Minimum Index of a Valid Split

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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.

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 return value if no valid split exists?
What are the possible ranges for the integer values within the input array?
Can the input array be empty or null?
Are duplicate numbers allowed within the input array, and if so, how should their frequency affect the determination of a valid split?
Could you define more precisely what constitutes a 'valid split'? Specifically, what does 'majority' mean when the left and right subarrays have the same value counts?

Brute Force Solution

Approach

The brute force approach to finding a valid split involves checking every possible split point in the given list. We systematically divide the list into two parts at each possible location and then check if that split meets our 'valid' criteria. If a split is valid, we remember it; otherwise, we move on to the next possible split location.

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

Imagine we can place a divider between elements in the list.
First, place the divider after the very first element and see if that creates a valid split according to the requirements.
If the split is valid, remember where this divider is.
Next, shift the divider one position to the right, placing it after the second element. Again, check if this new split is valid.
If this split is valid, compare where it occurs to previously found locations and keep the earlier one.
Continue moving the divider one position at a time, checking the validity of the split at each location, until you have tried placing the divider after every element (except the very last one).
After checking all the possible locations, if we have found a valid location for the divider, return where that divider was placed. If we have not, then it's impossible to find a location and we can record this

Code Implementation

def minimum_index_of_a_valid_split(numbers):
    list_length = len(numbers)
    if list_length < 2:
        return -1

    minimum_index = -1

    for split_index in range(list_length - 1):
        # Create the left and right sublists
        left_sublist = numbers[:split_index + 1]
        right_sublist = numbers[split_index + 1:]

        # Find the dominant element in the left sublist
        left_dominant = find_dominant(left_sublist)
        # Find the dominant element in the right sublist
        right_dominant = find_dominant(right_sublist)

        # Check if the split is valid.
        if left_dominant != -1 and right_dominant != -1 and left_dominant == right_dominant:
            # If no valid split has been found yet, record the current split index
            if minimum_index == -1:
                minimum_index = split_index

    return minimum_index

def find_dominant(sublist):
    list_length = len(sublist)
    if list_length == 0:
        return -1

    element_counts = {}
    for number in sublist:
        element_counts[number] = element_counts.get(number, 0) + 1

    # Find the element with a count greater than half of the sublist's length.
    for number, count in element_counts.items():
        if count * 2 > list_length:
            return number

    return -1

Big(O) Analysis

Time Complexity

O(n²) – The brute force approach iterates through all possible split points, which is up to n-1 splits. For each split, we need to iterate through the left and right subarrays to count occurrences of each number, which takes O(n) time in the worst case. Therefore, the overall time complexity is approximately (n-1) * O(n), which simplifies to O(n²).

Space Complexity

O(1) – The provided approach iterates through the input list, checking each possible split point. It only needs to store a variable to track the minimum index of a valid split found so far. No auxiliary data structures like arrays, hash maps, or recursion are used; therefore, the space used remains constant irrespective of the input list's size, N.

Optimal Solution

Approach

The goal is to find a place to split a list into two parts where one value is dominant in both parts. The key idea is to efficiently track how many times each value appears as we go through the list, so we can quickly check dominance at each possible split point.

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

First, figure out how many times each number appears in the entire list. We need this for later.
Now, walk through the list from beginning to end, one number at a time.
As you move, keep track of how many times each number has appeared so far in the first part of the list.
At each number, check if any number is dominant in the first part of the list. A number is dominant if it appears more than half the time in that part.
Also, check if any number is dominant in the remaining part of the list (from where you are to the end). You can figure this out using the counts from step 1 and step 3.
If you find a number that is dominant in both parts, you've found a valid split point. The first such split point you find is the one we want.
If you get to the end of the list without finding a valid split, then there isn't one.

Code Implementation

def find_minimum_index_of_a_valid_split(numbers) -> int:
    total_counts = {}
    for number in numbers:
        total_counts[number] = total_counts.get(number, 0) + 1

    left_counts = {}
    for index in range(len(numbers) - 1):
        number = numbers[index]
        left_counts[number] = left_counts.get(number, 0) + 1

        left_length = index + 1
        left_dominant = None
        for num, count in left_counts.items():
            if count * 2 > left_length:
                left_dominant = num
                break

        # If no dominant element exists in the left, continue
        if left_dominant is None:
            continue

        right_length = len(numbers) - left_length
        right_dominant = None
        for num in total_counts:
            right_count = total_counts[num] - left_counts.get(num, 0)
            if right_count * 2 > right_length:
                right_dominant = num
                break

        # Dominant elements must match for valid split
        if right_dominant is None:
            continue

        if left_dominant == right_dominant:

            # Found a valid split
            return index

    return -1

Big(O) Analysis

Time Complexity

O(n) – The algorithm first counts the frequency of each number in the list, which takes O(n) time. Then, it iterates through the list once (O(n)), maintaining counts of elements seen so far. Inside the loop, dominance checks are performed, which involve comparing counts. These checks take constant time, O(1), since they operate on a fixed number of counters. Therefore, the overall time complexity is dominated by the two linear iterations, resulting in O(n).

Space Complexity

O(N) – The algorithm uses a hash map (or dictionary) to store the frequency of each number in the entire list, as mentioned in step 1. In the worst-case scenario, where all numbers in the input list are distinct, the hash map will store N key-value pairs, where N is the number of elements in the input list. Therefore, the auxiliary space required scales linearly with the input size. This leads to a space complexity of O(N).

Edge Cases

Case	How to Handle
Empty or null input array	Return -1 immediately as a valid split is impossible.
Array with only one element	Return -1 as a split is impossible with only one element.
All elements in the array are the same	Iterate and check for valid split, if none exist, return -1.
Large input array (potential for integer overflow in count or size calculations)	Use long or appropriately sized data types to avoid integer overflow.
No valid split exists	After iterating through the entire array, if no valid split is found, return -1.
Array contains negative numbers	The dominant element calculation should correctly handle negative numbers using a hashmap to track frequencies.
Array with elements close to the maximum integer value	Ensure that frequency counts and majority value calculations do not overflow.
Multiple valid splits exist, but we need the minimum index	The first valid split index encountered during iteration should be returned, ensuring the minimum index is found.

Empty or null input array

How to Handle:

Return -1 immediately as a valid split is impossible.

Array with only one element

How to Handle:

Return -1 as a split is impossible with only one element.

All elements in the array are the same

How to Handle:

Iterate and check for valid split, if none exist, return -1.

Large input array (potential for integer overflow in count or size calculations)

How to Handle:

Use long or appropriately sized data types to avoid integer overflow.

No valid split exists

How to Handle:

After iterating through the entire array, if no valid split is found, return -1.

Array contains negative numbers

How to Handle:

The dominant element calculation should correctly handle negative numbers using a hashmap to track frequencies.

Array with elements close to the maximum integer value

How to Handle:

Ensure that frequency counts and majority value calculations do not overflow.

Multiple valid splits exist, but we need the minimum index

How to Handle:

The first valid split index encountered during iteration should be returned, ensuring the minimum index is found.