LeetCode in Kotlin
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
class Solution {
fun lengthOfLIS(nums: IntArray): Int {
if (nums.isEmpty()) {
return 0
}
val dp = IntArray(nums.size + 1)
// prefill the dp table
for (i in 1 until dp.size) {
dp[i] = Int.MAX_VALUE
}
val left = 1
var right = 1
for (curr in nums) {
var start = left
var end = right
// binary search, find the one that is lower than curr
while (start + 1 < end) {
val mid = start + (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
}
}