How do you solve Longest Increasing Subsequence II on LeetCode?

Longest Increasing Subsequence II (LeetCode #2407) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: The problem combines the concepts of subsequences and constraints on differences, making it a unique challenge.

What is the brute force solution for Longest Increasing Subsequence II?

The brute force approach for Longest Increasing Subsequence II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsequences of the array and then checking each one to see if it meets the criteria of being strictly increasing and having adjacent differences of at most k. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Longest Increasing Subsequence II?

Always clarify the problem constraints and requirements before diving into coding. Think about how you can reduce the problem's complexity using data structures. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Longest Increasing Subsequence II?

Failing to consider the constraints on differences between adjacent elements. Not efficiently using data structures to track maximum lengths, leading to unnecessary computations.

#2407

Longest Increasing Subsequence II

Hard

ArrayDivide and ConquerDynamic ProgrammingBinary Indexed TreeSegment TreeQueueMonotonic QueueDynamic ProgrammingBinary SearchSegment Trees

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal solution uses dynamic programming combined with a data structure (like a Segment Tree or Binary Indexed Tree) to efficiently track the longest increasing subsequence that satisfies the conditions. This approach reduces the time complexity significantly by avoiding the need to check all subsequences.

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Algorithm

3 steps

1Step 1: Initialize a DP array to store the maximum length of subsequences ending with each value.
2Step 2: Iterate through the nums array, and for each element, check previous elements to see if they can form a valid subsequence with the current element.
3Step 3: Use a data structure to efficiently query and update the maximum lengths for valid subsequences.

solution.py13 lines

1def longestIncreasingSubsequence(nums, k):
2    from sortedcontainers import SortedList
3    dp = SortedList()
4    for num in nums:
5        idx = dp.bisect_right(num)
6        if idx > 0:
7            max_length = dp[idx - 1] + 1
8        else:
9            max_length = 1
10        dp.add(num)
11        if idx < len(dp):
12            dp[idx] = max(max_length, dp[idx])
13    return len(dp)

ℹ

Complexity note: The time complexity is O(n log n) due to the use of a sorted data structure for efficient querying and updating of the maximum lengths. The space complexity is O(n) because we store the lengths of subsequences in a map or similar structure.

1The problem combines the concepts of subsequences and constraints on differences, making it a unique challenge.
2Dynamic programming can significantly reduce the complexity of problems involving subsequences.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.