354. Russian Doll Envelopes

You have a number of envelopes with widths and heights given as a pair of integers (w, h). One envelope can fit into another if and only if both the width and height of one envelope is greater than the width and height of the other envelope.

What is the maximum number of envelopes can you Russian doll? (put one inside other)

Note:
Rotation is not allowed.

Example:

Input: [[5,4],[6,4],[6,7],[2,3]]
Output: 3 
Explanation: The maximum number of envelopes you can Russian doll is 3 ([2,3] => [5,4] => [6,7]).

 

Approach #1: Using dyanmic programming.

class Solution {
public:
    int maxEnvelopes(vector<pair<int, int>>& envelopes) {
        if (envelopes.size() == 0) return 0;
        sort(envelopes.begin(), envelopes.end(), cmp);
        int ans = 0;
        vector<int> dp(envelopes.size(), 1);
        for (int i = envelopes.size()-1; i >= 0; --i) {
            //std::cout << envelopes[i].first << " " << envelopes[i].second << endl;
            for (int j = i+1; j < envelopes.size(); ++j) {
                if (envelopes[i].first < envelopes[j].first && envelopes[i].second < envelopes[j].second) {
                    dp[i] = max(dp[i], dp[j]+1);

                }
            }
            //std::cout << dp[i] << endl;
            ans = max(ans, dp[i]);
        }
        return ans;
    }
    static bool cmp(const pair<int, int>& a, const pair<int, int>& b) {
        if (a.first == b.first) return a.second < b.second;
        else return a.first < b.first;
    }
};

Runtime: 360 ms, faster than 8.88% of C++ online submissions for Russian Doll Envelopes.

Approach #2: Have the samiler way with 300. Longest Increasing Subsequence

 

class Solution {
public:
    int maxEnvelopes(vector<pair<int, int>>& envelopes) {
        if (envelopes.size() == 0) return 0;
        sort(envelopes.begin(), envelopes.end(), cmp);
        int ans = 0;
        vector<int> dp;
        
        for (auto envelope : envelopes) {
            auto it = lower_bound(dp.begin(), dp.end(), envelope.second);
            if (it == dp.end()) dp.push_back(envelope.second);
            else if (*it > envelope.second) *it = envelope.second;
        }

        return dp.size();
    }
    static bool cmp(const pair<int, int>& a, const pair<int, int>& b) {
        if (a.first == b.first) return a.second > b.second;
        else return a.first < b.first;
    }
};

Runtime: 16 ms, faster than 99.58% of C++ online submissions for Russian Doll Envelopes.