What is the brute force solution for All Divisions With the Highest Score of a Binary Array?

The brute force approach for All Divisions With the Highest Score of a Binary Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves calculating the division score for every possible index by explicitly counting zeros and ones in the left and right subarrays. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve All Divisions With the Highest Score of a Binary Array?

All Divisions With the Highest Score of a Binary Array uses the following concepts: Array. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for All Divisions With the Highest Score of a Binary Array?

Explain your thought process clearly while coding. Discuss time and space complexity as you develop your solution. Consider edge cases and test your solution with them.

What are common mistakes when solving All Divisions With the Highest Score of a Binary Array?

Failing to consider edge cases like empty arrays. Not keeping track of the maximum score correctly.

#2155

All Divisions With the Highest Score of a Binary Array

Medium

ArrayHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages prefix sums to efficiently calculate the number of zeros and ones without redundant counting. This reduces the time complexity significantly.

⚙️

Algorithm

5 steps

1Step 1: Calculate the total number of ones in the entire array.
2Step 2: Initialize counters for left zeros and a list for indices with the maximum score.
3Step 3: Loop through the array, updating the count of left zeros and calculating the right ones using the total count minus the left count.
4Step 4: For each index, compute the score and update the maximum score and the list of indices accordingly.
5Step 5: Return the list of indices that have the highest score.

solution.py19 lines

1# Full working Python code
2
3def maxDivScore(nums):
4    total_ones = sum(nums)
5    left_zeros = 0
6    max_score = -1
7    indices = []
8    for i in range(len(nums) + 1):
9        right_ones = total_ones - (sum(nums[:i]))
10        score = left_zeros + right_ones
11        if score > max_score:
12            max_score = score
13            indices = [i]
14        elif score == max_score:
15            indices.append(i)
16        if i < len(nums) and nums[i] == 0:
17            left_zeros += 1
18    return indices
19

ℹ

Complexity note: This complexity is achieved by calculating the total number of ones in a single pass and then using a single loop to calculate scores, avoiding redundant counting.

1Understanding how to efficiently count elements in subarrays is crucial.
2Using prefix sums can drastically reduce time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.