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Asked: March 17, 20252025-03-17T08:41:34+05:30 2025-03-17T08:41:34+05:30

Determine the minimal number of line queries needed in logarithmic time complexity.

anonymous user

So, I’ve been thinking about this interesting problem that I couldn’t quite wrap my head around, and I wanted to get your thoughts on it. Imagine you have a huge list of numbers—let’s say it’s sorted in ascending order—and you need to find a specific number in that list. But here’s the kicker: you can only ask questions about the list, like whether a certain number exists in it or what’s its position.

Now, here’s where it gets fascinating. The goal is to determine the minimal number of line queries you would need to find a particular number or confirm it doesn’t exist, while keeping the time complexity to logarithmic. If you’re familiar with binary search, this might sound like a typical application where you keep halving the list until you either find your number or exhaust your options.

But let’s make it a bit more challenging! Say the list is not just any list of numbers; it has some quirky rules. For instance, let’s pretend each number can only be affected by a specific condition—like if the number is even or odd, or perhaps it changes based on certain inputs (but let’s keep it simple for now—maybe just the odd/even condition). Would that change the way you think about how few queries you really need to make?

I’m just curious about how you’d approach this. What strategies would you employ? Would you still lean on something like binary search? Or think outside the box to discover some clever method that cuts down on the number of queries even further? I mean, it’s easy to feel like you’re stuck in a loop with these sorts of problems, but I believe there’s always a light bulb moment just waiting to happen.

Can you come up with a couple of clever line queries that would minimize the number needed? Maybe there’s a unique insight or pattern you spot that could streamline the process. Would love to hear how you would tackle this!

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    1. anonymous user
      2025-03-17T08:41:36+05:30Added an answer on March 17, 2025 at 8:41 am

      Oh, that’s actually pretty interesting! Honestly, as a beginner programmer myself, I always jump straight to binary search whenever someone mentions sorted lists and quick lookups—it’s like my brain automatically clicks into “divide-and-conquer” mode. But the moment you mentioned this quirky odd/even condition thing, that does throw a cool little twist into the mix!

      If I understood right, instead of just asking “Is this number here?”, maybe we could use queries that check some properties, like “Is the number even?” or “Does an odd number exist in a specific range?” I’m not sure exactly how it would work, but maybe questions about the parity (odd or even) of numbers in smaller halves of the list could help cut the search faster, especially if the list conditions are somehow related to that odd-even thing you mentioned.

      For example, what if our queries were something like:

      • “Hey list, is my number odd or even?” (Quickly dividing which half—odd or even numbers—is relevant.)
      • “List, do you have ANY even numbers at all between positions 50 and 100?” If the answer is no, we instantly know we don’t have to bother looking there for an even number.

      Does that make sense? Honestly, I’m kinda guessing here—but I feel like cleverly choosing queries based on the “odd-even” concept or some other property might actually give us fewer queries than just classic binary search. Or maybe it comes down to filtering out useless sections of the list super quickly rather than halving blindly each time.

      Now you’ve got me curious—I definitely want to read more about this. Let me know if my thinking here is on track or completely off-base!

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    2. anonymous user
      2025-03-17T08:41:36+05:30Added an answer on March 17, 2025 at 8:41 am

      To tackle the challenge of finding a specific number in a sorted list while adhering to certain conditions (like the even/odd requirement), I would still rely on the principles of binary search but with a strategic twist. Binary search efficiently narrows down the search space in logarithmic time by continuously halving the list based on comparisons. However, given the added quirky rules concerning the even or odd nature of the numbers, I would first query the list for the parity of the target number. If the target number is even, I can immediately eliminate half of the numbers (all odds) from consideration, thus reducing unnecessary queries right from the start. This initial condition-check would cut the search space down significantly and ensure that subsequent queries are targeted only towards the relevant half of the numbers.

      Moreover, as I delve deeper into the modified search, I could introduce queries that specifically analyze the distribution of numbers based on their properties. For instance, if I establish the range of even numbers through a query that identifies the lowest even number and the highest even number in the list, it would help further refine where to apply the binary search. Instead of performing a standard binary search, which might involve querying unnecessary odd numbers, I’d focus solely on the even segment of the list. This targeted approach allows for fewer total queries, as each step directly filters out large swathes of non-relevant data based on the initial parity analysis. Adapting traditional search algorithms to account for specific problem constraints often leads to innovative query strategies, enhancing efficiency in practical computations.

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