Two Cakes | Editorial by d

Two Cakes | Editorial by d_rushabh

As given in the problem statement there are two shops having same tire for cake so there will be 2*N elements (shops).
First of all scan the the tire offered by shop and the index of shop using  **vector pair**  (vector<pair<int,int> > in c++)
Now second step is sort this pair vector by vector.first means we are sorting array and keeping track of their original positions using pair of vector...
now initialize two pointers for person1 say Px and person2 say Py. (they both are at 0th position initially)
Px=Py=0
vector is sorted with respect to tire size so all two elements in sequence will be having same tire and their original index as second parameter.
Initialize  dis=0
Check for below..
1) if abs(Px-vec[i].second)+abs(Py-vec[i+1].second) <= abs(Px-vec[i+1].second)+abs(Py-vec[i].second (As we are finding minimum distance covered by both)
        then first person will take (i)th tire and 2nd will take (i+1)th tire
                    dis+=abs(Px-vec[i].second)+abs(Py-vec[i+1].second) 
2) else first person will take (i+1)th tire and 2nd will take (i)th tire
                    dis+=abs(Px-vec[i+1].second)+abs(Py-vec[i].second

**Repeat these two statements until all elements are covered that means i<(2*N)**
And print the distance

Two Cakes | Editorial by d_rushabh

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