How do I solve the Partition Array into Disjoint Intervals problem efficiently?

Use an array-driven approach with a single pass to maintain the left subarray's maximum and the right subarray's minimum, ensuring linear time complexity.

What is the time complexity of the Partition Array into Disjoint Intervals problem?

The time complexity is O(N) because the solution involves a single scan through the array, where N is the size of the array.

How do I partition the array such that the left subarray has smaller elements than the right?

Track the maximum value of the left subarray and the minimum value of the right subarray while iterating from left to right, and partition when the condition is satisfied.

Are there any edge cases I should be aware of for this problem?

Edge cases include arrays where the partition is at the very beginning or end of the array, or arrays where all elements are the same value.

Can this problem be solved using a divide-and-conquer approach?

While divide-and-conquer may work, the most efficient approach for this problem is a linear scan, which ensures the best time complexity of O(N).

#915

Medium

auto_awesome数组·driven

LeetCode 题解工作台

分割数组

给定一个数组 nums ，将其划分为两个连续子数组 left 和 right ，使得： left 中的每个元素都小于或等于 right 中的每个元素。 left 和 right 都是非空的。 left 的长度要尽可能小。在完成这样的分组后返回 left 的长度。用例可以保证存在这样的划分方…

数组

题目描述

给定一个数组 nums ，将其划分为两个连续子数组 left 和 right，使得：

left 中的每个元素都小于或等于 right 中的每个元素。
left 和 right 都是非空的。
left 的长度要尽可能小。

在完成这样的分组后返回 left 的长度 。

用例可以保证存在这样的划分方法。

示例 1：

输入：nums = [5,0,3,8,6]
输出：3
解释：left = [5,0,3]，right = [8,6]

示例 2：

输入：nums = [1,1,1,0,6,12]
输出：4
解释：left = [1,1,1,0]，right = [6,12]

提示：

2 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁶
可以保证至少有一种方法能够按题目所描述的那样对 nums 进行划分。

lightbulb

解题思路

方法一：前缀最大值 + 后缀最小值

划分后的两个子数组要满足题目要求，需要保证“数组前缀最大值”小于等于“数组后缀最小值”。

因此，我们可以先预处理出数组的后缀最小值，记录在 mi 数组中。

然后从前往后遍历数组，维护数组前缀的最大值 mx，当遍历到某个位置时，如果数组前缀最大值小于等于数组后缀最小值，那么当前位置就是划分的分界点，直接返回即可。

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

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class Solution:
    def partitionDisjoint(self, nums: List[int]) -> int:
        n = len(nums)
        mi = [inf] * (n + 1)
        for i in range(n - 1, -1, -1):
            mi[i] = min(nums[i], mi[i + 1])
        mx = 0
        for i, v in enumerate(nums, 1):
            mx = max(mx, v)
            if mx <= mi[i]:
                return i

speed

复杂度分析

指标	值
时间	O(N)
空间	O(1)

psychology

面试官常问的追问

外企场景

question_mark
Candidates should demonstrate proficiency with array-driven strategies and partitioning concepts.
question_mark
Watch for clear reasoning in determining the partition point based on left and right subarray comparisons.
question_mark
Efficient solutions with linear time complexity will be expected, especially for large input sizes.

warning

常见陷阱

外企场景

error
Not maintaining the left subarray's maximum or the right subarray's minimum properly during iteration.
error
Assuming the partition must be found at the first occurrence of an increasing element, without considering all array elements.
error
Overcomplicating the problem by using nested loops or extra space when a simple linear scan is sufficient.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Modify the problem to partition based on a different condition, such as comparing the sum of left and right subarrays.
arrow_right_alt
Extend the problem to support multiple partitions within the array, resulting in more than two subarrays.
arrow_right_alt
Change the partition condition so that the left subarray contains elements strictly greater than the right subarray.

help

常见问题

外企场景

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