Jobs

Interviews

Discussions

Count Subarrays With More Ones Than Zeros

Medium

Asked by:

12 views

Topics:

ArraysDynamic ProgrammingGreedy Algorithms

Given a binary array nums, return the number of subarrays having more 1's than 0's.

Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [0,1,1,0,1]
Output: 9
Explanation:
The subarrays having more 1's than 0's are:
- [1]
- [1, 1]
- [1, 1, 0]
- [1, 1, 0, 1]
- [1, 0, 1]
- [1, 0, 1]
- [0, 1]
- [1]
- [0, 1]

Example 2:

Input: nums = [0]
Output: 0
Explanation:
No subarrays have more 1's than 0's.

Constraints:

1 <= nums.length <= 10⁵
nums[i] is either 0 or 1.

Try coding it in LeetCode

Count Subarrays With More Ones Than Zeros

Medium

Asked by:

12 views

Topics:

ArraysDynamic ProgrammingGreedy Algorithms

Given a binary array nums, return the number of subarrays having more 1's than 0's.

Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [0,1,1,0,1]
Output: 9
Explanation:
The subarrays having more 1's than 0's are:
- [1]
- [1, 1]
- [1, 1, 0]
- [1, 1, 0, 1]
- [1, 0, 1]
- [1, 0, 1]
- [0, 1]
- [1]
- [0, 1]

Example 2:

Input: nums = [0]
Output: 0
Explanation:
No subarrays have more 1's than 0's.

Constraints:

1 <= nums.length <= 10⁵
nums[i] is either 0 or 1.

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 is the expected data type of the input array elements, and what values can they take (e.g., only 0 and 1, or other integers)?
Can the input array be empty or null? If so, what should the function return in these cases?
Are overlapping subarrays counted multiple times, or should I only consider non-overlapping subarrays?
Could you clarify what is meant by 'more ones than zeros'? Does this mean strictly greater, or greater than or equal to?
What is the maximum possible size of the input array?

Brute Force Solution

Approach

To solve this problem the brute force way, we'll look at every possible group of numbers within the entire list. For each group, we will count how many ones and zeros there are and then check if the group has more ones than zeros.

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

Start by considering the first number in the list as a group by itself.
Check if this group has more ones than zeros. If it does, count it.
Now, consider the first two numbers as a group. Check if this group has more ones than zeros. If it does, count it.
Continue this process, each time adding one more number to the group, until you've considered a group that contains the entire list of numbers starting from the beginning.
Next, start from the second number in the list, and repeat the same process: consider it as a group by itself, then with the number after it, and so on.
Keep doing this, moving the starting point one number further each time, until you've considered all possible groups within the list.
In the end, you'll have a count of all the groups that have more ones than zeros.

Code Implementation

def count_subarrays_with_more_ones(binary_array):
    subarray_count = 0
    array_length = len(binary_array)

    for start_index in range(array_length):
        for end_index in range(start_index, array_length):
            # Define the subarray
            current_subarray = binary_array[start_index:end_index + 1]
            ones_count = 0
            zeros_count = 0

            for number in current_subarray:
                if number == 1:
                    ones_count += 1
                else:
                    zeros_count += 1

            # The condition that determines if we increment the count
            if ones_count > zeros_count:

                subarray_count += 1

    return subarray_count

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible subarrays of the input array. The outer loop selects the starting index of the subarray, which ranges from 0 to n-1. The inner loop determines the end index of the subarray, starting from the selected starting index and going up to n-1. For each subarray, it counts the number of ones and zeros. Therefore, the number of operations is proportional to the number of possible subarrays, which is approximately n * n/2. Thus, the time complexity is O(n²).

Space Complexity

O(1) – The provided brute force approach iterates through all possible subarrays, but it doesn't use any auxiliary data structures like arrays, lists, or hash maps to store intermediate results. The only memory used is for a few counter variables to keep track of the number of ones, zeros, and the overall count of subarrays with more ones than zeros. Therefore, the space complexity remains constant regardless of the input array size N, resulting in O(1) space complexity.

Optimal Solution

Approach

The most efficient way to count these special sections is to transform the problem and then cleverly keep track of running totals. We turn the ones and zeros into pluses and minuses, and then use these totals to quickly identify the sections we need.

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

First, think of all the ones in the sequence as plus one, and all the zeros as minus one.
Now, calculate the running total as you go through the sequence. This means adding up all the pluses and minuses so far at each point.
Next, notice that a section has more ones than zeros if its total is a positive number. However, instead of checking every section individually, we look for a faster way.
Consider the running totals calculated so far. For each position, we need to find earlier positions where the running total was lower. A section from that earlier position to the current position will have a positive sum, meaning it has more ones than zeros.
Use the running totals to count the number of earlier positions with a lower running total. This efficiently tells you the number of valid sections that end at the current position.
Repeat this process for every position in the sequence. Sum up the counts obtained for each position to find the total number of sections with more ones than zeros.

Code Implementation

def count_subarrays(binary_array):
    array_length = len(binary_array)
    running_total = 0
    count = 0
    totals_frequency = {0: 1}

    for i in range(array_length):
        # Convert 0 to -1 and update the running total.
        if binary_array[i] == 0:
            running_total -= 1
        else:
            running_total += 1

        # Find earlier positions where total was lower.
        for previous_total in totals_frequency:
            if previous_total < running_total:

                # Add the frequency of lower totals to the count.
                count += totals_frequency[previous_total]

        # Update the frequency of the current total.
        if running_total in totals_frequency:
            totals_frequency[running_total] += 1
        else:
            totals_frequency[running_total] = 1

    return count

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the input array of size n, computing the running total for each element. For each running total, it then iterates through all previous running totals to count how many are smaller than the current running total. This nested iteration results in a check for each pair of elements in the array. Therefore, the number of operations approximates n * (n - 1) / 2, which simplifies to O(n²).

Space Complexity

O(N) – The space complexity stems from the running totals calculated for each position in the sequence, which implies storing these totals. While the prompt doesn't explicitly state storing ALL running totals, to efficiently 'count the number of earlier positions with a lower running total', we would likely need to store them or a derivative of them. Therefore, the space used grows linearly with the input size N, where N is the length of the input sequence, as we are implicitly or explicitly keeping track of N running totals. Thus, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Null or Empty Input Array	Return 0 immediately as no subarrays can exist.
Array with all Zeros	Return 0 since no subarray can have more ones than zeros.
Array with all Ones	Return n*(n+1)/2 where n is the length of the array, as every subarray is valid.
Array with alternating Zeros and Ones	The cumulative sum approach should correctly count the valid subarrays in this scenario.
Integer Overflow in Cumulative Sum	Use a data type with a larger range (e.g., long) for the cumulative sum to prevent overflow.
Large Input Array (Performance)	Ensure the algorithm has at least O(n log n) or better complexity to avoid time limit exceedance.
Array with very few ones compared to zeros	The algorithm should iterate and correctly determine valid subarrays, potentially requiring optimization for cumulative sum lookups.
Array containing only one element	If the single element is one return 1 otherwise return 0.

Null or Empty Input Array

How to Handle:

Return 0 immediately as no subarrays can exist.

Array with all Zeros

How to Handle:

Return 0 since no subarray can have more ones than zeros.

Array with all Ones

How to Handle:

Return n*(n+1)/2 where n is the length of the array, as every subarray is valid.

Array with alternating Zeros and Ones

How to Handle:

The cumulative sum approach should correctly count the valid subarrays in this scenario.

Integer Overflow in Cumulative Sum

How to Handle:

Use a data type with a larger range (e.g., long) for the cumulative sum to prevent overflow.

Large Input Array (Performance)

How to Handle:

Ensure the algorithm has at least O(n log n) or better complexity to avoid time limit exceedance.

Array with very few ones compared to zeros

How to Handle:

The algorithm should iterate and correctly determine valid subarrays, potentially requiring optimization for cumulative sum lookups.

Array containing only one element

How to Handle:

If the single element is one return 1 otherwise return 0.

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 data type of the input array elements, and what values can they take (e.g., only 0 and 1, or other integers)?
Can the input array be empty or null? If so, what should the function return in these cases?
Are overlapping subarrays counted multiple times, or should I only consider non-overlapping subarrays?
Could you clarify what is meant by 'more ones than zeros'? Does this mean strictly greater, or greater than or equal to?
What is the maximum possible size of the input array?

Brute Force Solution

Approach

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

Start by considering the first number in the list as a group by itself.
Check if this group has more ones than zeros. If it does, count it.
Now, consider the first two numbers as a group. Check if this group has more ones than zeros. If it does, count it.
Continue this process, each time adding one more number to the group, until you've considered a group that contains the entire list of numbers starting from the beginning.
Next, start from the second number in the list, and repeat the same process: consider it as a group by itself, then with the number after it, and so on.
Keep doing this, moving the starting point one number further each time, until you've considered all possible groups within the list.
In the end, you'll have a count of all the groups that have more ones than zeros.

Code Implementation

def count_subarrays_with_more_ones(binary_array):
    subarray_count = 0
    array_length = len(binary_array)

    for start_index in range(array_length):
        for end_index in range(start_index, array_length):
            # Define the subarray
            current_subarray = binary_array[start_index:end_index + 1]
            ones_count = 0
            zeros_count = 0

            for number in current_subarray:
                if number == 1:
                    ones_count += 1
                else:
                    zeros_count += 1

            # The condition that determines if we increment the count
            if ones_count > zeros_count:

                subarray_count += 1

    return subarray_count

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, think of all the ones in the sequence as plus one, and all the zeros as minus one.
Now, calculate the running total as you go through the sequence. This means adding up all the pluses and minuses so far at each point.
Next, notice that a section has more ones than zeros if its total is a positive number. However, instead of checking every section individually, we look for a faster way.
Consider the running totals calculated so far. For each position, we need to find earlier positions where the running total was lower. A section from that earlier position to the current position will have a positive sum, meaning it has more ones than zeros.
Use the running totals to count the number of earlier positions with a lower running total. This efficiently tells you the number of valid sections that end at the current position.
Repeat this process for every position in the sequence. Sum up the counts obtained for each position to find the total number of sections with more ones than zeros.

Code Implementation

def count_subarrays(binary_array):
    array_length = len(binary_array)
    running_total = 0
    count = 0
    totals_frequency = {0: 1}

    for i in range(array_length):
        # Convert 0 to -1 and update the running total.
        if binary_array[i] == 0:
            running_total -= 1
        else:
            running_total += 1

        # Find earlier positions where total was lower.
        for previous_total in totals_frequency:
            if previous_total < running_total:

                # Add the frequency of lower totals to the count.
                count += totals_frequency[previous_total]

        # Update the frequency of the current total.
        if running_total in totals_frequency:
            totals_frequency[running_total] += 1
        else:
            totals_frequency[running_total] = 1

    return count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or Empty Input Array	Return 0 immediately as no subarrays can exist.
Array with all Zeros	Return 0 since no subarray can have more ones than zeros.
Array with all Ones	Return n*(n+1)/2 where n is the length of the array, as every subarray is valid.
Array with alternating Zeros and Ones	The cumulative sum approach should correctly count the valid subarrays in this scenario.
Integer Overflow in Cumulative Sum	Use a data type with a larger range (e.g., long) for the cumulative sum to prevent overflow.
Large Input Array (Performance)	Ensure the algorithm has at least O(n log n) or better complexity to avoid time limit exceedance.
Array with very few ones compared to zeros	The algorithm should iterate and correctly determine valid subarrays, potentially requiring optimization for cumulative sum lookups.
Array containing only one element	If the single element is one return 1 otherwise return 0.

Null or Empty Input Array

How to Handle:

Return 0 immediately as no subarrays can exist.

Array with all Zeros

How to Handle:

Return 0 since no subarray can have more ones than zeros.

Array with all Ones

How to Handle:

Return n*(n+1)/2 where n is the length of the array, as every subarray is valid.

Array with alternating Zeros and Ones

How to Handle:

The cumulative sum approach should correctly count the valid subarrays in this scenario.

Integer Overflow in Cumulative Sum

How to Handle:

Use a data type with a larger range (e.g., long) for the cumulative sum to prevent overflow.

Large Input Array (Performance)

How to Handle:

Ensure the algorithm has at least O(n log n) or better complexity to avoid time limit exceedance.

Array with very few ones compared to zeros

How to Handle:

The algorithm should iterate and correctly determine valid subarrays, potentially requiring optimization for cumulative sum lookups.

Array containing only one element

How to Handle:

If the single element is one return 1 otherwise return 0.