Can this problem be solved using Binary Search
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Can this problem be solved using Binary Search
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NO. First test case, 100 can be represented with 1 banknote. And so does 1. Basically, you cannot compare it to the final answer.
What do u mean by final answer
For each test case, print one integer — the minimum positive number of burles s that cannot be represented with k or fewer banknotes.
You can use binary search in the array a so that you can say that the cost(10^b) = 10^(b-last) where last is the last ai such that ai <= b and use this to solve the problem(this is quite useless since n = 10 but actually optimize it 157598538), otherways I don't where to use binary search. If you meant binary search the answer, it is not possible since the cost(x) does not strictly increment, in a case like a: {0, 4} cost(10000) = 1, cost(9999) = 9999.
No. Binary search requires the searchspace to have a special property. If it can be done for some value x then it can be done also for x+1 and it's enough.
(The direction can be reversed, ie x -> x-1).
This problem clearly doesn't satisfy this.