<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 search(vector<int> dp, int st, int en, int tar) {
        while(st<=en) {
            int mid = (en+st)/2;
            if(dp[mid] >= tar) {
                en = mid-1;
            }
            else st = mid+1;
        }
        return st;
    }
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        vector<int> dp(n);
        dp[0] = nums[0];
        int len = 1;

        for(int i=1; i<n; i++) {
            if(nums[i] < dp[0]) {
                dp[0] = nums[i];
            }
            else if(nums[i] > dp[len-1]) {
                dp[len] = nums[i];
                len++;
            }
            else {
                int idx = search(dp, 0, len-1, nums[i]);
                dp[idx] = nums[i];
            }
        }
        return len;
    }
};
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