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$ curl repovive.com/roadmaps/pattern-22/dynamic-programming-1d/longest-increasing-subsequence-implementation

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$ curl repovive.com/roadmaps/pattern-22/dynamic-programming-1d/longest-increasing-subsequence-implementation

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LeetCode 300 Longest Increasing Subsequence - Implementation - Dynamic Programming - 1D | Pattern 22: LeetCode Interview Patterns | Repovive

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Pattern 22: LeetCode Interview Patterns24 sections · 736 units32h total

Open in Course

LeetCode 300 Longest Increasing Subsequence - Implementation

The approach

For each position, find best previous ending to extend.

def lengthOfLIS(nums): n = len(nums) dp = [1] * n

for i in range(1, n):
    for j in range(i):
        if nums[j] < nums[i]:
            dp[i] = max(dp[i], dp[j] + 1)

return max(dp)

$O(n^2)$ time, $O(n)$ space.