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?
function lengthOfLIS(nums: number[]): number {
if (nums === null || nums.length === 0) {
return 0
}
const dp: number[] = new Array(nums.length + 1).fill(0)
// Prefill the dp table
for (let i = 1; i < dp.length; i++) {
dp[i] = Number.MAX_SAFE_INTEGER
}
let left: number = 1
let right: number = 1
for (const curr of nums) {
let start: number = left
let end: number = right
// Binary search, find the one that is lower than curr
while (start + 1 < end) {
let mid: number = start + Math.floor((end - start) / 2)
if (dp[mid] > curr) {
end = mid
} else {
start = mid
}
}
// Update our dp table
if (dp[start] > curr) {
dp[start] = curr
} else if (curr > dp[start] && curr < dp[end]) {
dp[end] = curr
} else if (curr > dp[end]) {
dp[++end] = curr
right++
}
}
return right
}
export { lengthOfLIS }