Minimum Operations to Make Array Modulo Alternating II solution and explanation
The problem
You are given an integer array nums and an integer k.
In one operation, you can increase or decrease any element of nums by 1.
An array is called modulo alternating if there exist two distinct integers x and y (0 <= x, y < k) such that:
- For every even index
i,nums[i] % k == x - For every odd index
i,nums[i] % k == y
Return the minimum number of operations required to make nums modulo alternating.
Example 1:
Input: nums = [1,4,2,8], k = 3
Output: 2
Explanation:
- Let's choose
x = 1for even indices andy = 2for odd indices. - Perform the following operations:
- Increment
nums[1] = 4by 1, givingnums = [1, 5, 2, 8]. - Decrement
nums[2] = 2by 1, givingnums = [1, 5, 1, 8].
- Increment
- Now, for even indices,
nums[i] % k = 1, and for odd indices,nums[i] % k = 2. - Thus, the total number of operations required is 2.
Example 2:
Input: nums = [1,1,1], k = 3
Output: 1
Explanation:
- Incrementing
nums[1]by 1 givesnums = [1, 2, 1], which satisfies the condition withx = 1andy = 2. - Thus, the total number of operations required is 1.
Example 3:
Input: nums = [6,7,8], k = 2
Output: 0
Explanation:
The array already satisfies the condition with x = 0 and y = 1. Thus, no operations are required.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 1092 <= k <= 105
Approach overview
Problem Overview: You are given an array and must perform the minimum number of operations so that the sequence of nums[i] % k values alternates between two different remainders. Adjacent indices cannot share the same modulo value. Each operation changes an element to any value, effectively allowing you to control its remainder.
Approach 1: Brute Force Remainder Pair Enumeration (O(n * m^2) time, O(m) space)
First compute r[i] = nums[i] % k. The final array must alternate between two remainders a and b where a != b. Try every possible ordered pair of remainders from the modulo domain. For each pair, iterate through the array and count how many positions do not match the expected alternating pattern. The minimum mismatch count is the answer. This approach works but becomes expensive if the remainder domain is large because every candidate pair must be checked.
Approach 2: Frequency Counting on Even/Odd Indices (O(n) time, O(m) space)
Observe that alternating patterns split the array into two independent groups: even indices and odd indices. Count the frequency of each remainder separately for these two groups using a hash map. The optimal pattern keeps the most frequent remainder on even positions and the most frequent different remainder on odd positions. If the top remainders for both groups differ, use them directly. If they are equal, try the second-best option for one side. The number of operations equals n - (bestEvenFreq + bestOddFreq). This converts the problem into a greedy selection over frequency counts and runs in linear time.
Approach 3: Top-Two Frequency Optimization (O(n) time, O(m) space)
Instead of scanning every remainder candidate, track only the top two frequencies for even and odd indices. This guarantees you can resolve conflicts when the most frequent remainders are identical. The algorithm iterates once to compute counts and once more to determine the best pair combination. The logic mirrors problems like minimum operations to make array alternating and relies on counting patterns in an array combined with a simple greedy decision.
Recommended for interviews: The frequency-counting approach with top-two tracking is what interviewers expect. Brute force demonstrates the correct interpretation of the alternating modulo constraint, but the optimized counting method shows you recognize the independence of even and odd positions and can reduce the search space to constant candidates.
The written solution for this problem is still being prepared. Try it yourself in the meantime.
Try solving it yourselfComplexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Remainder Pair Enumeration | O(n * m^2) | O(m) | When the modulo domain is very small and simplicity matters |
| Even/Odd Frequency Counting | O(n) | O(m) | General case; efficiently determines optimal alternating remainders |
| Top-Two Frequency Optimization | O(n) | O(m) | Preferred interview solution with constant candidate evaluation |
Frequently asked questions
Is Minimum Operations to Make Array Modulo Alternating II easy or hard?
The problem is typically labeled Hard because it requires recognizing the alternating structure and reducing the search space using frequency analysis. Once the even/odd independence insight is clear, the implementation becomes a straightforward O(n) counting problem.
Minimum Operations to Make Array Modulo Alternating II Python/Java solution
Implement the algorithm by computing nums[i] % k, maintaining two frequency maps for even and odd indices, and extracting the top two frequencies from each. Combine the best non-equal remainder pair and compute operations as n - keptPositions. The logic is identical across Python, Java, and C++ with only syntax differences.
How to solve Minimum Operations to Make Array Modulo Alternating II in O(n)?
Convert each element to its remainder r[i] = nums[i] % k. Count frequencies separately for even and odd indices. Select the most frequent remainder for each side while ensuring they differ; if they match, consider the second-most frequent option. The minimum operations equal n minus the number of positions already matching the chosen alternating pattern.
What is the best approach for Minimum Operations to Make Array Modulo Alternating II?
The most efficient approach uses frequency counting on even and odd indices. Compute nums[i] % k for each position, track the most frequent remainders for even and odd groups, and choose two different remainders that maximize preserved positions. This greedy selection produces the minimum operations in O(n) time and O(m) space where m is the number of distinct remainders.
Is Minimum Operations to Make Array Modulo Alternating II asked at Google/Amazon/Meta?
Alternating array transformation and frequency-based greedy problems frequently appear in interviews at companies like Amazon, Google, and Meta. Variants involving modulo constraints or alternating patterns test understanding of counting techniques and greedy optimization.
What data structure is used in Minimum Operations to Make Array Modulo Alternating II?
Hash maps or frequency arrays are used to count occurrences of each remainder for even and odd positions. This structure allows constant-time updates and quick retrieval of the most frequent candidates required for the greedy selection.
What is the time complexity of Minimum Operations to Make Array Modulo Alternating II?
The optimal solution runs in O(n) time because the array is scanned once to compute modulo values and remainder frequencies. A constant number of candidate remainder pairs are evaluated afterward. Space complexity is O(m) for storing remainder counts in hash maps.