2058. Find the Minimum and Maximum Number of Nodes Between Critical Points | DSA | Hindi

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Problem:
2058. Find the Minimum and Maximum Number of Nodes Between Critical Points
Problem Statement:
A critical point in a linked list is defined as either a local maxima or a local minima.
A node is a local maxima if the current node has a value strictly greater than the previous node and the next node.
A node is a local minima if the current node has a value strictly smaller than the previous node and the next node.
Note that a node can only be a local maxima/minima if there exists both a previous node and a next node.
Given a linked list head, return an array of length 2 containing [minDistance, maxDistance] where minDistance is the minimum distance between any two distinct critical points and maxDistance is the maximum distance between any two distinct critical points. If there are fewer than two critical points, return [-1, -1].
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Опубликовано:

3 июл 2024

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Комментарии : 16

@shashwat_tiwari_st 26 дней назад

like target for this video is 80. Please do like if you understood the solution😄

@PiyushSharma-we8yd 26 дней назад

op bhaiya, aaj ka khud krliya tha, almost same approach bs aapka code kafi neat hai mere se😅😅

@RohitKumar-dz8dh 26 дней назад

Thanks 😊 , today I approached the problem same you did.

@kunalsarpal7564 25 дней назад

thankyou so much sir ❤❤❤❤❤❤❤❤

@varadpanchal1231 25 дней назад

One of best explanation

@ankitagrawal4477 25 дней назад

Thank you ! Today i solved these on my own😊

@GirjeshSharma-zv3xo 25 дней назад

Love from usa❤

@motivationalquotes6581 24 дня назад

slow aur acha padhate ho jisse dsa ke question easy lagne laga

@deluluvish 26 дней назад

thank you so much for making this video

@sankalpbarriar1029 26 дней назад

great work can you also make development related videos

@user-oi5ls4rs5g 26 дней назад

great

@aggarwalsachin4854 25 дней назад

hi sir, I am from upes!!

@shashwat_tiwari_st 25 дней назад

Hello there!

@ashwinmali4338 25 дней назад

I had solved but the!!!!....🥲🥲 the time complexity of the solution is O(N + K log K) import java.util.ArrayList; import java.util.Collections; class ListNode { int val; ListNode next; ListNode(int x) { val = x; } } class Solution { public int[] nodesBetweenCriticalPoints(ListNode head) { // Step 1: Convert linked list to array and determine size ListNode current = head; int size = 0; ArrayList values = new ArrayList(); while (current != null) { values.add(current.val); current = current.next; size++; } // Step 2: Handle edge case for size values.get(j) && values.get(j) < values.get(j + 1)) { mini.add(j); } } ArrayList result = new ArrayList(); result.addAll(maxi); result.addAll(mini); Collections.sort(result); int[] arr = new int[result.size()]; for (int i = 0; i < result.size(); i++) { arr[i] = result.get(i); } if (arr.length >= 2) { int minDiff = Integer.MAX_VALUE; // Find the minimum difference for (int i = 1; i < arr.length; i++) { int diff = arr[i] - arr[i - 1]; if (diff < minDiff) { minDiff = diff; } } // Find the maximum difference int maxDiff = arr[arr.length - 1] - arr[0]; return new int[] {minDiff, maxDiff}; } else { return new int[] {-1, -1}; } } }

@GirjeshSharma-zv3xo 25 дней назад

Please explain this question again unable to understand 🙏🏻

@shashwat_tiwari_st 25 дней назад

you have seen linked list playlist ? If no, please watch that first, because you should know, linked list traversal well and then only solve problems.