Efficient Algorithm for Longest Increasing Subsequence Problem
Explore dynamic programming algorithms for the Longest Increasing Subsequence (LIS) problem, understand the concept of maintaining the smallest integers ending a subsequence of each length, and learn how to efficiently compute and maintain this list for every prefix. Discover a more efficient solution that determines the smallest integer ending a longest increasing subsequence, enhancing your understanding of dynamic programming concepts.
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Presentation Transcript
Dynamic Programming (4) Longest Increasing Subsequence (LCS)
DP alg. for LIS The idea is to compute for each 1 ? ?: L[i] = The length of the longest increasing subsequence ending with ?? P[i] = The index of the element that precedes ??in the longest sequence that ends at ?? The length of the LIS is max 1 ? ??[?] ? ? = max ? ? ??< ?? + 1 ?<? ? 1 2 3 4 5 6 7 8 9 ? 2 4 10 9 7 5 6 8 8 L P
DP alg. for LIS The idea is to compute for each 1 ? ?: L[i] = The length of the longest increasing subsequence ending with ?? P[i] = The index of the element that precedes ?? in the longest sequence that ends at ?? The length of the LIS is max 1 ? ??[?] ? ? = max ? ? ??< ?? + 1 ?<? ? 1 2 3 4 5 6 7 8 9 ? 2 4 10 9 7 5 6 8 6 L 1 2 3 3 3 3 4 5 4 P 1 2 2 2 2 6 7 6
Running time ? ? ?2
A more efficient algorithm ? Back to our previous solution, where we wanted to know the smallest integer that ends a longest increasing subsequence I[j] = the smallest integer that ends an increasing subsequence of length j
Smallest integers ending subsequence of each length Back to our previous solution, where we wanted to know the smallest integer that ends a longest increasing subsequence I[j] = the smallest integer that ends an increasing subsequence of length j ? 1 2 3 4 5 6 7 8 9 ? 2 4 3 5 1 7 6 9 8 I
Lets try to compute this list for every prefix Back to our previous solution, where we wanted to know the smallest integer that ends a longest increasing subsequence I[j] = the smallest integer that ends an increasing subsequence of length j ? 1 2 3 4 5 6 7 8 9 ? 2 4 3 5 1 7 6 9 8 I [2] [2,4] [2,3] [2,3,5] [1,3,5] [1,3,5,7] [1,3,5,6] [1,3,5,6,9] [1,3,5,6,8]
How do we maintain the list I I ? Back to our previous solution, where we wanted to know the smallest integer that ends a longest increasing subsequence I[j] = the smallest integer that ends an increasing subsequence of length j ? 1 2 3 4 5 6 7 8 9 ? 2 4 3 5 1 7 6 9 8 I [2] [2,4] [2,3] [2,3,5] [1,3,5] [1,3,5,7] [1,3,5,6] [1,3,5,6,9] [1,3,5,6,8] The list I is sorted When traversing ?? find ? such that ? ? 1 < ?? ?[?] Update ? ? = ??
Smallest integers ending subsequence of each length Back to our previous solution, where we wanted to know the smallest integer that ends a longest increasing subsequence I[j] = the smallest integer that ends an increasing subsequence of length j ? 1 2 3 4 5 6 7 8 9 ? 2 4 10 9 7 5 6 8 3 I [2] [2,4] [2,4,10] [2,4,9] [2,4,7] [2,4,5] [2,4,5,6] [2,4,5,6,8] [2,3,5,6,8] The list I is sorted When traversing ?? find ? such that ? ? 1 < ?? ?[?] Update ? ? = ??
Efficient Implementation Maintain I such that you can find ? by a binary search ? ? log ? time
How do we find the list itself ? Try to find the algorithm by yourself
