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# Longest Increasing Subsequence LeetCode Solution

## Problem – Longest Increasing Subsequence

Given an integer array `nums`, return the length of the longest strictly increasing subsequence.

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?

### Longest Increasing Subsequence LeetCode Solution in Java

``````public int lengthOfLIS(int[] nums) {
int[] tails = new int[nums.length];
int size = 0;
for (int x : nums) {
int i = 0, j = size;
while (i != j) {
int m = (i + j) / 2;
if (tails[m] < x)
i = m + 1;
else
j = m;
}
tails[i] = x;
if (i == size) ++size;
}
return size;
}
// Runtime: 2 ms
``````

### Longest Increasing Subsequence LeetCode Solution in Python

``````def lengthOfLIS(self, nums):
tails =  * len(nums)
size = 0
for x in nums:
i, j = 0, size
while i != j:
m = (i + j) / 2
if tails[m] < x:
i = m + 1
else:
j = m
tails[i] = x
size = max(i + 1, size)
return size

# Runtime: 48 ms
``````

### Longest Increasing Subsequence LeetCode Solution in C++

``````int lengthOfLIS(vector<int>& nums) {
vector<int> res;
for(int i=0; i<nums.size(); i++) {
auto it = std::lower_bound(res.begin(), res.end(), nums[i]);
if(it==res.end()) res.push_back(nums[i]);
else *it = nums[i];
}
return res.size();
}
``````
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