Longest Increasing Subsequence - Amazon Top Interview Questions


Problem Statement :


Given an unsorted list of integers nums, return the longest strictly increasing subsequence of the array.

Bonus: Can you solve it in \mathcal{O}(n \log n)O(nlogn) time?

Constraints

n ≤ 1,000 where n is the length of nums

Example 1

Input

nums = [6, 1, 7, 2, 8, 3, 4, 5]

Output


5

Explanation

Longest increasing subsequence would be [1, 2, 3, 4, 5]

Example 2

Input

nums = [12, 5, 6, 25, 8, 11, 10]

Output

4

Explanation

One longest increasing subsequence would be [5, 6, 8, 11]



Solution :



title-img




                        Solution in C++ :

int solve(vector<int>& nums) {
    int n = nums.size();
    vector<int> dp;  // dp[i] = last digit in the list of size i+1, dp is increasing
    for (int& i : nums) {
        if (dp.empty()) {
            dp.push_back(i);
            continue;
        }
        auto it = lower_bound(dp.begin(), dp.end(), i);
        if (it == dp.end())
            dp.push_back(i);
        else {
            int idx = it - dp.begin();
            dp[idx] = i;
        }
    }
    return dp.size();
}
                    




                        Solution in Python : 
                            
class Solution:
    def solve(self, nums):
        if not nums:
            return 0

        def get_index(arr, target):
            l, r = 0, len(arr) - 1
            res = r + 1
            while l <= r:
                mid = (l + r) // 2
                if target < arr[mid]:
                    res = mid
                    r = mid - 1
                elif target > arr[mid]:
                    l = mid + 1
                else:
                    return mid
            return res

        dp = []
        res = 0
        for num in nums:
            idx = get_index(dp, num)
            if idx == res:
                dp.append(num)
                res += 1
            else:
                dp[idx] = num

        return res
                    


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