Minimum Array Changes to Make Differences Equal

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Arrays

You are given an integer array nums of size n where n is even, and an integer k.

You can perform some changes on the array, where in one change you can replace any element in the array with any integer in the range from 0 to k.

You need to perform some changes (possibly none) such that the final array satisfies the following condition:

There exists an integer X such that abs(a[i] - a[n - i - 1]) = X for all (0 <= i < n).

Return the minimum number of changes required to satisfy the above condition.

Example 1:

Input: nums = [1,0,1,2,4,3], k = 4

Output: 2

Explanation:
We can perform the following changes:

Replace nums[1] by 2. The resulting array is nums = [1,2,1,2,4,3].
Replace nums[3] by 3. The resulting array is nums = [1,2,1,3,4,3].

The integer X will be 2.

Example 2:

Input: nums = [0,1,2,3,3,6,5,4], k = 6

Output: 2

Explanation:
We can perform the following operations:

Replace nums[3] by 0. The resulting array is nums = [0,1,2,0,3,6,5,4].
Replace nums[4] by 4. The resulting array is nums = [0,1,2,0,4,6,5,4].

The integer X will be 4.

Constraints:

2 <= n == nums.length <= 10⁵
n is even.
0 <= nums[i] <= k <= 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 range of values within the input array?
Can the input array be empty or null? If so, what should I return?
Are we looking for the minimum number of *operations* (changing array elements), or the minimum number of *elements* that need to be changed?
By 'differences equal', do you mean the absolute difference between any two elements in the modified array should be the same, or specifically zero (all elements equal)?
If multiple solutions exist with the same minimum number of changes, is any one of them acceptable?

Brute Force Solution

Approach

The goal is to modify some numbers in a collection so that the difference between any two consecutive numbers is the same. A brute force approach means we systematically try every possible combination of changes to these numbers to see which gets us the desired outcome.

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

Consider the first number and try changing it to every possible value within some reasonable range.
For each of those changes, try changing the second number to every possible value within that same reasonable range.
Continue this process for every number in the collection, systematically exploring all possible combinations of values.
For each combination of changes to the numbers, check if the difference between each pair of consecutive numbers is equal.
Keep track of all the combinations that result in the differences being equal.
Among all the combinations that work, pick the one where you made the fewest changes to the original numbers. This is the solution that requires the minimum number of modifications.

Code Implementation

def min_array_changes_brute_force(numbers):
    array_length = len(numbers)
    min_changes = array_length + 1

    # Iterate through a reasonable range of values
    for first_value in range(-10, 11):
        for common_difference in range(-10, 11):
            changes_needed = 0
            
            # Check how many changes are needed for this sequence
            for index in range(array_length):
                expected_value = first_value + index * common_difference

                # Increment changes if current doesn't match expected
                if numbers[index] != expected_value:
                    changes_needed += 1

            # Update minimum if this sequence needs fewer changes
            if changes_needed < min_changes:
                min_changes = changes_needed

    if min_changes > array_length: 
        return -1
    else:
        return min_changes

Big(O) Analysis

Time Complexity

O(R^n * n) – The algorithm explores all possible combinations of values for each of the n numbers in the input array. Assuming a reasonable range R for each number's possible values, the algorithm iterates through R possibilities for each of the n elements. This leads to R^n combinations. For each of these R^n combinations, the algorithm checks if the differences between consecutive elements are equal, which takes O(n) time. Therefore, the overall time complexity is O(R^n * n).

Space Complexity

O(1) – The described brute force approach doesn't explicitly allocate any auxiliary data structures of significant size. It involves iterating and comparing, potentially storing a counter for the minimum number of changes. This counter requires constant space, independent of the input size N, where N is the number of elements in the input array. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The goal is to find the smallest number of changes needed to make all differences between consecutive numbers in a sequence equal. The clever idea is to realize that the final difference must be a value that already appears as a difference in the original sequence. By focusing on differences that already exist, we greatly limit our search.

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

First, calculate all the differences between consecutive numbers in the given sequence.
Next, count how many times each difference appears. We want to find the difference that occurs most often, because we can keep these differences without changing any values.
For each possible difference, determine how many changes it would take to make all differences equal to that value. This is simply the total number of differences minus the number of times the current difference appears.
Finally, find the smallest number of changes needed across all the possible differences. This is the minimum number of changes required to make all consecutive differences equal.

Code Implementation

def min_array_changes(numbers):
    number_count = len(numbers)
    if number_count <= 1:
        return 0

    differences = []
    for i in range(number_count - 1):
        differences.append(numbers[i+1] - numbers[i])

    difference_counts = {}
    for difference in differences:
        difference_counts[difference] = difference_counts.get(difference, 0) + 1

    # If there's only one difference, no changes are needed.
    if len(difference_counts) <= 1:
        return 0

    min_changes = len(differences)
    # Iterate through all possible differences and count required changes.
    for target_difference in difference_counts:
        changes = 0
        for difference in differences:
            if difference != target_difference:
                changes += 1

        # Keep track of the minimum number of changes.
        min_changes = min(min_changes, changes)

    return min_changes

Big(O) Analysis

Time Complexity

O(n) – The solution first calculates differences between consecutive numbers, which takes O(n) time. Then, it counts the frequency of each difference using a hash map. The size of the hash map is bound by the number of differences, which is O(n). Finally, it iterates through the unique differences (at most n) and calculates the changes needed for each, taking O(n) time in the worst case to iterate through all differences. Therefore, the overall time complexity is dominated by O(n).

Space Complexity

O(N) – The algorithm calculates differences between consecutive numbers, storing these differences in a data structure (implicitly, though could be a list or hash map) to count their occurrences. The size of this data structure grows linearly with the input size N, where N is the number of elements in the input sequence. Therefore, the space complexity is determined by the storage of these differences and their counts, leading to an auxiliary space of O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as no changes are needed to make differences equal for an empty array.
Array with a single element	Return 0 as no changes are needed when there's only one element.
All elements in the array are identical	Return 0 because the differences are already equal (all zero).
Array contains large numbers (potential integer overflow)	Use a 64-bit integer type (long or long long) to prevent integer overflow when calculating differences or frequencies.
Array with a wide range of values, including negative values	The solution should correctly handle negative differences and a wide span of numbers when calculating frequencies.
Array has extreme size causing potential memory issues	Consider using a more memory-efficient data structure like a hashmap or frequency counter with bounds or sampling the input array
No valid solution exists, making differences all equal requires more changes than feasible (e.g. array with diverse elements)	The algorithm inherently finds the minimum changes, so if many changes are computed, it indicates a highly diverse array.
Array contains zero values	Zero values should be handled correctly in difference calculations without causing division by zero or logical errors.

Null or empty input array

How to Handle:

Return 0 immediately as no changes are needed to make differences equal for an empty array.

Array with a single element

How to Handle:

Return 0 as no changes are needed when there's only one element.

All elements in the array are identical

How to Handle:

Return 0 because the differences are already equal (all zero).

Array contains large numbers (potential integer overflow)

How to Handle:

Use a 64-bit integer type (long or long long) to prevent integer overflow when calculating differences or frequencies.

Array with a wide range of values, including negative values

How to Handle:

The solution should correctly handle negative differences and a wide span of numbers when calculating frequencies.

Array has extreme size causing potential memory issues

How to Handle:

Consider using a more memory-efficient data structure like a hashmap or frequency counter with bounds or sampling the input array

No valid solution exists, making differences all equal requires more changes than feasible (e.g. array with diverse elements)

How to Handle:

The algorithm inherently finds the minimum changes, so if many changes are computed, it indicates a highly diverse array.

Array contains zero values

How to Handle:

Zero values should be handled correctly in difference calculations without causing division by zero or logical errors.