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auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Longest Increasing Subsequence

Solve the Longest Increasing Subsequence problem using dynamic programming and binary search to efficiently find the subsequence length.

array binary search dynamic programming

Problem Statement

Given an integer array nums, return the length of the longest strictly increasing subsequence (LIS). A subsequence is derived by deleting some or no elements from the array, without changing the relative order of the remaining elements.

For example, in the array [10,9,2,5,3,7,101,18], the LIS is [2,3,7,101], with a length of 4. The goal is to determine the most efficient algorithm for finding this subsequence's length for arrays up to 2500 elements long.

Examples

Example 1

Input: nums = [10,9,2,5,3,7,101,18]

Output: 4

The longest increasing subsequence is [2,3,7,101], therefore the length is 4.

Example 2

Input: nums = [0,1,0,3,2,3]

Output: 4

Example details omitted.

Example 3

Input: nums = [7,7,7,7,7,7,7]

Output: 1

Example details omitted.

Constraints

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

Solution Approach

Dynamic Programming (DP)

A direct DP approach involves constructing a table to store the length of the LIS up to each element. This can be done by iterating over the array and comparing each element to those before it, updating the LIS length accordingly. While simple, the time complexity of O(n^2) can be improved with binary search.

Binary Search Optimization

Binary search can be used in combination with a dynamic array that stores potential LIS elements in sorted order. By maintaining this array, each new element can be placed in the correct position via binary search, reducing the time complexity to O(n log n).

Space Optimization

In the O(n log n) approach, the space complexity can be optimized to O(n) using a list to track the subsequence, instead of a full DP table. This is particularly useful when memory is a concern for larger input arrays.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the DP approach is O(n^2), where n is the length of the input array. By incorporating binary search, the time complexity improves to O(n log n). Space complexity for both methods can be O(n), depending on the specific implementation, though optimized space solutions exist for the binary search approach.

What Interviewers Usually Probe

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

Common Pitfalls or Variants

Common pitfalls

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

Follow-up variants

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

FAQ

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.

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Solution

Solution 1

#### Python3

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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)

Solution 2

#### Python3

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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)

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