How can dynamic programming help solve the Longest Increasing Subsequence problem?

Dynamic programming allows us to break the problem into subproblems by storing the LIS length for each element, which avoids redundant computations and optimizes the solution.

Why is binary search important for improving LIS solution performance?

Binary search helps maintain a sorted subsequence and allows for efficient insertion of new elements, reducing the time complexity of the LIS algorithm from O(n^2) to O(n log n).

How does the Longest Increasing Subsequence relate to state transition dynamic programming?

The LIS problem uses state transition dynamic programming to decide whether to include an element in the subsequence based on the previous subsequences' values, effectively solving the problem iteratively.

What are the main challenges when solving the Longest Increasing Subsequence problem?

Challenges include optimizing the time complexity for large arrays, managing the space complexity of dynamic programming tables, and ensuring the correct order of elements in the subsequence.

How does GhostInterview support solving problems like Longest Increasing Subsequence?

GhostInterview offers a step-by-step guide to mastering algorithms like LIS, helping you implement and optimize solutions using dynamic programming and binary search for improved interview performance.

#300

Medium

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

最长递增子序列

给你一个整数数组 nums ，找到其中最长严格递增子序列的长度。子序列是由数组派生而来的序列，删除（或不删除）数组中的元素而不改变其余元素的顺序。例如， [3,6,2,7] 是数组 [0,3,1,6,2,2,7] 的子序列。示例 1：输入： nums = [10,9,2,5,3,7,10…

数组二分查找动态规划

题目描述

给你一个整数数组 nums ，找到其中最长严格递增子序列的长度。

子序列 是由数组派生而来的序列，删除（或不删除）数组中的元素而不改变其余元素的顺序。例如，[3,6,2,7] 是数组 [0,3,1,6,2,2,7] 的子序列。

示例 1：

输入：nums = [10,9,2,5,3,7,101,18]
输出：4
解释：最长递增子序列是 [2,3,7,101]，因此长度为 4 。

示例 2：

输入：nums = [0,1,0,3,2,3]
输出：4

示例 3：

输入：nums = [7,7,7,7,7,7,7]
输出：1

提示：

1 <= nums.length <= 2500
-10⁴ <= nums[i] <= 10⁴

进阶：

你能将算法的时间复杂度降低到 O(n log(n)) 吗?

lightbulb

解题思路

方法一：动态规划

我们定义 $f[i]$ 表示以 $nums[i]$ 结尾的最长递增子序列的长度，初始时 $f[i] = 1$ ，答案为 $f[i]$ 的最大值。

对于 $f[i]$ ，我们需要枚举 $0 \le j \lt i$ ，如果 $nums[j] \lt nums[i]$ ，则 $f[i] = \max(f[i], f[j] + 1)$ 。

最后的答案即为 $f[i]$ 的最大值。

时间复杂度 $O(n^2)$ ，空间复杂度 $O(n)$ 。其中 $n$ 为数组长度。

1

2

3

4

5

6

7

8

9

10

class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        n = len(nums)
        f = [1] * n
        for i in range(1, n):
            for j in range(i):
                if nums[j] < nums[i]:
                    f[i] = max(f[i], f[j] + 1)
        return max(f)

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
Check if the candidate can explain the differences between the O(n^2) and O(n log n) solutions.
question_mark
Assess the candidate’s understanding of how binary search integrates with dynamic programming to optimize the problem.
question_mark
Evaluate how well the candidate discusses space optimization and its impact on performance in practical scenarios.

warning

常见陷阱

外企场景

error
Failing to recognize the need for binary search to reduce time complexity from O(n^2) to O(n log n).
error
Not maintaining the correct relative order when constructing the LIS array, leading to incorrect results.
error
Overcomplicating the solution with unnecessary optimizations when a simpler approach would suffice for smaller inputs.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Find the longest decreasing subsequence (LDS) by reversing the input array and applying the LIS algorithm.
arrow_right_alt
Determine the longest subsequence that does not allow adjacent elements to differ by more than a constant value.
arrow_right_alt
Generalize to k-longest increasing subsequences instead of just one, which requires advanced dynamic programming techniques.

help

常见问题

外企场景

继续练习

#354 俄罗斯套娃信封问题 #410 分割数组的最大值 #718 最长重复子数组