How do you solve Longest Increasing Subsequence on LeetCode?

Longest Increasing Subsequence (LeetCode #300) 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 longest increasing subsequence can be efficiently found using binary search.

What is the brute force solution for Longest Increasing Subsequence?

The brute force approach for Longest Increasing Subsequence is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible subsequences of the array and determining which ones are strictly increasing. This method is straightforward but inefficient due to the exponential number of subsequences. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Longest Increasing Subsequence?

Longest Increasing Subsequence uses the following concepts: Array, Binary Search, Dynamic Programming. The recommended patterns to study are: Binary Search, Dynamic Programming, Greedy Algorithms.

What are the interview tips for Longest Increasing Subsequence?

Explain your thought process clearly when discussing approaches. Be prepared to discuss time and space complexities. Practice coding the binary search part separately to gain confidence.

What are common mistakes when solving Longest Increasing Subsequence?

Not using binary search correctly, leading to O(n²) complexity. Confusing subsequences with substrings.

#300

Longest Increasing Subsequence

Medium

ArrayBinary SearchDynamic ProgrammingBinary SearchDynamic ProgrammingGreedy Algorithms

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 approach uses a combination of dynamic programming and binary search to efficiently find the longest increasing subsequence. By maintaining a list that represents the smallest tail of all increasing subsequences found so far, we can quickly determine where to place new elements.

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Algorithm

5 steps

1Step 1: Initialize an empty list 'tails' to store the smallest tail for all increasing subsequences.
2Step 2: Iterate through each number in the input array.
3Step 3: Use binary search to find the index of the current number in 'tails'.
4Step 4: If the number is larger than all elements in 'tails', append it. Otherwise, replace the found index with the current number.
5Step 5: The length of 'tails' will be the length of the longest increasing subsequence.

solution.py12 lines

1# Full working Python code
2import bisect
3
4def lengthOfLIS(nums):
5    tails = []
6    for num in nums:
7        index = bisect.bisect_left(tails, num)
8        if index == len(tails):
9            tails.append(num)
10        else:
11            tails[index] = num
12    return len(tails)

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Complexity note: The time complexity is O(n log n) due to the binary search operation for each element in the array, while the space complexity is O(n) for storing the 'tails' list.

1The longest increasing subsequence can be efficiently found using binary search.
2Maintaining a list of smallest tails helps in optimizing the search for subsequences.

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