Given a sequence of intervals [Li, Ri], try to transform them one by one using L+, L-, R+, R- in the shortest operation sequence. Tie breaker is the lexicographical order.
EXPLANATION:
Actually, you only need to consider how to transform [A, B] to [C, D].
The first key point is that, the length of shortest operation sequence equals to |A - C| + |B - D|. This is almost straightforward, because we can always move one end towards its target.
The second one is that we can enumerate the next step (4 operations) and greedily choose the minimum one (both length and lexicographical order). This is because the lexicographical order is determined by the first different character.
The only trick is to remember that you must avoid L = R during this process.
This algorithm’s time complexity is O(Len), where Len is the length of output.
AUTHOR’S AND TESTER’S SOLUTIONS:
Author’s solution can be found here.
Tester’s solution can be found here.
@admin I came here to know the trick to avoid L = R with maintaining lexicographical order and what I get is The only trick is to remember that you must avoid L = R during this process. Really disappointing from codechef side.
@crucifix, considering that lexicographical order of the operations is: L+, L-, R+, R- you can transform the interval following this order. L==R only occurs for the first operation. If you follow these operations in the proper order it’s impossible to have a conflict for the second operation since R is already bigger than L and it’s impossible for the 3rd and 4th operations since L has already been transformed in one of the previous operations and the case L==R has been dealt with in the first operation. I think this is more understandable by the following pseudo code to transform from one interval to the other:
total = number of operations done so far
answer = string of operations done so far
L = current value of L
R = current value of R
nextL = L value of the next interval
nextR = R value of the next interval
while(L < nextL) { // first operation "L+"
// check for the special case first since we are increasing L
if(L == R-1) {
R = R+1; // increase R because if L < nextL then surely R < nextR
total = total + 1;
add string "R+" to answer;
}
L = L+1;
total = total + 1;
add string "L+" to answer;
}
// for the remaining loops we don't have to check the special case
while(L > nextL) {
L = L-1;
total = total + 1;
add string "L-" to answer;
}
while(R < nextR) {
R = R+1;
total = total+1;
add string "R+" to answer;
}
while(R > nextR) {
R = R-1;
total = total+1;
add string "R-" to answer
}
@admin - In the tester’s solution, inside the condition l>pl he has checked for this condition while(pl!=pr-1)
{
ans+=“L+”;
pl++;
if(pl==l) break;
}
But he has not used this condition in the (l<=pl) condition. Why? Because in that case also pl can become equal to pr which we do not want.
This is quite simple honestly. You only need to enumerate all four operations, and first check whether they lead to L = R. And then, for all valid operations, check whether they are shortest by judging the distance (formula given in the editorial). In the end, choose the lexicographical smallest one among the operations which are still valid after 2-step checks.
@paramvi that cannot happen because the input [4 3] is not valid. Check the constraints on the problem statement page: l<r. The code I wrote is strongly based on that assumption and that’s why we only have to check for the special case in the first while block.