Problem
Given an integer array nums, return **the length of the longest *strictly increasing subsequence***.
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 {number[]} nums
* @return {number}
*/
var lengthOfLIS = function(nums) {
var arr = [nums[0]];
for (var i = 1; i < nums.length; i++) {
if (nums[i] > arr[arr.length - 1]) {
arr.push(nums[i]);
} else {
var index = binarySearch(arr, nums[i]);
arr[index] = nums[i];
}
}
return arr.length;
};
var binarySearch = function(arr, num) {
var left = 0;
var right = arr.length - 1;
while (left < right) {
var mid = left + Math.floor((right - left) / 2);
if (arr[mid] > num) {
right = mid;
} else if (arr[mid] === num) {
return mid;
} else {
left = mid + 1;
}
}
return left;
};
Explain:
nope.
Complexity:
- Time complexity : O(n * log(n)).
- Space complexity : O(n).