Problem Description

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

The array nums is called alternating if it satisfies two conditions:

1. Elements that are 2 positions apart must be equal: nums[i - 2] == nums[i] for all valid indices where 2 <= i <= n - 1
2. Adjacent elements must be different: nums[i - 1] != nums[i] for all valid indices where 1 <= i <= n - 1

In simpler terms, an alternating array follows a pattern where:

• All elements at even positions (indices 0, 2, 4, ...) have the same value
• All elements at odd positions (indices 1, 3, 5, ...) have the same value
• The value at even positions must be different from the value at odd positions

For example, [3, 1, 3, 1, 3] is alternating because all even positions contain 3, all odd positions contain 1, and 3 ≠ 1.

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

Your task is to find the minimum number of operations required to transform the given array into an alternating array.

For instance, if nums = [3, 1, 3, 2, 4, 3], you could make it alternating by changing index 3 from 2 to 1 and index 4 from 4 to 3, resulting in [3, 1, 3, 1, 3, 1] with just 2 operations.

Intuition

To make the array alternating with minimum operations, we need to think about what values should appear at even positions and odd positions. Since all even positions must have the same value and all odd positions must have the same value (but different from even positions), we want to maximize the number of elements that are already in the correct position.

The key insight is that we should choose the most frequent values already present at even and odd positions respectively. Why? Because if a value appears frequently at even positions, keeping it means fewer changes are needed at those positions.

Let's break down the strategy:

1. Count the frequency of each value at even positions (indices 0, 2, 4, ...)
2. Count the frequency of each value at odd positions (indices 1, 3, 5, ...)
3. Find the most frequent value at even positions and the most frequent value at odd positions

However, there's a catch: what if the most frequent value at even positions is the same as the most frequent value at odd positions? We can't use the same value for both since adjacent elements must be different.

This leads us to track not just the most frequent value at each position type, but also the second most frequent. If the most frequent values are the same for even and odd positions, we have two choices:

• Use the most frequent value at even positions and the second most frequent at odd positions
• Use the second most frequent value at even positions and the most frequent at odd positions

We choose whichever option preserves more existing elements (requires fewer changes).

The total operations needed equals n (total elements) minus the number of elements we can keep unchanged. By maximizing the elements we keep, we minimize the operations required.

Learn more about Greedy patterns.

Solution Approach

The implementation uses a helper function f(i) that analyzes elements at positions starting from index i with a step of 2 (i.e., nums[i::2]). This allows us to separately analyze even positions (when i=0) and odd positions (when i=1).

Helper Function f(i):

1. Uses Counter to count the frequency of each value at positions i, i+2, i+4, ...
2. Tracks two variables k1 and k2 representing the most frequent and second most frequent values
3. Iterates through the counter to find these top two values:
   • If a value's count is greater than the current most frequent (cnt[k1] < v), it becomes the new most frequent, and the previous most frequent becomes second
   • If a value's count is greater than the current second most frequent but not the first (cnt[k2] < v), it becomes the new second most frequent
4. Returns a tuple: (k1, cnt[k1], k2, cnt[k2]) containing the most frequent value, its count, second most frequent value, and its count

Main Algorithm:

1. Call f(0) to get the top two values and their counts at even positions, stored in tuple a

   • a[0]: most frequent value at even positions
   • a[1]: count of most frequent value at even positions
   • a[2]: second most frequent value at even positions
   • a[3]: count of second most frequent value at even positions

2. Call f(1) to get the top two values and their counts at odd positions, stored in tuple b

   • Similar structure as tuple a but for odd positions

3. Check if the most frequent values are different (a[0] != b[0]):

   • If different: We can use both most frequent values. The minimum operations = n - (a[1] + b[1])
   • This means we keep a[1] elements at even positions and b[1] elements at odd positions unchanged

4. If the most frequent values are the same (a[0] == b[0]):

   • We have two options:
     • Keep the most frequent at even positions and second most frequent at odd positions: a[1] + b[3] elements unchanged
     • Keep the second most frequent at even positions and most frequent at odd positions: a[3] + b[1] elements unchanged
   • Choose the option that keeps more elements: n - max(a[1] + b[3], a[3] + b[1])

The time complexity is O(n) for counting elements, and space complexity is O(n) for storing the frequency counts.

Example Walkthrough

Let's walk through the solution with nums = [2, 4, 2, 4, 2, 1].

Step 1: Analyze even positions (indices 0, 2, 4)

• Elements at even positions: [2, 2, 2]
• Using f(0):
  • Count frequencies: {2: 3}
  • Most frequent value k1 = 2 with count 3
  • Second most frequent k2 = 0 with count 0 (default, no other values)
  • Returns: (2, 3, 0, 0)

Step 2: Analyze odd positions (indices 1, 3, 5)

• Elements at odd positions: [4, 4, 1]
• Using f(1):
  • Count frequencies: {4: 2, 1: 1}
  • Most frequent value k1 = 4 with count 2
  • Second most frequent k2 = 1 with count 1
  • Returns: (4, 2, 1, 1)

Step 3: Determine minimum operations

• Check if most frequent values are different: 2 ≠ 4 ✓
• Since they're different, we can use both:
  • Keep value 2 at all even positions (3 elements unchanged)
  • Keep value 4 at odd positions (2 elements unchanged)
  • Total elements we can keep: 3 + 2 = 5
  • Minimum operations: 6 - 5 = 1

Result: We need only 1 operation - change index 5 from 1 to 4, giving us [2, 4, 2, 4, 2, 4].

Another example with conflict: nums = [3, 1, 3, 3, 3]

Step 1: Even positions (indices 0, 2, 4)

• Elements: [3, 3, 3]
• f(0) returns: (3, 3, 0, 0)

Step 2: Odd positions (indices 1, 3)

• Elements: [1, 3]
• f(1) returns: (1, 1, 3, 1) or (3, 1, 1, 1) depending on iteration order

Let's say f(1) returns (3, 1, 1, 1) (3 appears once, 1 appears once)

Step 3: Handle conflict

• Most frequent values are the same: 3 == 3
• Two options:
  • Use 3 at even, 1 at odd: Keep 3 + 1 = 4 elements
  • Use 0 at even, 3 at odd: Keep 0 + 1 = 1 element
• Choose first option (keeps more): 5 - 4 = 1 operation

Result: Change index 3 from 3 to 1, giving us [3, 1, 3, 1, 3].

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the array nums.

The algorithm processes each element in the array exactly once when creating the Counter objects. The function f is called twice - once for even indices f(0) and once for odd indices f(1). Each call to f:

• Creates a Counter from half of the array elements using nums[i::2], which takes O(n/2) time
• Iterates through the Counter items once to find the two most frequent elements, which takes O(unique_elements) time, and in the worst case O(n/2) time

Since we call f twice and each call processes approximately n/2 elements, the total time is O(n/2) + O(n/2) = O(n).

Space Complexity: O(n), where n is the length of the array nums.

The space is primarily consumed by:

• The Counter object in function f, which stores at most n/2 elements (either all even-indexed or all odd-indexed elements)
• The function is called twice, but the Counter objects are not stored simultaneously - they are created, used, and then discarded
• The variables a and b store tuples of fixed size (4 elements each), which is O(1)

In the worst case, when all elements at even (or odd) positions are unique, the Counter would store n/2 key-value pairs, resulting in O(n) space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall: Not Handling Edge Cases with Array Length ≤ 2

The Problem: The current implementation doesn't properly handle arrays with length 1 or 2. When nums has:

• Length 1: The array is already alternating (trivially), requiring 0 operations
• Length 2: If the two elements are different, it's already alternating (0 operations); if they're the same, we need 1 operation

The helper function find_top_two_frequent will attempt to access nums[start_index::2], which works but the Counter logic with uninitialized keys (frequency_map[0] when 0 is not in the map) will cause a KeyError.

Example of the Issue:

nums = [1]  # Length 1
# find_top_two_frequent(1) tries to access nums[1::2] which is empty
# frequency_map[most_frequent_val] with most_frequent_val=0 causes KeyError

nums = [1, 1]  # Length 2, same values
# The function might not handle this correctly

Solution: Add special handling for small arrays at the beginning of the main function:

class Solution:
    def minimumOperations(self, nums: List[int]) -> int:
        n = len(nums)
      
        # Handle edge cases
        if n == 1:
            return 0
        if n == 2:
            return 0 if nums[0] != nums[1] else 1
      
        def find_top_two_frequent(start_index: int) -> Tuple[int, int, int, int]:
            # Use defaultdict or handle empty slices
            frequency_map = Counter(nums[start_index::2])
          
            # Handle case where there are no elements at these positions
            if not frequency_map:
                return (0, 0, 0, 0)
          
            # Initialize with actual values from the map to avoid KeyError
            most_frequent_val = 0
            second_frequent_val = 0
            most_frequent_count = 0
            second_frequent_count = 0
          
            for value, count in frequency_map.items():
                if count > most_frequent_count:
                    second_frequent_val = most_frequent_val
                    second_frequent_count = most_frequent_count
                    most_frequent_val = value
                    most_frequent_count = count
                elif count > second_frequent_count:
                    second_frequent_val = value
                    second_frequent_count = count
          
            return (most_frequent_val, most_frequent_count, 
                    second_frequent_val, second_frequent_count)
      
        # Rest of the implementation remains the same...

Alternative Pitfall: Integer Overflow in Other Languages

While not an issue in Python, if implementing this in languages like C++ or Java, counting operations with large arrays could potentially cause integer overflow. Always use appropriate data types (like long in Java or long long in C++) when dealing with counts and sums.