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Copy file name to clipboardExpand all lines: leetcode/medium/300_longest_increasing_subsequence.md
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@@ -36,60 +36,57 @@ class Solution:
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return max(dp, default=0)
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```
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## Dynamic Programming with Binary Search Solution
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## Binary Search Solution
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- Run-time: O(Nlog(N))
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- Space: O(N)
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- N = Number of elements in array
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Slightly modifying the previous approach.
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We can use a dynamic programming 1d array of 0s but instead store the subsequence themselves instead of what the previous longest subsequence were.
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The 1d array can be thought of as keeping a sorted array within itself, there will be a subset of indexes that would represent the longest subsequence.
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If we instead used a sorted array to keep a subsequence, we can decide whether or not we can create a larger subsequence with binary search.
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For each n, we can binary search the sorted array for the left-most number that is larger or equal to n.
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If that number is found, we can replace that number with n.
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If no number is found, we can append n to the end of the array, this basically means we can create a larger subsequence.
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With this sorted array, we can then binary search it to find where a given n should be placed.
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We want to find a number that is greater than or equal to n, then replace that number in the array.
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If a number isn't found, we can increase the range of indexes in the sorted array by 1 and add the new n at the end.
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By increasing the sorted array by 1, it shows that there is a longer subsequence.
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This won't exactly tell us the 'actual' subsequence because of the replacement and ordering but works since the question only cares about what is the longest subsequence.
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Since we replace numbers in the sorted array, the sorted array does not represent the 'actual' longest subsequence.
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We need to replace numbers in the array because we can potentially create a larger subsequence.
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For example, given an input of [1,2,3,99,4,5,6], when we reach number 5, if we didn't replace 99 in the sorted array, we can never increase the subsequence by 1 when the sorted array was [1,2,3,99] instead of [1,2,3,4].
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```
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Input: [1,3,6,7,9,4,10,5,6]
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[0, 0, 0, 0, 0, 0, 0, 0, 0]
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[1, 0, 0, 0, 0, 0, 0, 0, 0]
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[1, 3, 0, 0, 0, 0, 0, 0, 0]
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[1, 3, 6, 0, 0, 0, 0, 0, 0]
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[1, 3, 6, 7, 0, 0, 0, 0, 0]
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[1, 3, 6, 7, 9, 0, 0, 0, 0]
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[1, 3, 4, 7, 9, 0, 0, 0, 0]
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[1, 3, 4, 7, 9, 10, 0, 0, 0]
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[1, 3, 4, 5, 9, 10, 0, 0, 0]
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[1, 3, 4, 5, 6, 10, 0, 0, 0]
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[]
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[1]
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[1, 3]
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[1, 3, 6]
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[1, 3, 6, 7]
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[1, 3, 6, 7, 9]
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[1, 3, 4, 7, 9]
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[1, 3, 4, 7, 9, 10]
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[1, 3, 4, 5, 9, 10]
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[1, 3, 4, 5, 6, 10]
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```
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```
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class Solution(object):
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def lengthOfLIS(self, nums):
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def binary_search_insertion_idx(n):
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left, right = 0, end_idx
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left, right = 0, len(sorted_subseq) - 1 if len(sorted_subseq) != 0 else -1
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