2020-10-04 19:32:08 +05:30
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/*
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2021-12-08 03:50:23 -08:00
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Problem:
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Given two sequences, find the length of longest subsequence present in both of them.
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A subsequence is a sequence that appears in the same relative order, but not necessarily contiguous.
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For example, “abc”, “abg”, “bdf”, “aeg”, ‘”acefg”, .. etc are subsequences of “abcdefg”
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Our Solution:
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We use recursion with tabular memoization.
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Time complexity: O(M x N)
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Solving each subproblem has a cost of O(1). Again, there are MxN subproblems,
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and so we get a total time complexity of O(MxN).
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Space complexity: O(M x N)
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We need to store the answer for each of the MxN subproblems.
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Improvement:
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It's possible to optimize space complexity to O(min(M, N)) or time to O((N + r)log(N))
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where r is the number of matches between the two sequences. Try to figure out how.
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References:
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[wikipedia](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem)
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[leetcode](https://leetcode.com/problems/longest-common-subsequence/)
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2020-10-04 19:32:08 +05:30
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*/
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2021-12-08 03:50:23 -08:00
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/**
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* Finds length of the longest common subsequence among the two input string
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* @param {string} str1 Input string #1
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* @param {string} str2 Input string #2
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* @returns {number} Length of the longest common subsequence
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*/
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function longestCommonSubsequence (str1, str2) {
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const memo = new Array(str1.length + 1).fill(null)
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.map(() => new Array(str2.length + 1).fill(null))
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function recursive (end1, end2) {
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if (end1 === -1 || end2 === -1) {
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return 0
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}
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if (memo[end1][end2] !== null) {
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return memo[end1][end2]
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}
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if (str1[end1] === str2[end2]) {
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memo[end1][end2] = 1 + recursive(end1 - 1, end2 - 1)
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return memo[end1][end2]
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2020-10-04 19:32:08 +05:30
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} else {
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2021-12-08 03:50:23 -08:00
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memo[end1][end2] = Math.max(
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recursive(end1 - 1, end2),
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recursive(end1, end2 - 1)
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)
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return memo[end1][end2]
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2020-10-04 19:32:08 +05:30
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}
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}
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2021-12-08 03:50:23 -08:00
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return recursive(str1.length - 1, str2.length - 1)
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}
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2020-10-04 19:32:08 +05:30
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2021-10-10 17:00:21 +02:00
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export { longestCommonSubsequence }
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