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300. Longest Increasing Subsequence
Medium
Given an integer array nums, return the length of the longest strictly increasing subsequence.
A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7] is a subsequence of the array [0,3,1,6,2,2,7].
Example 1:
Input: nums = [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
Example 2:
Input: nums = [0,1,0,3,2,3]
Output: 4
Example 3:
Input: nums = [7,7,7,7,7,7,7]
Output: 1
Constraints:
1 <= nums.length <= 2500-104 <= nums[i] <= 104
Follow up: Can you come up with an algorithm that runs in O(n log(n)) time complexity?
Solution
# @param {Integer[]} nums
# @return {Integer}
def length_of_lis(nums)
return 0 if nums.nil? || nums.empty?
dp = Array.new(nums.length + 1, Float::INFINITY)
left = 1
right = 1
nums.each do |curr|
start_idx = left
end_idx = right
# Binary search: find the index where dp[index] is lower than curr
while start_idx + 1 < end_idx
mid_idx = start_idx + (end_idx - start_idx) / 2
if dp[mid_idx] > curr
end_idx = mid_idx
else
start_idx = mid_idx
end
end
# Update the dp table
if dp[start_idx] > curr
dp[start_idx] = curr
elsif curr > dp[start_idx] && curr < dp[end_idx]
dp[end_idx] = curr
elsif curr > dp[end_idx]
dp[end_idx + 1] = curr
right += 1
end
end
right
end