Minimum Operations to Make the Array Alternating

You are given a 0-indexed array nums consisting of n positive integers.

The array nums is called alternating if:

• nums[i - 2] == nums[i], where 2 <= i <= n - 1.
• nums[i - 1] != nums[i], where 1 <= i <= n - 1.

In one operation, you can choose an index i and change nums[i] into any positive integer.

Return the minimum number of operations required to make the array alternating.

For example:

nums = [3,1,3,2,4,3] should return 3 because one way to make the array alternating is by converting it to [3,1,3,1,3,1]. The number of operations required in this case is 3.

nums = [1,2,2,2,2] should return 2 because one way to make the array alternating is by converting it to [1,2,1,2,1]. The number of operations required in this case is 2.

Let's start by outlining a brute-force approach to solve this problem, followed by a more optimized solution.

Brute-Force Approach

1. Count occurrences: For each position (even and odd), count the occurrences of each number.
2. Iterate through all possibilities: Try every possible pair of numbers for even and odd positions.
3. Calculate operations: Calculate the number of operations required for each pair.
4. Find minimum: Find the minimum number of operations among all pairs.

def min_operations_brute_force(nums):
    n = len(nums)
    if n <= 1:
        return 0

    even_counts = {}
    odd_counts = {}

    for i in range(0, n, 2):
        even_counts[nums[i]] = even_counts.get(nums[i], 0) + 1
    for i in range(1, n, 2):
        odd_counts[nums[i]] = odd_counts.get(nums[i], 0) + 1

    even_nums = list(even_counts.keys())
    odd_nums = list(odd_counts.keys())

    min_ops = float('inf')

    for even_num in even_nums:
        for odd_num in odd_nums:
            if even_num != odd_num:
                ops = 0
                for i in range(0, n, 2):
                    if nums[i] != even_num:
                        ops += 1
                for i in range(1, n, 2):
                    if nums[i] != odd_num:
                        ops += 1
                min_ops = min(min_ops, ops)

    return min_ops

Optimized Approach

1. Count occurrences: Count the occurrences of each number at even and odd positions.
2. Find most frequent: Find the most and second most frequent numbers at even and odd positions.
3. Calculate operations:
   • If the most frequent numbers are different, the operations needed are n - count_even_most - count_odd_most.
   • If the most frequent numbers are the same, consider using the second most frequent numbers to minimize operations.

from collections import Counter

def min_operations(nums):
    n = len(nums)
    even_nums = nums[::2]
    odd_nums = nums[1::2]

    even_counts = Counter(even_nums)
    odd_counts = Counter(odd_nums)

    even_most_common = even_counts.most_common(2)
    odd_most_common = odd_counts.most_common(2)

    if even_most_common[0][0] != odd_most_common[0][0]:
        return n - even_most_common[0][1] - odd_most_common[0][1]
    else:
        if len(even_most_common) == 1 and len(odd_most_common) == 1:
            return min(len(even_nums), len(odd_nums))
        elif len(even_most_common) == 1:
            return n - even_most_common[0][1] - odd_most_common[1][1]
        elif len(odd_most_common) == 1:
            return n - even_most_common[1][1] - odd_most_common[0][1]
        else:
            return min(n - even_most_common[0][1] - odd_most_common[1][1],
                       n - even_most_common[1][1] - odd_most_common[0][1])

Example

nums = [3, 1, 3, 2, 4, 3]
print(min_operations(nums))

nums = [1, 2, 2, 2, 2]
print(min_operations(nums))

Big(O) Run-time Analysis

The optimized approach uses Counter to count occurrences, which takes O(n) time. Finding the most common elements takes O(1) because we are only retrieving the top 2. Therefore, the overall time complexity is O(n).

Big(O) Space Usage Analysis

The space complexity is O(n) because, in the worst case, the Counter objects even_counts and odd_counts can store up to n/2 distinct elements each.

Edge Cases

1. Empty array: If the array is empty, return 0.
2. Array with one element: If the array has only one element, return 0.
3. All elements are the same: The algorithm should correctly handle cases where all elements are the same.
4. Even and odd positions have same most frequent number: The algorithm considers second most frequent numbers to handle this case optimally.