Medium — Dynamic Programming | Array | Binary Search

The Problem

Find the length of the longest strictly increasing subsequence in an array of integers. A subsequence maintains the relative order of elements but doesn’t need to be contiguous.

Approach

Use dynamic programming where dp[i] represents the length of the longest increasing subsequence ending at index i. For each element, check all previous elements and extend their subsequences if the current element is larger.

Time: O(n²) · Space: O(n)

Code

class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        if not nums:
            return 0

        # Initialize a dynamic programming array dp with all values set to 1.
        dp = [1] * len(nums)

        # Iterate through the array to find the longest increasing subsequence.
        for i in range(len(nums)):
            for j in range(i):
                if nums[i] > nums[j]:
                    dp[i] = max(dp[i], dp[j] + 1)

        # Return the maximum value in dp, which represents the length of the longest increasing subsequence.
        return max(dp)

Watch It Run

Try it yourself: Open TraceLit and step through every line.

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Built with TraceLit — the visual algorithm tracer for LeetCode practice.

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