The space complexity is dominated by the frequency counting step where, for each group of numbers, we store the frequency of each distinct number encountered. In the worst case, all N numbers in the input array are distinct, requiring us to store up to N frequencies. Therefore, the auxiliary space used for frequency counting grows linearly with the input size N, resulting in a space complexity of O(N).

Optimal Solution

Approach

The goal is to find the biggest difference between how often even numbers appear and how often odd numbers appear. We can keep track of these counts and update the maximum difference we've seen so far as we process each number.

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

First, create two separate counts: one for even numbers and one for odd numbers. Start both counts at zero.
Then, go through the numbers one by one.
For each number, check if it is even or odd.
If the number is even, increase the even count by one. If it's odd, increase the odd count by one.
After processing each number, calculate the difference between the even count and the odd count (even count minus odd count).
Keep track of the largest difference you've encountered so far. If the current difference is larger than the largest difference you've seen, update the largest difference.
After going through all of the numbers, the largest difference you've tracked is the answer.

Code Implementation

def maximum_difference_even_odd_frequency(numbers):
    even_number_frequency = 0
    odd_number_frequency = 0

    maximum_frequency_difference = 0

    for number in numbers:
        # Need to check if the number is even or odd
        if number % 2 == 0:
            even_number_frequency += 1

        else:
            odd_number_frequency += 1

        # This is the key calculation of the difference
        current_frequency_difference = \
            even_number_frequency - odd_number_frequency

        # Update the maximum difference if needed
        if current_frequency_difference > \
                maximum_frequency_difference:

            maximum_frequency_difference = \
                current_frequency_difference

    return maximum_frequency_difference

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array once, where n is the size of the array. Inside the loop, it performs constant time operations: checking if a number is even or odd, incrementing a counter, and calculating/comparing the difference. These operations do not depend on the input size. Therefore, the time complexity is directly proportional to the number of elements in the input array, resulting in a linear time complexity of O(n).

Space Complexity

O(1) – The provided solution uses two integer variables to store the counts of even and odd numbers, and another integer variable to store the maximum difference encountered so far. These variables take up a constant amount of space, irrespective of the number of input elements. Therefore, the auxiliary space used by the algorithm does not depend on the input size, N. Consequently, the space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 if the input array is null or empty to avoid null pointer exceptions and represent no difference.
Array with all elements having the same frequency parity (all even or all odd)	If all frequencies are even or all are odd, the maximum difference will be negative if odd or 0 if even; correctly returning the computed difference handles this scenario.
Array containing only one element	Frequency would be odd; return the element's frequency (1) if it's positive, or handle it based on problem specification if negative, since there are no even frequencies to subtract.
Large input array size (scaling concerns)	Use a hash map (dictionary) to store frequencies for O(n) time complexity to ensure scalability.
Presence of negative numbers in the array	Hash map will correctly count negative numbers as different elements, and the frequency calculation remains valid.
Large frequency values leading to integer overflow	Consider using long integers (64-bit) for frequency counts if array sizes could lead to integer overflow.
Array containing zeros	Zero is treated as any other number and its frequency is tracked correctly by the hash map.
Array with a mix of very small and very large numbers	The algorithm treats the numbers independently during frequency counting and difference calculation, so the magnitude doesn't inherently cause an issue; ensure the frequency counters can handle large counts.

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 is the expected return value if the array is empty or if there are no even or odd frequency elements?
Are the array elements guaranteed to be integers, or could they be other data types like floating-point numbers?
How large can the input array be, and what is the range of possible values for the elements within the array?
If multiple pairs of elements yield the same maximum difference, is any one of them acceptable as a valid solution?
Can you provide an example input and its corresponding expected output to illustrate the problem more clearly?

Brute Force Solution

Approach

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

Consider all possible starting points in the set of numbers.
For each starting point, create a group of numbers by adding one number at a time.
For each group of numbers, count how many times each distinct number appears. This gives us the frequencies.
Separate the numbers that appeared an even number of times from the numbers that appeared an odd number of times.
Add up all the even frequencies. This is the total even frequency.
Add up all the odd frequencies. This is the total odd frequency.
Subtract the total odd frequency from the total even frequency. This is the difference for that group.
Compare this difference to the largest difference we've found so far. If it's bigger, remember it.
Repeat steps 2-8 for every starting point.
After checking every possible group, the largest difference we remembered is our answer.

Code Implementation

def maximum_difference_between_even_and_odd_frequency_brute_force(numbers):
    maximum_difference = float('-inf')

    for start_index in range(len(numbers)):
        for end_index in range(start_index, len(numbers)):
            sub_array = numbers[start_index : end_index + 1]
            
            frequency_map = {}
            for number in sub_array:
                frequency_map[number] = frequency_map.get(number, 0) + 1
            
            even_frequency_sum = 0
            odd_frequency_sum = 0

            #Iterate through the map to aggregate even and odd frequencies
            for number, frequency in frequency_map.items():
                if frequency % 2 == 0:
                    even_frequency_sum += frequency
                else:
                    odd_frequency_sum += frequency

            difference = even_frequency_sum - odd_frequency_sum

            # Update maximum difference if necessary.
            if difference > maximum_difference:
                maximum_difference = difference

    return maximum_difference

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, create two separate counts: one for even numbers and one for odd numbers. Start both counts at zero.
Then, go through the numbers one by one.
For each number, check if it is even or odd.
If the number is even, increase the even count by one. If it's odd, increase the odd count by one.
After processing each number, calculate the difference between the even count and the odd count (even count minus odd count).
Keep track of the largest difference you've encountered so far. If the current difference is larger than the largest difference you've seen, update the largest difference.
After going through all of the numbers, the largest difference you've tracked is the answer.

Code Implementation

def maximum_difference_even_odd_frequency(numbers):
    even_number_frequency = 0
    odd_number_frequency = 0

    maximum_frequency_difference = 0

    for number in numbers:
        # Need to check if the number is even or odd
        if number % 2 == 0:
            even_number_frequency += 1

        else:
            odd_number_frequency += 1

        # This is the key calculation of the difference
        current_frequency_difference = \
            even_number_frequency - odd_number_frequency

        # Update the maximum difference if needed
        if current_frequency_difference > \
                maximum_frequency_difference:

            maximum_frequency_difference = \
                current_frequency_difference

    return maximum_frequency_difference

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 if the input array is null or empty to avoid null pointer exceptions and represent no difference.
Array with all elements having the same frequency parity (all even or all odd)	If all frequencies are even or all are odd, the maximum difference will be negative if odd or 0 if even; correctly returning the computed difference handles this scenario.
Array containing only one element	Frequency would be odd; return the element's frequency (1) if it's positive, or handle it based on problem specification if negative, since there are no even frequencies to subtract.
Large input array size (scaling concerns)	Use a hash map (dictionary) to store frequencies for O(n) time complexity to ensure scalability.
Presence of negative numbers in the array	Hash map will correctly count negative numbers as different elements, and the frequency calculation remains valid.
Large frequency values leading to integer overflow	Consider using long integers (64-bit) for frequency counts if array sizes could lead to integer overflow.
Array containing zeros	Zero is treated as any other number and its frequency is tracked correctly by the hash map.
Array with a mix of very small and very large numbers	The algorithm treats the numbers independently during frequency counting and difference calculation, so the magnitude doesn't inherently cause an issue; ensure the frequency counters can handle large counts.