My thinking processing
Note\nThis post is inspired by lee215\'s post and the code is almost the same as lee215\'s solution.\nI just want to share my thinking processing.\n\n### Naive 
### Note\nThis post is inspired by lee215\'s post and the code is almost the same as [lee215\'s solution](https://leetcode.com/problems/minimum-moves-to-pick-k-ones/solutions/4886430/java-c-python-minimum-distance-of-points).\nI just want to share my thinking processing.\n\n### Naive Approach and Initial Observations:\n\n1. **Observation:** \n   - We start by considering a naive greedy approach, where we choose each index as Dylan index.\n   - One can aim to minimize moves by prioritizing:\n     - Picking up ones at the current index if available.\n     - Picking up adjacent ones with one operation.\n     - Utilizing `Operation1` to pick up additional ones efficiently.\n\n2. **Challenges and Reflections**\n   - The primary challenge lies in efficiently identifying the nearest ones to the current index (`i`).\n   - The naive approach leads to a quadratic time complexity (`O(n^2)`), which is impractical for larger inputs.\n   - If trying every index does not seem to lead to a linear solution, what about choose the ones to be merged?\n\n###  What if you know there are `x` `Operation1` used in the optimal solution?\n\n1. **Optimization Strategy:**\n   - We explore the scenario where we know the count of `Operation1` (`x`) in the optimal solution.\n   - To pick up the remaining ones (`y = k - x`) without using `Operation1`, we select `y` consecutive ones.\n   - From high school math we know, given a line formed by connecting two point `A` and `C`, whatever point `B` you choose on the line `AC`, `AB`+`BC` == `AC`\n   - By choosing the median index among these consecutive ones, we balance the operations required.\n   - The sliding-window approach, along with prefix sum arrays, facilitates efficient computation of required operations.\n\n### Determining the Range of `x` or `y`:\n\n1. **Equations and Constraints:**\n   - We establish equations and constraints governing the relationship between `x` and `y`:\n     - `eq1`: `x + y = k`, representing the total count of ones picked up using `Operation1` (`x`) and the other ones picked up without using `Operation1` (`y`).\n     - `eq2: x >= 0`\n     - `eq3: x <= k`\n     - `eq4: y >= 0`\n     - `eq5: y <= k`\n     - `eq6: x <= maxChanges`: Constraint from `maxChanges`.\n\n2. **Inference Process:**\n   - Combining `eq6` with `eq1`, we obtain `eq7: k - maxChanges <= y `, representing a lower bound for `y`.\n   - Combining `eq7` with `eq4`, we derive `eq8: max(0, k - maxChanges <= y`, refining the lower bound for `y`.\n   - Considering Dylan can pick up at most three ones with less than two operations, we establish `eq9: y <= max(0, k - maxChanges) + 3`.\n   - Acknowledging the maximum number of ones Dylan can pick up is limited by the sum of elements in `nums`, we introduce `eq10: y <= sum(nums)`.\n   \n3. **Final Range of `y`:**\n   - Combining all equations, we establish the final range of `max(0, k-maxChanges) <= y <= min(max(0, k-maxChanges), sum(nums), k)`.\n   - Time complexity: `(O(4 * n)` = `O(n)`\n\n\n# Code\n```\nclass Solution:\n    def minimumMoves(self, nums: List[int], k: int, maxChanges: int) -> int:\n        nums = [i for i, val in enumerate(nums) if val != 0]\n        n = len(nums)\n        prefix_sum = [0] + list(accumulate(nums))\n        res = inf\n        for m in range(max(0, k-maxChanges), min(n, max(0, k-maxChanges)+3, k) + 1):\n            for i in range(n - m + 1):\n                cur = (prefix_sum[i + m] - prefix_sum[i + m - m // 2]) - (prefix_sum[i + m // 2] - prefix_sum[i])\n                res = min(res, cur + (k - m) * 2)\n        return res\n        \n```
['Python3']