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Commit eff2f61

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Time: 709 ms (74.54%), Space: 133 MB (51.66%) - LeetHub
1 parent 429b0e5 commit eff2f61

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class Solution {
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public:
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/*
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For every event, I've two options. Whether I choose that event or not.
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If choose, then I add its value to my answer. If not choosen I move forward to the next event.
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Now if I select a event, my next event has to have start time > end time of choosen event. Also of all the possible next events I should choose the one with max value
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To select the next event with start time > end time of choosen event, I can do a Binary search.
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What to select the Max value I can find the value of all legal subsets possible, and select the maximum of all the values.
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*/
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vector<int> start;
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int dp[100002][2];
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int solve(vector<vector<int>>& events, int i, int choose){
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if(i>=events.size()){
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return 0;
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}
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if(choose>=2){ // since I cannot chose more than two elemnts.
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return 0;
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}
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if(dp[i][choose]!=-1){
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return dp[i][choose];
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}
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int idx = upper_bound(start.begin(),start.end(),events[i][1])-start.begin(); // binary search to find the next event's index
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return dp[i][choose]=max(events[i][2]+solve(events,idx,choose+1),solve(events,i+1,choose)); //either I select an event, or I dont
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}
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int maxTwoEvents(vector<vector<int>>& events) {
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for(int i=0;i<100002;i++){
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for(int j=0;j<2;j++){
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dp[i][j]=-1;
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}
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}
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int n = events.size();
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//sort elemts acc to start time
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sort(events.begin(),events.end());
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start.clear();
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for(int i=0;i<n;i++){
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start.push_back(events[i][0]);
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}
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return solve(events,0,0);
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}
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};

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