Want a more optimised program

#1

i got the same question in hackerRank and solved it using the concept provided here…but its not the optimised one…the code didnt execute within time limit

#2

O(Nlogn) solution using binary search
The basic idea is present in the majority of solutions shared for this task, I have only tried to implement it in a manner as concise as possible without damaging the code readability.

The idea is that as you iterate the sequence, you keep track of the minimum value a subsequence of given length might end with, for all so far possible subsequence lengths. So dp[i] is the minimum value a subsequence of length i+1 might end with. Having this info, for each new number we iterate to, we can determine the longest subsequence where it can be appended using binary search. The final answer is the length of the longest subsequence we found so far.

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();
}
#3

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