Minimum Operations to Make the Array Alternating

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Arrays

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.

Example 1:

Input: nums = [3,1,3,2,4,3]
Output: 3
Explanation:
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.
It can be proven that it is not possible to make the array alternating in less than 3 operations.

Example 2:

Input: nums = [1,2,2,2,2]
Output: 2
Explanation:
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.
Note that the array cannot be converted to [2,2,2,2,2] because in this case nums[0] == nums[1] which violates the conditions of an alternating array.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵

Try coding it in LeetCode

Minimum Operations to Make the Array Alternating

Medium

Asked by:

4 views

Topics:

Arrays

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.

Example 1:

Input: nums = [3,1,3,2,4,3]
Output: 3
Explanation:
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.
It can be proven that it is not possible to make the array alternating in less than 3 operations.

Example 2:

Input: nums = [1,2,2,2,2]
Output: 2
Explanation:
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.
Note that the array cannot be converted to [2,2,2,2,2] because in this case nums[0] == nums[1] which violates the conditions of an alternating array.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵

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 are the constraints on the size of the array `nums`?
What is the range of values that the elements in `nums` can take? Can they be negative, zero, or floating point numbers?
If there are multiple ways to achieve the minimum number of operations, is any one of them acceptable?
What should I return if the input array is empty or null?
Can you provide a more concrete example of an array that requires multiple operations to become alternating?

Brute Force Solution

Approach

The brute force strategy for making an array alternating involves exploring every possible combination of changes to the array's elements. We essentially try out different modifications, count the operations needed for each, and ultimately pick the one requiring the fewest operations that achieves the alternating pattern. This approach guarantees finding the absolute best answer by trying everything.

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

Consider every possible value for each spot in the array.
For each potential value in each spot, calculate how many changes it would take to transform the array into an alternating pattern.
An alternating pattern means that elements at even positions have the same value and elements at odd positions have the same value, and these two values are different.
Evaluate how many operations were needed to produce the alternating pattern for a selected combination of numbers.
Repeat the process of testing different numbers and checking operations for all possible combinations.
Keep track of the number of operations required for each scenario explored.
Finally, find the scenario from all the possibilities we explored that took the minimum number of operations.

Code Implementation

def min_operations_brute_force(numbers):
    array_length = len(numbers)
    unique_numbers = sorted(list(set(numbers)))
    minimum_operations = array_length

    # Iterate through all possible combinations for even and odd indices
    for even_index_number in unique_numbers:
        for odd_index_number in unique_numbers:
            # Ensure even and odd indexed numbers are different
            if even_index_number != odd_index_number:
                operations = 0
                # Count operations required for current combination

                for index in range(array_length):
                    if index % 2 == 0 and numbers[index] != even_index_number:
                        operations += 1
                    elif index % 2 != 0 and numbers[index] != odd_index_number:
                        operations += 1

                minimum_operations = min(minimum_operations, operations)

    # If no valid alternating pattern found return array length
    if minimum_operations == array_length:
        return array_length
    else:
        return minimum_operations

Big(O) Analysis

Time Complexity

O(k^n) – The provided brute force approach considers every possible value for each position in the array. If we assume there are k possible values for each of the n positions in the array, then we need to explore k * k * ... * k (n times) combinations. For each of these k^n combinations, we then need to verify if the array is alternating which takes O(n) time in the worst case. Therefore, the overall time complexity becomes O(n * k^n). Simplifying, the total number of operations is dominated by k^n, giving us a final time complexity of O(k^n) where k is the number of possible values each element can take and n is the size of the input array.

Space Complexity

O(1) – The brute force strategy, as described, does not utilize any auxiliary data structures that scale with the input array's size. It only calculates operation counts for different value combinations in place. Thus, the space complexity is constant because the algorithm does not use additional space that depends on the size, N, of the input array; it only involves storing a few counters.

Optimal Solution

Approach

The core idea is to analyze the frequency of numbers at even and odd positions separately. By understanding which numbers appear most often in these positions, we can minimize the changes needed to create an alternating pattern. It's like figuring out the most popular colors for even/odd slots and using those as a guide.

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

First, count how many times each different number appears in the even positions (0, 2, 4, etc.).
Then, do the same for the odd positions (1, 3, 5, etc.).
Find the number that shows up most often in the even positions and the number that shows up most often in the odd positions.
If the most frequent number in the even positions is *different* from the most frequent number in the odd positions, then the answer is simply the total number of elements minus the sum of those two frequencies. This means you keep the most frequent numbers and change the rest.
If the most frequent number is the *same* in both even and odd positions, things get a little trickier. In this case, you need to consider the second most frequent number in *both* even and odd positions.
Calculate two possibilities: one where you keep the most frequent number in the even positions and the *second* most frequent in the odd positions, and another where you keep the second most frequent in the even positions and the most frequent in the odd positions.
Take the minimum of those two possibilities. The minimum number represents the least amount of changes you need to make to get the alternating pattern.

Code Implementation

def minimum_operations_to_make_array_alternating(numbers):
    array_length = len(numbers)
    even_positions = {}
    odd_positions = {}

    for i in range(0, array_length, 2):
        even_positions[numbers[i]] = even_positions.get(numbers[i], 0) + 1

    for i in range(1, array_length, 2):
        odd_positions[numbers[i]] = odd_positions.get(numbers[i], 0) + 1

    even_sorted = sorted(even_positions.items(), key=lambda item: item[1], reverse=True)
    odd_sorted = sorted(odd_positions.items(), key=lambda item: item[1], reverse=True)

    even_most_frequent = even_sorted[0][0] if even_sorted else None
    even_most_frequent_count = even_sorted[0][1] if even_sorted else 0
    even_second_most_frequent = even_sorted[1][0] if len(even_sorted) > 1 else None
    even_second_most_frequent_count = even_sorted[1][1] if len(even_sorted) > 1 else 0

    odd_most_frequent = odd_sorted[0][0] if odd_sorted else None
    odd_most_frequent_count = odd_sorted[0][1] if odd_sorted else 0
    odd_second_most_frequent = odd_sorted[1][0] if len(odd_sorted) > 1 else None
    odd_second_most_frequent_count = odd_sorted[1][1] if len(odd_sorted) > 1 else 0

    # Most frequent numbers are different
    if even_most_frequent != odd_most_frequent:
        return array_length - even_most_frequent_count - odd_most_frequent_count

    # Most frequent numbers are the same,
    # need to consider the second most frequent
    else:
        option1 = array_length - even_most_frequent_count - odd_second_most_frequent_count
        option2 = array_length - even_second_most_frequent_count - odd_most_frequent_count

        # Choose optimal scenario to minimize operations.
        return min(option1, option2)

Big(O) Analysis

Time Complexity

O(n) – We iterate through the input array nums of size n to count the frequencies of numbers in even and odd positions separately. Finding the most frequent numbers in even and odd positions involves iterating through frequency maps, which in the worst case has a size proportional to n (if all numbers are distinct). Finally, the comparisons and calculations to determine the minimum operations take constant time. Therefore, the dominant operation is the initial frequency counting, resulting in a time complexity of O(n).

Space Complexity

O(N) – The algorithm uses hash maps (or dictionaries) to count the frequency of each number at even and odd positions. In the worst case, all numbers in the input array are distinct, leading to two hash maps potentially storing N/2 key-value pairs each, where N is the length of the input array. Additionally, the variables for storing the most frequent and second most frequent numbers consume constant space. Therefore, the auxiliary space used is proportional to N, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
Null or empty array input	Return 0 if the array is null or empty, as no operations are needed on an empty array.
Array with only one element	Return 0, as a single-element array is trivially alternating.
Array with two identical elements	Return 1, as one operation is needed to make them different.
Array with two different elements	Return 0, as it is already alternating.
Array with all elements identical	The result would be length of the array divided by 2, rounded up.
Large array with extreme value differences, potentially causing integer overflow in counting	Use a 64-bit integer type (long in Java/C++) for counts to prevent integer overflow.
Array with a large number of distinct elements	The HashMap or dictionary used to count frequencies could consume significant memory; ensure sufficient memory is available.
Array where the number of operations required is very high approaching the input size	The algorithm should still correctly identify the need for many changes and not get stuck in local minima.