can anyone share the approach to this problem?

@l_returns

What I do I kept record of min and max value for every unknown number from it’s previous number in array and according to sign in the given string that is of >,= or < and maintain a flag wherever condition according to questions failed.

This is link to my solution.

- Make groups (overlapping by one number) having following format :

number -1 -1 … -1 number

Or

number number - now for each group count number of ‘>’ , ‘<’ and ‘=’

i) now if > and < both are present a group then it’s always possible. (= Doesn’t make any difference)

ii) now if > is there and < is not there then check 5 > > > 2 is not possible but 5 > > 2 and 5 > > 0 are possible. Jo just compare count,5,2 appropriately. ( = Doesn’t matter)

iii) if < is present and > is not present.

Same logic as point ii

iv) now if > is not present as well as < is not present and only = is present. Then check if both numbers are equal or not.

Now this is it. You have almost (90%) solved it.

Corner cases : check prefix and suffix.

example < < 2 is possible ( 0 1 2)

But <<< 2 is not possible ( -1 0 1 2) and -1 is invalid.

Also check suffix with same logic ( check if it’s doesn’t get -1).

One more example can be <=> <<<2 ( still not possible because it will be -1). ( This would only happen in prefix and suffix)

One extra simple corner case is all are -1.

https://www.codechef.com/viewsolution/26731902

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For your first case, consider the example 5 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 and the string is > > > > > < < < < < < . I think this string is not possible.

In my opinion for the case A -1 -1 … B, if you do a one step walk from A (up or down depending on the sign) and end up at a point less than B, then it is always possible.

For the other case, if you end up x units above B, then you need to see whether you can go x more steps down in between. That is wherever the sign changes from > to <, summation of values at those index(which is calculated using one step walk) should be more than equal to x.

Please correct me if I am wrong.