Auto-solve daily LeetCode problem using GPT-5-mini

Author: jeremymanning

problems/3634/gpt5-mini.md

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+# [Problem 3634: Minimum Removals to Balance Array](https://leetcode.com/problems/minimum-removals-to-balance-array/description/?envType=daily-question)
+
+## Initial thoughts (stream-of-consciousness)
+I can't share internal stream-of-consciousness reasoning, but here's a concise, non-sensitive summary of the idea that comes to mind: sort the array and use a two-pointer (sliding window) approach to find the largest subarray where max <= k * min. The minimum removals equals total length minus the size of that largest valid subarray.
+
+## Refining the problem, round 2 thoughts
+- After sorting, any valid remaining array (non-empty) with min at index i and max at index j must satisfy nums[j] <= nums[i] * k. Because the array is sorted, for a fixed i we can expand j as far right as this inequality holds; when it fails, increment i.
+- We want to maximize the number of elements we keep (i.e., the window length j-i+1), so answer = n - max_window_len.
+- Two-pointer sliding window is O(n) after sorting, so total complexity O(n log n) due to sort.
+- Edge cases: arrays of size 1 return 0 (already balanced). k >= 1 per constraints. Python integers handle large products safely.
+
+## Attempted solution(s)
+```python
+from typing import List
+
+class Solution:
+    def minimumRemovals(self, nums: List[int], k: int) -> int:
+        n = len(nums)
+        if n <= 1:
+            return 0
+        nums.sort()
+        i = 0
+        max_keep = 1  # at least one element can be kept
+        for j in range(n):
+            # increase left pointer until window [i..j] is valid: nums[j] <= nums[i] * k
+            while i <= j and nums[j] > nums[i] * k:
+                i += 1
+            # window length is j - i + 1
+            curr_len = j - i + 1
+            if curr_len > max_keep:
+                max_keep = curr_len
+        return n - max_keep
+```
+- Approach notes: Sort nums and use sliding window / two pointers to maintain the largest contiguous block where max <= k * min. The sorted property ensures that for fixed left i, the rightmost valid j is monotonic as i increases, enabling O(n) scanning after sort.
+- Time complexity: O(n log n) for sorting + O(n) for the two-pointer scan => O(n log n) overall.
+- Space complexity: O(1) extra (ignoring input sort in-place) or O(n) if the language's sort requires extra space.