{ "data": { "question": { "questionId": "3958", "questionFrontendId": "3634", "categoryTitle": "Algorithms", "boundTopicId": 3738803, "title": "Minimum Removals to Balance Array", "titleSlug": "minimum-removals-to-balance-array", "content": "
You are given an integer array nums and an integer k.
An array is considered balanced if the value of its maximum element is at most k times the minimum element.
You may remove any number of elements from nums without making it empty.
Return the minimum number of elements to remove so that the remaining array is balanced.
\n\nNote: An array of size 1 is considered balanced as its maximum and minimum are equal, and the condition always holds true.
\n\n\n
Example 1:
\n\nInput: nums = [2,1,5], k = 2
\n\nOutput: 1
\n\nExplanation:
\n\nnums[2] = 5 to get nums = [2, 1].max = 2, min = 1 and max <= min * k as 2 <= 1 * 2. Thus, the answer is 1.Example 2:
\n\nInput: nums = [1,6,2,9], k = 3
\n\nOutput: 2
\n\nExplanation:
\n\nnums[0] = 1 and nums[3] = 9 to get nums = [6, 2].max = 6, min = 2 and max <= min * k as 6 <= 2 * 3. Thus, the answer is 2.Example 3:
\n\nInput: nums = [4,6], k = 2
\n\nOutput: 0
\n\nExplanation:
\n\nnums is already balanced as 6 <= 4 * 2, no elements need to be removed.\n
Constraints:
\n\n1 <= nums.length <= 1051 <= nums[i] <= 1091 <= k <= 105给你一个整数数组 nums 和一个整数 k。
如果一个数组的 最大 元素的值 至多 是其 最小 元素的 k 倍,则该数组被称为是 平衡 的。
你可以从 nums 中移除 任意 数量的元素,但不能使其变为 空 数组。
返回为了使剩余数组平衡,需要移除的元素的 最小 数量。
\n\n注意:大小为 1 的数组被认为是平衡的,因为其最大值和最小值相等,且条件总是成立。
\n\n\n\n
示例 1:
\n\n输入:nums = [2,1,5], k = 2
\n\n输出:1
\n\n解释:
\n\nnums[2] = 5 得到 nums = [2, 1]。max = 2, min = 1,且 max <= min * k,因为 2 <= 1 * 2。因此,答案是 1。示例 2:
\n\n输入:nums = [1,6,2,9], k = 3
\n\n输出:2
\n\n解释:
\n\nnums[0] = 1 和 nums[3] = 9 得到 nums = [6, 2]。max = 6, min = 2,且 max <= min * k,因为 6 <= 2 * 3。因此,答案是 2。示例 3:
\n\n输入:nums = [4,6], k = 2
\n\n输出:0
\n\n解释:
\n\nnums 已经平衡,因为 6 <= 4 * 2,所以不需要移除任何元素。\n\n
提示:
\n\n1 <= nums.length <= 1051 <= nums[i] <= 1091 <= k <= 105nums and use two pointers i and j so that the window's minimum is nums[i] and maximum is nums[j].",
"Expand j while nums[j] <= k * nums[i] to maximize the balanced window; answer = n - (j - i + 1)."
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