All Divisions With the Highest Score of a Binary Array

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You are given a 0-indexed binary array nums of length n. nums can be divided at index i (where 0 <= i <= n) into two arrays (possibly empty) nums_left and nums_right:

nums_left has all the elements of nums between index 0 and i - 1 (inclusive), while nums_right has all the elements of nums between index i and n - 1 (inclusive).
If i == 0, nums_left is empty, while nums_right has all the elements of nums.
If i == n, nums_left has all the elements of nums, while nums_right is empty.

The division score of an index i is the sum of the number of 0's in nums_left and the number of 1's in nums_right.

Return all distinct indices that have the highest possible division score. You may return the answer in any order.

Example 1:

Input: nums = [0,0,1,0]
Output: [2,4]
Explanation: Division at index
- 0: nums_left is []. nums_right is [0,0,1,0]. The score is 0 + 1 = 1.
- 1: nums_left is [0]. nums_right is [0,1,0]. The score is 1 + 1 = 2.
- 2: nums_left is [0,0]. nums_right is [1,0]. The score is 2 + 1 = 3.
- 3: nums_left is [0,0,1]. nums_right is [0]. The score is 2 + 0 = 2.
- 4: nums_left is [0,0,1,0]. nums_right is []. The score is 3 + 0 = 3.
Indices 2 and 4 both have the highest possible division score 3.
Note the answer [4,2] would also be accepted.

Example 2:

Input: nums = [0,0,0]
Output: [3]
Explanation: Division at index
- 0: nums_left is []. nums_right is [0,0,0]. The score is 0 + 0 = 0.
- 1: nums_left is [0]. nums_right is [0,0]. The score is 1 + 0 = 1.
- 2: nums_left is [0,0]. nums_right is [0]. The score is 2 + 0 = 2.
- 3: nums_left is [0,0,0]. nums_right is []. The score is 3 + 0 = 3.
Only index 3 has the highest possible division score 3.

Example 3:

Input: nums = [1,1]
Output: [0]
Explanation: Division at index
- 0: nums_left is []. nums_right is [1,1]. The score is 0 + 2 = 2.
- 1: nums_left is [1]. nums_right is [1]. The score is 0 + 1 = 1.
- 2: nums_left is [1,1]. nums_right is []. The score is 0 + 0 = 0.
Only index 0 has the highest possible division score 2.

Constraints:

n == nums.length
1 <= n <= 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 size of the input array? Should I optimize for very large arrays?
Can the binary array be empty, or contain values other than 0 and 1?
If multiple divisions have the same highest score, in what order should I return their indices?
Could you define more precisely how the "score" for a division is calculated when the array is divided into two sub-arrays?
If all elements in the array are 0, or all are 1, what constitutes the 'highest score', and how should I handle such cases?

Brute Force Solution

Approach

We need to find the best place to split a series of zeros and ones. The brute force method involves checking every possible split to see which one gives us the highest score. We calculate the score for each possible split and keep track of the best ones.

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

Imagine you can cut the series of zeros and ones at different points.
Start by cutting before the first number. Count all the zeros that are to the left of the cut and all the ones that are to the right of the cut. Add those two counts together. This is your score for that cut.
Now, move the cut to after the first number. Again, count all the zeros to the left and all the ones to the right. Add them together. This is your new score.
Keep moving the cut one position at a time, calculating the score for each position.
As you calculate the scores, remember the highest score you've seen so far. Also, remember all the cut positions that resulted in that highest score.
After you've tried every possible cut position, look at the positions that gave you the highest score. Those are your best cuts.

Code Implementation

def all_divisions_with_highest_score_brute_force(numbers):
    maximum_score = -1
    indices_with_maximum_score = []

    for division_index in range(len(numbers) + 1):
        # Initialize counts for zeros on the left and ones on the right
        zeros_on_left = 0
        ones_on_right = 0

        # Count zeros to the left of the division
        for index_left in range(division_index):
            if numbers[index_left] == 0:
                zeros_on_left += 1

        # Count ones to the right of the division
        for index_right in range(division_index, len(numbers)):
            if numbers[index_right] == 1:
                ones_on_right += 1

        current_score = zeros_on_left + ones_on_right

        # Update the maximum score and indices if necessary
        if current_score > maximum_score:
            # Found a new maximum, so reset
            maximum_score = current_score
            indices_with_maximum_score = [division_index]
        elif current_score == maximum_score:
            # Found another index with the maximum, so add
            indices_with_maximum_score.append(division_index)

    return indices_with_maximum_score

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array nums of size n to calculate the score for each possible division. For each division point, it counts the number of zeros to the left and ones to the right. However, to avoid nested loops and achieve O(n), these counts can be efficiently computed using prefix sums or by maintaining running counts of zeros and ones. Therefore the time complexity is directly proportional to the size of the input array, resulting in O(n) time complexity.

Space Complexity

O(1) – The provided explanation describes an iterative process that calculates scores at each division point and keeps track of the highest score seen so far and the corresponding division indices. Only a few variables are used to store the current score, the maximum score found so far, and a list to store the indices with the maximum score. The space required for these variables does not depend on the size of the input array; therefore, the auxiliary space complexity is constant.

Optimal Solution

Approach

The key to efficiently finding the best division point in the binary array is to avoid repeatedly counting zeros and ones for each potential division. We can achieve this by calculating the prefix sums of both zeros and ones, and then using these to compute the score for each division point very quickly.

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

First, count all the ones in the entire array. This is like having a total count that you can use as a reference.
Next, go through the array from left to right, keeping track of how many zeros you've seen so far and how many ones you've seen so far.
For each possible dividing point, figure out the score. The score is the number of zeros to the left of the divider plus the number of ones to the right of the divider. We can get the ones to the right by substracting our running ones count from the total ones count.
Compare the score at each divider to the highest score found so far. If the current score is higher, replace the highest score and start a new list of dividers with that score.
If the score is equal to the highest score, add the current divider to the list of dividers with the highest score.
In the end, you will have a list of all the divider positions that give the highest possible score.

Code Implementation

def find_best_division(binary_array):
    total_ones = sum(binary_array)
    max_score = -1
    indices_with_max_score = []
    zeros_count = 0
    ones_count = 0

    for i in range(len(binary_array) + 1):
        # Calculates the score for the current division.
        score = zeros_count + (total_ones - ones_count)

        if score > max_score:
            # A new higher score found.
            max_score = score
            indices_with_max_score = [i]
        elif score == max_score:
            indices_with_max_score.append(i)

        # Avoid index out of bounds error at the last iteration
        if i < len(binary_array):
            if binary_array[i] == 0:
				#Update the count of zeros seen so far.
                zeros_count += 1
            else:
                ones_count += 1

    return indices_with_max_score

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array nums once to count the total number of ones, which takes O(n) time. Then, it iterates through the array again to calculate the score for each division point. Inside this loop, only constant-time operations (addition, subtraction, comparison) are performed. Thus, the second loop also takes O(n) time. The overall time complexity is therefore O(n) + O(n), which simplifies to O(n).

Space Complexity

O(N) – The algorithm uses a list to store the indices corresponding to the highest score. In the worst-case scenario, every division point could result in the same highest score, requiring the storage of all indices from 0 to N. Therefore, the space used by the algorithm scales linearly with the size of the input array, N, due to the space needed for this list. All other variables take constant space.

Edge Cases

Case	How to Handle
Null or empty input array	Return an empty list because no divisions can be evaluated.
Array containing only zeros	All divisions will have the same score (the array length), so return a list of all indices.
Array containing only ones	The scores increase linearly from 0 to n, so only index n will be returned.
Large input array exceeding memory constraints	Optimize the solution to use O(1) extra space, avoiding storing large intermediate arrays.
Array with alternating 0s and 1s	Carefully consider how the prefix sums of 0s and 1s interact to correctly compute the scores.
Input array is very long, leading to potential integer overflow issues when calculating score	Use a 64-bit integer type to store scores, or normalize them if needed.
Multiple divisions with the same highest score	The algorithm needs to collect and return all indices where the highest score is achieved.
Array with a single element	Return 0 if the element is 1 and 1 if the element is 0.

Null or empty input array

How to Handle:

Return an empty list because no divisions can be evaluated.

Array containing only zeros

How to Handle:

All divisions will have the same score (the array length), so return a list of all indices.

Array containing only ones

How to Handle:

The scores increase linearly from 0 to n, so only index n will be returned.

Large input array exceeding memory constraints

How to Handle:

Optimize the solution to use O(1) extra space, avoiding storing large intermediate arrays.

Array with alternating 0s and 1s

How to Handle:

Carefully consider how the prefix sums of 0s and 1s interact to correctly compute the scores.

Input array is very long, leading to potential integer overflow issues when calculating score

How to Handle:

Use a 64-bit integer type to store scores, or normalize them if needed.

Multiple divisions with the same highest score

How to Handle:

The algorithm needs to collect and return all indices where the highest score is achieved.

Array with a single element

How to Handle:

Return 0 if the element is 1 and 1 if the element is 0.