What is the main pattern in Minimum Operations to Exceed Threshold Value I?

The pattern is a direct array count. Since each move removes the smallest value, the answer is simply the number of elements in nums that are less than k.

Why does counting values below k give the minimum number of operations?

Any value below k keeps the array invalid, and the allowed operation eventually removes it because smaller values are always deleted first. Each such value must disappear once, so the count is both necessary and sufficient.

Do I need sorting or a min-heap for LeetCode 3065?

No. Sorting or heap operations simulate the removal order, but this problem only asks for the minimum count. The order does not change how many below-threshold elements must be removed.

What happens when k is already small enough for every element?

Then the answer is 0. During the scan, no element satisfies num < k, so no operation is needed.

What is the exact time and space complexity for this solution?

The solution runs in O(n) time because it scans the array once, and it uses O(1) extra space because it only stores a counter.

#3065

Easy

auto_awesome数组·driven

LeetCode 题解工作台

超过阈值的最少操作数 I

给你一个下标从 0 开始的整数数组 nums 和一个整数 k 。一次操作中，你可以删除 nums 中的最小元素。你需要使数组中的所有元素都大于或等于 k ，请你返回需要的最少操作次数。示例 1：输入： nums = [2,11,10,1,3], k = 10 输出： 3 解释：第一次操…

数组

题目描述

给你一个下标从 0 开始的整数数组 nums 和一个整数 k 。

一次操作中，你可以删除 nums 中的最小元素。

你需要使数组中的所有元素都大于或等于 k ，请你返回需要的最少操作次数。

示例 1：

输入：nums = [2,11,10,1,3], k = 10
输出：3
解释：第一次操作后，nums 变为 [2, 11, 10, 3] 。
第二次操作后，nums 变为 [11, 10, 3] 。
第三次操作后，nums 变为 [11, 10] 。
此时，数组中的所有元素都大于等于 10 ，所以我们停止操作。
使数组中所有元素都大于等于 10 需要的最少操作次数为 3 。

示例 2：

输入：nums = [1,1,2,4,9], k = 1
输出：0
解释：数组中的所有元素都大于等于 1 ，所以不需要对 nums 做任何操作。

示例 3：

输入：nums = [1,1,2,4,9], k = 9
输出：4
解释：nums 中只有一个元素大于等于 9 ，所以需要执行 4 次操作。

提示：

1 <= nums.length <= 50
1 <= nums[i] <= 10⁹
1 <= k <= 10⁹
输入保证至少有一个满足 nums[i] >= k 的下标 i 存在。

lightbulb

解题思路

方法一：遍历计数

我们只需要遍历一遍数组，统计小于 $k$ 的元素个数即可。

时间复杂度 $O(n)$ ，其中 $n$ 为数组长度。空间复杂度 $O(1)$ 。

1

2

3

4

class Solution:
    def minOperations(self, nums: List[int], k: int) -> int:
        return sum(x < k for x in nums)

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Notice that the operation always removes the current minimum, so ask whether simulation is actually necessary.
question_mark
A strong solution explains why counting num < k is equivalent to counting required deletions.
question_mark
The key insight is proving that elements already at least k never block the answer for this problem.

warning

常见陷阱

外企场景

error
Sorting the array first adds extra work even though the answer depends only on how many values are below k.
error
Using <= k instead of < k is wrong because values equal to k already satisfy the threshold.
error
Simulating removals can hide the simple invariant that every below-threshold element must be deleted exactly once.

swap_horiz

进阶变体

外企场景

arrow_right_alt
What changes if the operation can remove any element, not just the smallest, before all values reach k?
arrow_right_alt
How would the approach differ in Minimum Operations to Exceed Threshold Value II, where elements are combined instead of simply removed?
arrow_right_alt
What if you had to return the actual removed values in order rather than only the minimum count?

help

常见问题

外企场景

继续练习

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