Introduction

In this blog post, we will explore how to solve the LeetCode problem "2155" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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) numsleft and numsright: numsleft has all the elements of nums between index 0 and i - 1 (inclusive), while numsright has all the elements of nums between index i and n - 1 (inclusive). If i == 0, numsleft is empty, while numsright has all the elements of nums. If i == n, numsleft has all the elements of nums, while numsright is empty. The division score of an index i is the sum of the number of 0's in numsleft and the number of 1's in numsright. 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: numsleft is []. numsright is [0,0,1,0]. The score is 0 + 1 = 1. - 1: numsleft is [0]. numsright is [0,1,0]. The score is 1 + 1 = 2. - 2: numsleft is [0,0]. numsright is [1,0]. The score is 2 + 1 = 3. - 3: numsleft is [0,0,1]. numsright is [0]. The score is 2 + 0 = 2. - 4: numsleft is [0,0,1,0]. numsright 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: numsleft is []. numsright is [0,0,0]. The score is 0 + 0 = 0. - 1: numsleft is [0]. numsright is [0,0]. The score is 1 + 0 = 1. - 2: numsleft is [0,0]. numsright is [0]. The score is 2 + 0 = 2. - 3: numsleft is [0,0,0]. numsright 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: numsleft is []. numsright is [1,1]. The score is 0 + 2 = 2. - 1: numsleft is [1]. numsright is [1]. The score is 0 + 1 = 1. - 2: numsleft is [1,1]. numsright 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.

Explanation

Here's the breakdown of the solution:

• Prefix Sums: Utilize prefix sums to efficiently calculate the number of zeros to the left of each index and the number of ones to the right.
• Calculate Scores: Iterate through possible division indices and compute the division score using the prefix sums.

• Find Maximum and Indices: Track the maximum score encountered and store all indices that achieve this maximum score.

• Runtime Complexity: O(n), where n is the length of the input array.

• Storage Complexity: O(n) for the prefix sums, but can be optimized to O(1) by avoiding storing the prefix array, and O(k) for storing indices with maximum score, where k is the number of such indices. In the worst case, k could be n.

Code

    def max_score_indices(nums):
    n = len(nums)
    max_score = -1
    indices = []

    for i in range(n + 1):
        left_zeros = 0
        right_ones = 0

        # Calculate zeros in the left subarray
        for j in range(i):
            if nums[j] == 0:
                left_zeros += 1

        # Calculate ones in the right subarray
        for j in range(i, n):
            if nums[j] == 1:
                right_ones += 1

        score = left_zeros + right_ones

        if score > max_score:
            max_score = score
            indices = [i]
        elif score == max_score:
            indices.append(i)

    return indices