Solution Explanation for Minimum Operations to Make the Array Alternating

This problem asks for the minimum number of operations needed to transform a given array into an alternating array. An array is considered alternating if:

1. nums[i - 2] == nums[i] for 2 <= i <= n - 1 (elements at even distance are equal).
2. nums[i - 1] != nums[i] for 1 <= i <= n - 1 (adjacent elements are different).

The optimal solution involves a greedy approach combined with counting element frequencies. We leverage the fact that an alternating array will have all even-indexed elements equal to some value, and all odd-indexed elements equal to a different value.

Algorithm:

1. Count Frequencies: We separately count the frequencies of elements at even and odd indices. This is efficiently done using a hash map (or dictionary) in most programming languages.

2. Find Top Two Frequencies: For both even and odd indices, we identify the two most frequent elements and their counts. This helps determine the best candidates for making the array alternating. If there's a tie for the second most frequent element, any of them can be chosen.

3. Calculate Minimum Operations: Two cases arise:

   • Different Top Elements: If the most frequent element at even indices is different from the most frequent element at odd indices, we can directly construct an alternating array using these elements. The number of operations needed is the total number of elements minus the sum of the counts of these two most frequent elements.

   • Same Top Elements: If the most frequent elements are the same for both even and odd indices, we need to consider two possibilities:

     • Use the most frequent element at even indices and the second most frequent element at odd indices.
     • Use the second most frequent element at even indices and the most frequent element at odd indices. We choose the option that requires fewer operations.

Time and Space Complexity:

• Time Complexity: O(N), where N is the length of the input array. This is dominated by the single pass to count frequencies and the subsequent steps, which all take linear time.
• Space Complexity: O(N) in the worst case. This is due to the hash maps used to store element frequencies. In the best case (few unique elements), the space usage could be much less.

Code Implementation (Python):

from collections import Counter
 
class Solution:
    def minimumOperations(self, nums: List[int]) -> int:
        def get_top_two(counts):
            sorted_counts = sorted(counts.items(), key=lambda item: item[1], reverse=True)
            if len(sorted_counts) < 2:
                return sorted_counts[0][0], sorted_counts[0][1], 0, 0  #Handle case with only one element
            return sorted_counts[0][0], sorted_counts[0][1], sorted_counts[1][0], sorted_counts[1][1]
 
 
        even_counts = Counter(nums[::2])
        odd_counts = Counter(nums[1::2])
 
        even_top1, even_cnt1, even_top2, even_cnt2 = get_top_two(even_counts)
        odd_top1, odd_cnt1, odd_top2, odd_cnt2 = get_top_two(odd_counts)
 
        if even_top1 != odd_top1:
            return len(nums) - (even_cnt1 + odd_cnt1)
        else:
            return len(nums) - max(even_cnt1 + odd_cnt2, even_cnt2 + odd_cnt1)

The code in other languages (Java, C++, Go, TypeScript) follows a very similar structure, adapting the data structures and syntax appropriately. The core logic of frequency counting and the conditional operation calculation remains consistent across all implementations.