Problem:- BXRED Problem - CodeChef Code:- #include<bits/stdc++.h> using namespace std; int main(){ int n; cin>>n; int arr[n]; for(int i=0;i<n;i++)cin>>arr[i]; sort(arr,arr+n,greater<int>()); if(n==1) cout<<"1"<<endl; else{ //map<int,int> m; bool …

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Done,can u tell the problem in the code also?

Yeah, greedy is not the optimal solution for this problem. Consider this test-case: ▶ TEST-CASE correct answer is 2 ( [3,6] and [15,5]) but your code outputs 3

Greedy can be used but before that do sorting and divide the array into two equal halves(or 1 element more in case of odd) and try to make pairs from one half to the other half iteratively. By this, we will end up getting all the possible pairing and note to keep taking into account the unpaired num…

yeah i got that solution accepted,but can u again explain why greedy with 2 halves works ? PS:-Thx for reply

Because that’s the best way to make pair I thought. Notice that if you can put the box at index i at the top of index j after sorting you can always put index i at the top of any other box greater than index j. But to avoid wastage of bigger boxes I put at the top of first possible in second half. S…

Help in BOX REDUCTION(BXRED)

help

cubefreak777 January 13, 2021, 1:50pm 2

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