Problem
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 <= 2500-10^4 <= nums[i] <= 10^4
Follow up: Can you come up with an algorithm that runs in O(n log(n)) time complexity?
Solution (Java)
class Solution {
public int lengthOfLIS(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int[] dp = new int[nums.length + 1];
// prefill the dp table
for (int i = 1; i < dp.length; i++) {
dp[i] = Integer.MAX_VALUE;
}
int left = 1;
int right = 1;
for (int curr : nums) {
int start = left;
int end = right;
// binary search, find the one that is lower than curr
while (start + 1 < end) {
int 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;
}
}
Solution (Javascript)
/**
* @param {number[]} nums
* @return {number}
*/
var lengthOfLIS = function(nums) {
// Create dp array
const dp = Array.from(nums, () => 1);
// Max subsequence length
let max = 1
// Check all increasing subsequences up to the current ith number in nums
for (let i = 1; i < nums.length; i++) {
// Keep track of subsequence length in the dp array
for (let j = 0; j < i; j++) {
// Only change dp value if the numbers are increasing
if (nums[i] > nums[j]) {
// Set the value to be the larget subsequence length
dp[i] = Math.max(dp[i], dp[j] + 1)
// Check if this subsequence is the largest
max = Math.max(dp[i], max)
}
}
}
return max;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).