Introduction

In this blog post, we will explore how to solve the LeetCode problem "2780" 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

An element x of an integer array arr of length m is dominant if more than half the elements of arr have a value of x. You are given a 0-indexed integer array nums of length n with one dominant element. You can split nums at an index i into two arrays nums[0, ..., i] and nums[i + 1, ..., n - 1], but the split is only valid if: 0 <= i < n - 1 nums[0, ..., i], and nums[i + 1, ..., n - 1] have the same dominant element. Here, nums[i, ..., j] denotes the subarray of nums starting at index i and ending at index j, both ends being inclusive. Particularly, if j < i then nums[i, ..., j] denotes an empty subarray. Return the minimum index of a valid split. If no valid split exists, return -1. Example 1: Input: nums = [1,2,2,2] Output: 2 Explanation: We can split the array at index 2 to obtain arrays [1,2,2] and [2]. In array [1,2,2], element 2 is dominant since it occurs twice in the array and 2 2 > 3. In array [2], element 2 is dominant since it occurs once in the array and 1 2 > 1. Both [1,2,2] and [2] have the same dominant element as nums, so this is a valid split. It can be shown that index 2 is the minimum index of a valid split. Example 2: Input: nums = [2,1,3,1,1,1,7,1,2,1] Output: 4 Explanation: We can split the array at index 4 to obtain arrays [2,1,3,1,1] and [1,7,1,2,1]. In array [2,1,3,1,1], element 1 is dominant since it occurs thrice in the array and 3 2 > 5. In array [1,7,1,2,1], element 1 is dominant since it occurs thrice in the array and 3 2 > 5. Both [2,1,3,1,1] and [1,7,1,2,1] have the same dominant element as nums, so this is a valid split. It can be shown that index 4 is the minimum index of a valid split. Example 3: Input: nums = [3,3,3,3,7,2,2] Output: -1 Explanation: It can be shown that there is no valid split. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 109 nums has exactly one dominant element.

Explanation

Here's the breakdown of the solution:

• Find the dominant element: Iterate through the array and use a dictionary (or collections.Counter) to efficiently determine the dominant element of the entire nums array.
• Iterate through potential splits: Iterate from index 0 up to n-2. For each index i, create virtual subarrays nums[0...i] and nums[i+1...n-1]. Determine the dominant elements for each subarray.

• Check for valid split: If both subarrays have the same dominant element, and it matches the dominant element of the original array, then i is a valid split. Return the smallest such index. If no valid split is found, return -1.

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

• Storage Complexity: O(n) in the worst-case scenario where the counts dictionary (or Counter) stores all unique elements of the array.

Code

    from collections import Counter

def minimumIndex(nums):
    n = len(nums)

    # Find the dominant element in the original array
    counts = Counter(nums)
    dominant_element = None
    for num, count in counts.items():
        if count * 2 > n:
            dominant_element = num
            break

    for i in range(n - 1):
        # Split the array into two subarrays: nums[0...i] and nums[i+1...n-1]
        left_subarray = nums[:i+1]
        right_subarray = nums[i+1:]

        # Find the dominant element in the left subarray
        left_counts = Counter(left_subarray)
        left_dominant = None
        left_len = len(left_subarray)
        if left_len > 0:
            for num, count in left_counts.items():
                if count * 2 > left_len:
                    left_dominant = num
                    break

        # Find the dominant element in the right subarray
        right_counts = Counter(right_subarray)
        right_dominant = None
        right_len = len(right_subarray)

        if right_len > 0:
            for num, count in right_counts.items():
                if count * 2 > right_len:
                    right_dominant = num
                    break

        # Check if the split is valid
        if left_dominant == dominant_element and right_dominant == dominant_element and left_dominant == right_dominant:
            return i

    return -1