最长递增和递减子序列（自上而下记忆）

Question

Question - Given an array of integers, A of length N, find the length of longest subsequence which is first increasing then decreasing. Input:[1, 11, 2, 10, 4, 5, 2, 1]

Output: 6

Explanation:[1 2 10 4 2 1] is the longest subsequence.

I wrote a top-down approach. I have five arguments - vector A(containing the sequence), start index(denoting the current index), previous value, large(denoting maximum value in current subsequence) and map(m) STL.

For the backtrack approach I have two cases -

1. element is excluded - In this case we move to next element(start+1). prev and large remains same.

2. element is included - having two cases

   a. if current value(A[start]) is greater than prev and prev == large then this is the case of increasing sequence. Then equation becomes 1 + LS(start+1, A[start], A[start]) ie prev becomes current element(A[start]) and largest element also becomes A[start].

   b.if current value (A[start]) is lesser than prev and current (A[start]) < large then this is the case of decreasing sequence. Then equation becomes 1 + LS(start+1, A[start], large) ie prev becomes current element(A[start]) and largest element remains same ie large.

Base Cases -

1. if current index is out of the array ie start == end then return 0.

2. if sequence is decreasing and then increasing then return 0. ie if(current> previous and previous < maximum value) then return 0.

This is not an optimized approach approach as map.find() is itself a costly operation. Can someone suggest optimized top-down approach with memoization.

int LS(const vector<int> &A, int start, int end, int prev, int large, map<string, int>&m){

    if(start == end){return 0;}
    if(A[start] > prev && prev < large){
        return 0;
    }

    string key = to_string(start) + '|' + to_string(prev) + '|' + to_string(large);

    if(m.find(key) == m.end()){
        int excl = LS(A, start+1, end, prev, large, m);
        int incl = 0;
        if(((A[start] > prev)&&(prev==large))){
            incl = 1 + LS(A, start+1, end, A[start],A[start], m); 
        }else if(((A[start]<prev)&&(A[start]<large))){
            incl = 1+ LS(A, start+1, end, A[start], large, m);
        }  

        m[key] = max(incl, excl);
    }

    return m[key];
}

int Solution::longestSubsequenceLength(const vector<int> &A) {
        map<string, int>m;
        return LS(A, 0, A.size(), INT_MIN, INT_MIN, m);
}

Answer 1

Not sure about top-down but it seems we could use the classic LIS algorithm to just approach each element from "both sides" as it were. Here's the example with each element as the rightmost and leftmost, respectively, as we iterate from both directions. We can see three instances of a valid sequence of length 6:

[1, 11, 2, 10, 4, 5, 2, 1]

 1 11       11 10 4 2 1
 1 2                2 1
 1 2 10        10 4 2 1
 1 2 4            4 2 1
 1 2 4 5          5 2 1
 1 2                2 1