Watch 3 video solutions for Minimum Operations to Make Elements Within K Subarrays Equal, a hard level problem involving Array, Hash Table, Math. This walkthrough by Programming Live with Larry has 351 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
You are given an integer array nums and two integers, x and k. You can perform the following operation any number of times (including zero):
nums by 1.Return the minimum number of operations needed to have at least k non-overlapping subarrays of size exactly x in nums, where all elements within each subarray are equal.
Example 1:
Input: nums = [5,-2,1,3,7,3,6,4,-1], x = 3, k = 2
Output: 8
Explanation:
nums[1] and use 2 operations to subtract 2 from nums[3]. The resulting array is [5, 1, 1, 1, 7, 3, 6, 4, -1].nums[5] and use 2 operations to subtract 2 from nums[6]. The resulting array is [5, 1, 1, 1, 7, 4, 4, 4, -1].[1, 1, 1] (from indices 1 to 3) and [4, 4, 4] (from indices 5 to 7) are equal. Since 8 total operations were used, 8 is the output.Example 2:
Input: nums = [9,-2,-2,-2,1,5], x = 2, k = 2
Output: 3
Explanation:
nums[4]. The resulting array is [9, -2, -2, -2, -2, 5].[-2, -2] (from indices 1 to 2) and [-2, -2] (from indices 3 to 4) are equal. Since 3 operations were used, 3 is the output.
Constraints:
2 <= nums.length <= 105-106 <= nums[i] <= 1062 <= x <= nums.length1 <= k <= 152 <= k * x <= nums.lengthProblem Overview: You are given an array and a parameter k. Elements that belong to the same repeating k-based subarray pattern must become equal using the minimum number of operations. The goal is to compute the minimum total operations required to make all elements inside each of these k-linked groups equal.
Approach 1: Brute Force Group Equalization (O(n * m log m) time, O(m) space)
Start by grouping indices that belong to the same k-relative position. For example, indices i, i+k, i+2k... form one group. For every group, try converting all elements to each candidate value from the group and compute the total cost. This requires iterating through the group multiple times and sorting or evaluating distances repeatedly. The approach demonstrates the correct grouping idea but becomes expensive when groups grow large.
Approach 2: Sort + Median Minimization (O(n log n) time, O(n) space)
The key observation is that the minimum number of operations to make all values equal in a set occurs when you convert everything to the median. First partition the array into groups based on index % k using a hash table or array of lists. Sort each group and select its median. Then compute the sum of absolute differences between each element and that median. Sorting dominates the complexity, but this drastically reduces the work compared with brute force.
Approach 3: Heap-Based Median Maintenance (O(n log k) time, O(k) space)
Instead of sorting every group, maintain the median dynamically using two heaps (a max heap for the lower half and a min heap for the upper half). As elements of a group are processed, rebalance the heaps and track the median in O(log m) time. The total cost is computed using prefix sums or accumulated differences. This technique is common in problems involving minimum absolute deviation and appears alongside priority queues and sliding window median patterns.
Approach 4: DP + Group Cost Precomputation (O(n log n) time, O(n) space)
If constraints require evaluating multiple segment configurations, you can precompute the equalization cost of each k-group and reuse it through dynamic programming. Each group’s optimal cost is determined using the median trick, then DP aggregates these results to compute the final minimal operations. This structure is useful when the problem includes additional constraints or overlapping decisions.
Recommended for interviews: The median-based grouping approach is the expected solution. Identifying that indices spaced by k must be processed together shows the key insight, and using the median to minimize absolute differences demonstrates strong algorithmic intuition. Brute force confirms the grouping idea, but the optimized median solution shows mastery of array optimization techniques.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Group Equalization | O(n * m log m) | O(m) | Useful for understanding the grouping structure during early problem exploration |
| Sort + Median Minimization | O(n log n) | O(n) | General solution when groups are processed independently |
| Two Heaps Median Tracking | O(n log k) | O(k) | Efficient when maintaining medians dynamically or when group sizes are large |
| DP with Precomputed Group Costs | O(n log n) | O(n) | When the problem introduces additional constraints requiring reuse of group costs |