So if I were to just look at this, I think the first approach I would go with would be a linear search through where we have a peak behind elements. In your case, you really did well, because you give me an efficient solution and overall your communication of great. 5 years ago. [4,2,3] Output: True Explanation: You could modify the first 4 to 1 to get a non-decreasing array. : For sure. I'll do the easy things first, if not num, I'm going to assume that counts because it's not, it's a non decreasing array, because it's not an array. I don't need to find all peaks. Once unsuspended, seanpgallivan will be able to comment and publish posts again. It's not strictly increasing or strictly decreasing like it has the possibility of equal values on occasion, like how will bin search perform form in that case? code of conduct because it is harassing, offensive or spammy. So if this was our case, so let's say we had a case where p one is less than P two or is less than mid. Yeah. So, so our P two is valid for M? Explanation: You could modify the first 4 to 1 to get a non-decreasing array. Are modern compilers passing parameters in registers instead of on the stack? I'll be honest, this particular statement here is a little bit confusing to me, because so I think nums, negative one is equal to last elements of array. : No look at this case? But it kinda, you know, it might reflect as a negative thing, if let's say you say something wrong. And so by making this change, it now assures us that we are going to be O(1) right? Alright, so let's see here. Go through. Technically it's o n. So that wouldn't actually improve efficiency at all, if you did this kind of thing. Right away, we can see that there are three possible scenarios for each of the surrounding elements. But at the same time, there are some things that you said that were a bit wrong. Because that's usually generally pretty fast. Find the non decreasing order array from given array - GeeksforGeeks Find the non decreasing order array from given array pawan_asipu Read Discuss Courses Practice Given an array A [] of size N / 2, the task is to construct the array B [] of size N such that: B [] is sorted in non-decreasing order. : So if we we're to do this, so let's get this case. But which one is better? : Right, or was there a more specific term? With that in mind, we can think of the different possible scenarios faced when we find an incidence of decreasing value. This article is being improved by another user right now. And. So. Templates let you quickly answer FAQs or store snippets for re-use. And that's, that's an interesting thing for me to think about here. And I'm curious, how would you have interpreted that whole kind of segway on your end? : Oh, sorry. And that's one of those things, it's kind of hard to pan out. If you're doing the slicing, that would lead to a very inefficient code. The split point. : And what if they are racy, has you call it has the possibility of equal values like it's not strictly non increasing or non decreasing? So if the array is [4,2,3], then the answer will be true. Non-decreasing Array - Given an array nums with n integers, your task is to check if it could become non-decreasing by modifying at most one element. Alright, so let's see. Here, our endpoints make sense. According to my logic when the pointer is at 2 for [-1,4,2,3],the if block will be executed at i-2(-1) is less than i(2) and the value of 4 will chnage to 2 but apparently that does not happen. : This is not very helpful, but no. : Alright say, we don't have an iterator per se, but instead of a stream, think of it as we actually have an array, it's really huge. So if you have time, I'll advise you check them out, just in case they might give you good practice anyway. : Yeah that's totally fine. : Yeah, that is a good point. Made with love and Ruby on Rails. And so, so long as this is the case, and I know that I keep up I have to keep doing what. : I hear ya. Well, I use count to remember how many descending pairs in the array, if count>=2, obviously we cannot modify a single element to obtain a non-descreaing array, so we return False.When count equals one, we check if we can modify nums[i] (the first one in the descending pair, by decreasing it) or nums[i+1] (the second one in the descending pair, by increasing it), if in both situations, we . So it's just good to be familiar with it. Non-decreasing Array | Leetcode 665 Facebook Coding Interview question, google coding interview question, leetcode, Non-decreasing Array, Non-decreasing Array C++, Non-decreasing Array Java, Non-decreasing Array python, Non-decreasing Array solution, 665. Like regardless of where we're looking. Affordable solution to train a team and make them project ready. Yeah, it keeps you keep speaking then it cuts out and comes back in but should be back now. Non-decreasing Array | LeetCode 665 | C++, Java, Python Six, so previous element is three current element is six, this is false. And you actually get you have your iterators you had infered initially. Leetcode: 665. Non-decreasing Array - DEV Community You keep checking even if count is already 2 or greater though, which is very inefficient for long series. So paste it right here. Actually, I don't know if I've actually fully thought about that before, I suppose. So I didn't really have any other deep extension. It's continuously decreasing, so my slope is going up, but this is the highest point. rev2023.7.27.43548. So let's get to this. : I hear you. So we're setting this remove the true part here. I won't interrupt you on this, I'll just give you a chance to practice. You can assume in most interviewers, interviews, you're not gonna be told what the ideal runtime is. And let's see. 665. Non-decreasing Array (python+cpp) - Katastros : Let's see, specifically, let's look at line 22. One thing you did mention is, when we're talking about binary search things that you're going to look for something that you liked, did you find that? : Let me ask you this. : Yeah when it comes by its giving a really windy noise but when it's not like running right past you, it's fine. : All right. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. No, no, actually not. And I'm just gonna set that for the moment and a peak. Yeah. So just out of curiosity, when would you use binary search in a general sense? I'm not sure why. Maximize count of Decreasing Consecutive Subsequences from an Array So this is three went to six and then to seven, right? I suppose you could parallel parallelize this a little bit, right? Okay. Add Two Numbers 3. It's telling me the direction of the peak. Example 1: And, you know, you can find discrepancies here and there but you're only allowed to change it once and so you're supposed to return to Rule either if the array is non decreasing, or if a single change is sufficient to make it non decreasing and false otherwise. And the end should always be greater than or equal to the start. And then otherwise, lets see previous element is going to be zero, just kind of pass that first element and then current elements and getting past that first first element again or set up a removed variable, initiate this false. So you're targeting L5. So I'm hoping that so far, you know, maybe you can just jump into this. And my technical feedback on this will be one, I really appreciated the fact that you started by writing, or writing a wrapping function. We can define an array is non-decreasing if it satisfies this rule: array [i] <= array [i + 1] for every i (1 <= i < n). Now, the array becomes {1, 1, 1, 1, 3}Input: arr[] = {1, 3, 1}Output: No. Give me honest feedback. So we're still going to be at an O(n). Given an array nums with n integers, your task is to check if it could become non-decreasing by modifying at most one element. And the gist of it is given an array of integers, purely integers, you're essentially allowed to change just one element to make sure that it's non decreasing, like non decreasing I.E. So it's going to be three and then we're going to look at two uh we're going to look at one is the current element, and then previous element is still greater than current okay. Added python solutions and corrected 128 to medium ( #263) last year. So peak element greater than its neighbours. You did a great job in my opinion. It's still going to be three. If you're familiar with Amazon's levels, I'm kind of am assuming you might be since you're in Seattle, but yeah, just kind of general. All right, then let's see. Eventually we'll have zero removals available to us and then we're gonna have to return false. Leetcode Merge Sorted Array problem solution - Programmingoneonone But then when we try to do like a binary search, we select midpoint and go left to right, that gets trickier because you'd end up with numbers that don't make sense. Can I use the door leading from Vatican museum to St. Peter's Basilica? And then we have three different peaks here, right? Binary Search Tree to Greater Sum Tree. So think through whatever you're thinking and then code it out after the opposite approach might also be effective. Like it might have equal values so strictly increasing or strictly decreasing in this case? By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Now we have a very huge array that, you know, if we were to process it serially, it's probably going to kill us. But let's see. : That's a good question. Making statements based on opinion; back them up with references or personal experience. "Sibi quisque nunc nominet eos quibus scit et vinum male credi et sermonem bene", Sci fi story where a woman demonstrating a knife with a safety feature cuts herself when the safety is turned off, "Pure Copyleft" Software Licenses? Let me see if. There'll be a peak somewhere at the end or somewhere in the middle. We can simply convert it to the non-decreasing array if we make the 4 to 1, then the array will be [1,2,3], To solve this, we will follow these steps , If arr has 2 or less elements, then return ture, for i in range 0 to the number of elements in arr 2, if ans is non 0, then return false, otherwise ans := True, if arr[i - 1] > arr[i + 1], then arr[i + 1] := arr[i], Let us see the following implementation to get a better understanding , Enjoy unlimited access on 5500+ Hand Picked Quality Video Courses. Yeah I'm just I'm just gonna do a quick analysis on this. Below table shows a cople of simple examples with 1 and two elements in the array Below table shows a couple of examples with 3 elements. Actually, this is the opposite here it is the opposite. I'm going to come back to this because there's a little bit of trickiness here when it comes to what is previous elements at this point. : Okay, I'll keep that off. So yeah, so like I said, if I were going to go through with this, I think I would just kind of take an iterative approach to start, keep peaking behind, keep peaking ahead, as we move through, check if we are in the middle of those two, we are then cool, that's a peak, and we can just quit out right away and return that particular index. python - Remove Duplicates from Sorted Array - Stack Overflow : Yeah, I guess. Ask Question Asked 4 years, 4 months ago Modified 4 years, 4 months ago Viewed 2k times 1 I have the following list distances = ['5.67', '8.91', '6.48', '9.32', '13.38', '14.99'] I have did some research and have tried sorted (distances , key = float) and I have also tried distances.sort () And so you catching the split point being an issue is really, really good. So we're given an array of numbers and just check if it could become non decreasing. Are self-signed SSL certificates still allowed in 2023 for an intranet server running IIS? If there are multiple peaks, your return any of the peaks, that's nice, don't have to find all of them. Can a lightweight cyclist climb better than the heavier one by producing less power? The Python Sort List Array Method - Ascending and Descending Explained with Examples. Longest substring without repeating characters. It shows some sort of thought process which is always gonna be better than just jumping to the efficient solution. So let's say let's say so this is a false case. Because like, the worst case scenario, is everything's fine. There isn't anything wrong with updating a list in-place though, it can be very efficient or possibly even a requirement for very long lists. : So think through it this way. Once unpublished, this post will become invisible to the public and only accessible to seanpgallivan. FYI, congrats on actually identifying the weakness even without me asking, that's actually a huge bluff. They can either be larger than a (x >= a), in between a and b (b < y < a) or smaller than b (z <= b): Scenarios XAB and ABZ can quickly be determined to trigger a return of false because in both cases the err count will increment to 2: Things get more complicated, however, once the values are staggered. In the case of ZABX, we can either move a down or b up in order to achieve increasing values, while in YABX we can only move b up and in ZABY we can only move a down: In the final scenario, YABY, there is no possible way to fix the array with one modification, even though there's only one initial incidence of a descending value: With this all in mind, we can write our function to return false if we see err > 1 or if we see the YABY scenario. All right, so overall doing very well so far. Right? Previous elements staying at three, then we come up here. 665. Non-decreasing Array - LeetCode-python - GitBook So good job. And to me, that's a huge, huge plus, that's something that tells me you actually know how to structure your communication. And, you know, do have a couple of practice questions particularly. So I'm glad I was able to think back to college days. . That was very nice of you to go through it, what you did. May 2021 Leetcode ChallengeLeetcode - Non-decreasing Array #665Difficulty: Medium Where the array in question generally, I'd say is sorted in some fashion, like it has some sort of order to it that lets you bisect it. Yep. acknowledge that you have read and understood our. As I'm looking at this, and wondering if we could do something. Turned into a function, consider this: The last one prints False, even though [2, 2, 2, 3] would work, but your solution will only try [4, 4, 4, 3] and then fail when it needs to change the final 3 as well. That's actually a huge plus. This is also another question they've asked me in the past. So maybe give me a quick overview of what you're targeting with the interviews or what levels you're targeting to interview for, and what exactly you'd like to get out of this. Do you have any other inquiries actually, say say back to the streaming idea. And again, this is kind of a little bit weird, because I know it's going to be binary search but when I look at this, and I look at these statements nothing here tells me that the array is necessarily ordered in any way like I look at it and I think, all right, peak element where it's greater than its neighbours, we can have 010, then we can have negative five, we have four, negative six, and then we can go back to do with like, 7, 8, 7. 665. Nope. Like in this case, you might say O(1) and while you know, I wouldn't punish you, but L5 I would expect a more efficient approach.