Minimum Operations to Make Subarray Elements Equal - Solution & Explanation
Problem Statement
You are given an integer array nums and an integer k. You can perform the following operation any number of times:
- Increase or decrease any element of
numsby 1.
Return the minimum number of operations required to ensure that at least one subarray of size k in nums has all elements equal.
Example 1:
Input: nums = [4,-3,2,1,-4,6], k = 3
Output: 5
Explanation:
- Use 4 operations to add 4 to
nums[1]. The resulting array is[4, 1, 2, 1, -4, 6]. - Use 1 operation to subtract 1 from
nums[2]. The resulting array is[4, 1, 1, 1, -4, 6]. - The array now contains a subarray
[1, 1, 1]of sizek = 3with all elements equal. Hence, the answer is 5.
Example 2:
Input: nums = [-2,-2,3,1,4], k = 2
Output: 0
Explanation:
-
The subarray
[-2, -2]of sizek = 2already contains all equal elements, so no operations are needed. Hence, the answer is 0.
Constraints:
2 <= nums.length <= 105-106 <= nums[i] <= 1062 <= k <= nums.length
Approach Overview
Problem Overview: You are given an integer array and a fixed window size k. For every subarray of length k, compute the minimum number of operations required to make all elements equal, where one operation increments or decrements a value by 1. The goal is to return the minimum cost among all possible windows.
Approach 1: Sort Every Window (Brute Force) (Time: O(n * k log k), Space: O(k))
Iterate over every contiguous subarray of length k. Copy the window, sort it, and pick the median element as the target value because the median minimizes the sum of absolute differences. Compute the cost by summing |nums[i] - median| for each element in the window. This method is straightforward but inefficient since sorting happens for every window. It works for small constraints but becomes too slow when n or k grows.
Approach 2: Sliding Window with Ordered Set / Two Heaps (Time: O(n log k), Space: O(k))
Use a sliding window of size k and maintain the median dynamically using two balanced structures. A common implementation uses two heaps (max‑heap for the left half and min‑heap for the right half) or an ordered multiset. The median stays at the top of the left heap. Track the sum of elements in both halves so the total cost can be computed in constant time using the formula derived from median distance.
When the window expands, insert the new value into the appropriate heap and rebalance to keep sizes valid. When the window slides forward, remove the outgoing element and rebalance again. Using maintained sums, compute the cost as the distance from the median to all elements on both sides. Each insert or remove operation costs O(log k), giving an overall complexity of O(n log k). This pattern frequently appears in problems involving median maintenance with heap or ordered structures over an array.
Recommended for interviews: Start by explaining why the median minimizes the sum of absolute differences in a window. Mention the brute force approach to show baseline understanding, then move to the sliding window with heaps or ordered set. Interviewers expect the O(n log k) solution because it demonstrates strong control over window maintenance, median tracking, and efficient data structures.
Solution
According to the problem description, we need to find a subarray of length k and make all elements in the subarray equal with the minimum number of operations. That is, we need to find a subarray of length k such that the minimum number of operations required to make all elements in the subarray equal to the median of these k elements is minimized.
We can use two ordered sets l and r to maintain the left and right parts of the k elements, respectively. l is used to store the smaller part of the k elements, and r is used to store the larger part of the k elements. The number of elements in l is either equal to the number of elements in r or one less than the number of elements in r, so the minimum value in r is the median of the k elements.
The time complexity is O(n times log k), and the space complexity is O(k). Here, n is the length of the array nums.
Detailed Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Sort Every Window (Brute Force) | O(n * k log k) | O(k) | Small inputs or when demonstrating the median insight first |
| Sliding Window with Ordered Set / Two Heaps | O(n log k) | O(k) | General optimal solution when the window moves across the array |
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