# Longest Increasing Subsequence LeetCode Solution | Easy Approach Share:

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`

Time: O(n)
Space: O(n)

### C++

``````class Solution {
public:
int lengthOfLIS(vector<int>& nums) {
if (nums.empty())
return 0;

// dp[i] := length of LIS ending at nums[i]
vector<int> dp(nums.size(), 1);

for (int i = 1; i < nums.size(); ++i)
for (int j = 0; j < i; ++j)
if (nums[j] < nums[i])
dp[i] = max(dp[i], dp[j] + 1);

return *max_element(begin(dp), end(dp));
}
};
``````

### Java

``````
class Solution {
public int lengthOfLIS(int[] nums) {
if (nums.length == 0)
return 0;

// dp[i] := length of LIS ending at nums[i]
int[] dp = new int[nums.length];
Arrays.fill(dp, 1);

for (int i = 1; i < nums.length; ++i)
for (int j = 0; j < i; ++j)
if (nums[j] < nums[i])
dp[i] = Math.max(dp[i], dp[j] + 1);

return Arrays.stream(dp).max().getAsInt();
}
}
``````

### Python

`````` class Solution:
def lengthOfLIS(self, nums: List[int]) -> int:
if not nums:
return 0

# dp[i] := LIS ending at nums[i]
dp =  * len(nums)

for i in range(1, len(nums)):
for j in range(i):
if nums[j] < nums[i]:
dp[i] = max(dp[i], dp[j] + 1)

return max(dp)

``````

#### Watch Tutorial

Checkout more Solutions here