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Asked: 2024-09-24T20:56:59+05:30 In:

You are given an array of integers and a positive integer k. Your task is to determine the number of distinct integers that appear in every contiguous subarray of size k. You need to output the counts of unique integers for each of these subarrays as you slide the window from the start of the array to the end. The subarrays will replicate the same sequence as you traverse through the array, ensuring that each count corresponds to each starting position of the window. Your function should efficiently handle the input to provide the results within a reasonable time frame.

I’ve been working with arrays recently, and I stumbled upon an interesting problem that got my mind racing. Here’s the scoop: imagine you have an array filled with integers, and you also have a positive integer \( k \). Your mission, should you choose to accept it, is to figure out how many distinct integers pop up in every single contiguous subarray of size \( k \).

Let’s say we have the following array: \([1, 2, 2, 1, 3, 4, 1, 2]\), and let’s say \( k = 3 \). You would first look at the initial three elements: \([1, 2, 2]\). So, the distinct integers here are \( 1 \) and \( 2 \)—that gives us a count of 2. Then, as you move your window one step to the right, you’d look at \([2, 2, 1]\). The unique integers are still \( 1 \) and \( 2\) (count remains 2).

Next up, you’d slide the window to \([2, 1, 3]\); now we’ve got \( 1 \), \( 2 \), and \( 3 \) peeking in—so that’s a count of 3. Keep sliding the window right, and repeat the count process for each subarray of size \( k \) until you reach the end of the original array.

So, the big question here is how to efficiently calculate the number of distinct integers for each of these overlapping windows without getting bogged down in unnecessary calculations. I mean, we don’t want our function to take ages, right?

If you’re interested in giving this a shot, think about how you can use data structures like hashmaps or sets to keep track of counts as you slide that window around. It could be a fun challenge to see if you can find a way to optimize the process, rather than recalculating everything from scratch for each new window.

What are your thoughts? Have you faced something similar before? Would love to hear how you might approach this problem!

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    2 Answers


    1. The problem of counting distinct integers in every contiguous subarray of size \( k \) is a classic sliding window problem that can be efficiently solved using hashmaps (or dictionaries in Python). As you slide the window across the array, you can maintain a count of the occurrences of each integer in the current window. When you move the window to the right, you’ll decrement the count of the integer that is left behind and increment the count of the new integer that comes into the window. This allows you to keep an up-to-date count of distinct integers without having to recount everything from scratch for each new window.

      Here’s a step-by-step approach you could take: First, initialize a hashmap to store the counts of each integer in the first window of size \( k \). As you iterate through the array, slide the window by adding the next integer and removing the integer that’s no longer in the window. After each slide, you simply check the size of the hashmap to find the number of distinct integers, which allows for efficient updates and checks. The time complexity for this approach is \( O(n) \) where \( n \) is the length of the array, making it much more efficient compared to recalculating distinct counts for every window from scratch. Using data structures like hashmaps optimizes both the speed and the performance of your algorithm, ensuring that you only deal with the current state of the window.







    2. Distinct Integers in Subarrays

      Finding Distinct Integers in Subarrays

      Okay, so this sounds super interesting! I’ve never really tackled something like this before, but it’s got my brain buzzing. Here’s what I’m thinking:

      We need to figure out how many distinct integers show up in each contiguous subarray of size k. I mean, when you slide the window over the elements in the array, it feels kind of daunting to start counting all over again each time. But maybe we can make this easier!

      First, let’s look at our array: [1, 2, 2, 1, 3, 4, 1, 2] with k = 3. Starting with the first three elements, we have [1, 2, 2]. The distinct integers are 1 and 2, which gives us a count of 2.

      As we move to the next subarray [2, 2, 1], it’s still 2 distinct ones. But then when we get to [2, 1, 3], wow, we’ve got three distinct integers now: 1, 2, and 3! So the count changes to 3.

      So, I guess the tricky part here is how to keep track of all these unique numbers without having to start from scratch every time we move the window. Maybe using a set would help? If we keep adding new numbers as they come into the window and removing the old ones that slide out, that might save us a ton of work!

      It’s a little intimidating, but I think tracking counts with a data structure like a hashmap could help keep things organized. I just need to wrap my head around how to manage that as the windows overlap. Anyway, it’s definitely a cool challenge to think about!


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