<p>Given an integer array <code>nums</code>, return <em>the length of the longest <strong>strictly increasing </strong></em><span data-keyword="subsequence-array"><em><strong>subsequence</strong></em></span>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [10,9,2,5,3,7,101,18]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,1,0,3,2,3]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [7,7,7,7,7,7,7]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2500</code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><b>Follow up:</b>&nbsp;Can you come up with an algorithm that runs in&nbsp;<code>O(n log(n))</code> time complexity?</p>

class Solution {
public:

    int helper(const vector<int>& nums, int currIndex, int prevIndex, int n) {
    if (currIndex >= n) return 0;
    int notTake = helper(nums, currIndex + 1, prevIndex, n);

    int take = 0;
    if (prevIndex == -1 || nums[currIndex] > nums[prevIndex]) {
        take = 1 + helper(nums, currIndex + 1, currIndex, n);
    }
    return max(take, notTake);
}

int lengthOfLIS(const vector<int>& nums) {

    int n = nums.size();

    // Initialize DP table with 0s
    vector<vector<int>> dp(n, vector<int>(n, 0));
    
    // Fill the DP table
    for (int i = n - 1; i >= 0; --i) {
        for (int j = i - 1; j >= -1; --j) {
            // Case: not taking the current element
            int notTake = (i + 1 < n) ? dp[i + 1][j + 1] : 0;

            // Case: taking the current element
            int take = 0;
            if (j == -1 || nums[i] > nums[j]) {
                take = 1 + ((i + 1 < n) ? dp[i + 1][i + 1] : 0);
            }

            // Take the maximum of taking or not taking the current element
            dp[i][j + 1] = max(take, notTake);
        }
    }

    return dp[0][0];
//   int n = nums.size();
//     if (n == 0) return 0;

//     vector<int> dp(n, 1);
//     int maxLength = 1;


//     for (int i = 1; i < n; ++i) {
//         for (int j = 0; j < i; ++j) {
//             if (nums[i] > nums[j]) {
//                 dp[i] = max(dp[i], dp[j] + 1);
//             }
//         }
//         if (dp[i] > maxLength) {
//             maxLength = dp[i];
//         }
//     }

//     return maxLength;
//     // return helper(nums, 0, -1, n);
}
};
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