[Hdp] lc1713. The minimum number of operations to get the subsequence (LIS+LCS optimization+weekly contest 222_4)

1. Source of the subject

Link: 1713. Get the minimum number of operations for the subsequence

2. Topic analysis

Foreword:

• [Linear dp] The longest ascending subsequence (template problem + the longest ascending subsequence model)

A very important knowledge point. When the array elements will not be repeated when the LCSproblem can be transformed into LISthe problem. LISGreedy can reduce the time complexity from $O\left(n^2\right)$Get$O(n-LOGn-)$ , comprising the processing10^5capability of the data range.

But LISthe greedy wording design to two points, find many of the border. It is recommended to directly back off the board!

• Time complexity : $O\left(nlogn\right)$。
• Space complexity : $O\left(n\right)$

Code:

class Solution {
    
    
public:
    int minOperations(vector<int>& target, vector<int>& arr) {
    
    
        unordered_map<int, int> hash;
        for (int i = 0; i < target.size(); i ++ ) 
            hash[target[i]] = i;
        
        vector<int> a;
        for (int i = 0; i < arr.size(); i ++ ) 
            if (hash.count(arr[i]))
                a.push_back(hash[arr[i]]);
        
        int len = 0;
        vector<int> q(a.size() + 1);
        for (int i = 0; i < a.size(); i ++ ) {
    
    
            int l = 0, r = len;
            while (l < r) {
    
    
                int mid = l + r + 1 >> 1;
                if (q[mid] < a[i]) l = mid;
                else r = mid - 1;
            }
            len = max(len, r + 1);
            q[r + 1] = a[i];
        }
        return target.size() - len;
    }
};