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There are multiple ways to handle the case when the key(the searched value) is not found by a binary search. Or any search on elements accessible by indices. What you see is the Java approach, the app …

Here is a simple algorithm that runs in $O(N^2)$ time and $O(N)$ space, assuming that $f(\emptyset)$, $f(\{1\})$, $f(\{2\})$, $\cdots$, $f(\{N\})$ are given in an array. The starting idea is about th …

Short answer Let $d=h_1-h_2$. Do a binary search for the largest value of $d$ in interval $[0,A]$ so that none of $n$ bulbs in the garland will be lower than the ground. The wanted minimum value of t …

Here is a cleaner and better way to solve the problem. # Return the smallest index where the element is bigger than `A[start_index]`. # If `len(A)` is returned, no element is bigger than `A[start_inde …

The number of queries in the question is assumed to mean the exact upper bound of the number of queries in the worst case. So, only the worst case is considered except in exercise 3. That is, the wan …

You can apply binary expanding search starting from both ends of the given array. For ease of statement, I will assume $n=2^k$ for some integer $k\ge0$. Otherwise, a boundary check can be included in …

You are almost there. "If yes then an index $i$ that satisfies $a[i]<a[i+1]$, will surely appear on the right side"; so continue searching on the right side. If no, continue searching on the left side …

A similar problem, named "Black Hole", appears as one of the problems of 2019 Russian Olympiad of schoolchildren in computer science. The problem asks for a program that interacts with a jury program …

The "Algorithm 1 Pivot selection" of that paper is rather sloppy. The critical mistake is, as you have noted, "the while loop in the algorithm is exited too early due to the stride being reduced down …

Assume the polygon is a convex polygon; otherwise there might not be an $O(\log n)$ algorithm to find the maximum of X-coordinates or Y-coordinates. First, let us handle the case of X-coordinate. Le …

Yes, it is pointless and absurd to implement an algorithm to "guess the number" using the most common kind of Turing machine, whose head can read any cell on the tape, since, as you pointed out, there …

It can be checked easily whether the product of the first $K$ primes is larger than $N$. If yes, there is no such number that is not larger than $N$ and is the product of $K$ consecutive primes. Other …