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2

The answer is circular buffer. Such structure could be implemented using array or linked list, but you cannot call explicit array-based circular buffer as linked-list. FSA cannot be "executed" infinitely (type 3 machine only uses fixed input and reads it ones), unless your FSA definition covers Turing Machines.


0

Hint, try to determine which element of $R$ should be mapped to 0. Then try to determine which element of $R$ should be mapped to 1. You will get the idea. Here is the pseudocode. Input: array $P[0], P[1], \cdots, P[n-1]$, which is a permutation of $0, 1, \cdots, n-1$. Output: array $R[0], R[1], \cdots, R[n-1]$, which is a permutation of $0, 1, \cdots, ...


1

All you're looking for is some mapping between the elements of $P$ and their indexes, that can be found in $O(1)$ for each element of $P$. Suppose given $p_i \in P$, you can use the fact that $p_i \in \{0, \dots n-1\}$, and that no $p_j \in P$ exists s.t $p_j = p_i$ (other than $j=i$). How will you do it? Set $R[x]$ to the index of element $x \in P$. Since ...


4

This is something you will encounter over and over, not just in science but also in engineering, in law, in programming, and generally in jargon. If there is a definition for a term, then that term means exactly what the definition says it means. No more. No less. In particular, you may have an intuitive notion of what the term means in English, but this is ...


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You can use quickselect with median of medians to find a good pivot so that the total running time is linear in the worst case.


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The algorithm from Geeks starts by finding the maximum element in the middle column (not the 1D-peak). Let us denote this element by $p_m$ and the middle column by $c_m$. A peak found on the $\min$-side is not guaranteed to be a peak in the initial matrix; consider the case where the peak is found in the cell directly adjacent to $p_m$. A peak found on the ...


2

The point is that we are not trying to find the global maximum element of the array, only a peak (that is, a local maximum). Let's say that you look at the middle column and find its global maximum, and this maximum is not a peak, because the value to its left is larger. Then you know that the left half of the matrix contains a value that is larger than ...


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The example in the question The minimum number of moves needed for the example in the question, from [0, 1, 2, 3, 4, 5] to [2, 1, 4, 0, 5, 3] is 3. // [0, 1, 2, 3, 4, 5], where 2, 4, 5 are good. move(2, 1); // [0, 2, 1, 3, 4, 5], where 2, 1, 4, 5 are good. move(0, 4); // [2, 1, 3, 4, 0, 5], where 2, 1, 4, 0, 5 are good. move(2, 5); // [2, 1, 4, ...


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