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Its working for all cases 👍

Great explanation, but it would be great if you use dark theme on leetcode

3 is not minrightnums2 , 9 is

such a sweet and simple solution, thank q mam.

Can you please explain more at timestamp 4:45 that why did you just took the string till index 7?

Can you please explain more at timestamp 4:45 that why did you not take the whole length and just till index 7?

keep it up didiiiii ; U and SashCode channel doing great job ; i request u too pls come up with the series of daily leetcode potd solutionss

Thanks a lot mam!!!

it is best and simple than above class Solution { public void merge(int[] nums1, int m, int[] nums2, int n) { for (int j = 0, i = m; j < n; j++) { nums1[i] = nums2[j]; i++; } Arrays.sort(nums1); } }

great explanation simple approach for those in new in dsa

But it is not working on [5,1,3] array

class Solution { public int lengthOfLongestSubstring(String label) { int end = 0; int maxLength = 0; List<Character> list = new ArrayList<>(); while (end < label.length()) { boolean isContained = list.contains(label.charAt(end)); if (!isContained) { list.add(label.charAt(end)); end++; maxLength = Math.max(maxLength, list.size()); } else { list.removeFirst(); } } return maxLength; } } i think the start variable unnecessary

the indentation and the brackets opening and closing are not taken care of , and also this code is partially complete , this doesn't pass all test cases

Guys whoever is getting wrong ans, just check my solution it is accepted by leetcode: class Solution { public int[] searchRange(int[] nums, int target) { int[] result = new int[2]; result[0] = findFirstPosition(nums, target); result[1] = findLastPosition(nums, target); return result; } private int findFirstPosition(int[] nums, int target) { int left = 0; int right = nums.length - 1; int firstPosition = -1; while (left <= right) { int mid = left + (right - left) / 2; if (nums[mid] == target) { firstPosition = mid; right = mid - 1; // keep searching to the left } else if (nums[mid] < target) { left = mid + 1; } else { right = mid - 1; } } return firstPosition; } private int findLastPosition(int[] nums, int target) { int left = 0; int right = nums.length - 1; int lastPosition = -1; while (left <= right) { int mid = left + (right - left) / 2; if (nums[mid] == target) { lastPosition = mid; left = mid + 1; // keep searching to the right } else if (nums[mid] < target) { left = mid + 1; } else { right = mid - 1; } } return lastPosition; } }

public int[] searchRange(int[] nums, int target) { int[] arr = new int[2]; arr[0] = arr[1] = -1; for (int i = 0; i < nums.length; i++) { if (nums[i] == target) { arr[0] = i; break; } } if (arr[0] != -1) { for (int i = arr[0]; i < nums.length; i++) { if (nums[i] == target) { arr[1] = i; } else { break; } } } return arr; } For anyone facing issue try this solution

Please do continue with the series, In entire you tube this is the only channel where I am able to understand the solution and co-relate with my approach. I understand because of medical condition you took a break, but please try uploading the videos as soon as possible please Amrita it's my humble request. And if you are not well, Get well soon and then upload the videos.

Essentially this not in constant space. Because in the for loop, you are creating another copy for the array.

I just have a silly questions?? How much time the loop run ???

what if array=["Thik","kuchh nhi","Thik h"] as per your code output is => Thik but output will be "" (blank)

maam what if there is two missing number then this code will be failed

Hello.. Thankyou for this video. Can you explain please why can't we use overflow condition like if (rev*10 > Integer.MAX_VALUE || rev*10 < Integer.MIN_VALUE)

regex kese kaam kr rha h maam @Technosage

legend

Tq😊

here's a single word 0ms code for it class Solution { public int strStr(String haystack, String needle) { return haystack.indexOf(needle); } }

thanks

Outstanding explanation in such a ez way <3 !

"Great explanation on removing duplicates from a sorted array! 𝑰 𝒓𝒆𝒄𝒆𝒏𝒕𝒍𝒚 𝒎𝒂𝒅𝒆 𝒂 𝒗𝒊𝒅𝒆𝒐 𝒐𝒏 𝒕𝒉𝒆 𝒔𝒂𝒎𝒆 𝒕𝒐𝒑𝒊𝒄 , exploring different approaches. It's always interesting to see how others tackle this problem."

PROBLEM SOLVED!

why visited = -1?

Good explanation.!!

great explanation mam..thank you so much

mam your code is left the last element always so return count is always less one here is the write code int pointer = 0; for (int i=0; i<nums.length-1; i++){ if (nums[i] != nums[i+1]){ nums[++pointer] = nums[i+1]; } } return nums.length ==0 ? 0 : pointer+1;

34:57 when you are in this screen is there a way to use vmware esxi gui on the same server? or it has to be managed yes or yes from that ip address in a browser?

NICE SUPER EXCELLENT MOTIVATED

you dint verified the case that majority element should be more the n/2 ????

NICE SUPER EXCELLENT MOTIVATED

amazing solution

Tnx, changing broadband speed options really worked well

Is there a way to automate it through CLI ?

why u have stop the series you may complete it all plzzzzzzzzzz

class Solution { public int search(int[] nums, int target) { if(nums.length == 0 || nums == null){ return -1; } int n= nums.length; int i; for( i=0;i<n;i++){ if(nums[i] == target){ return i; } } return -1; } }

I'm working with a large dataset and the data is stored in json files. Json file Holds data,path is correct too,not parsing the data through any api still it gives the same error. Even i tried multiple ways. Why is this happening?

Thank you

Please share your code series of java

Why are you not solving in Leetcode , In Leetcode this code didn't pass all test cases

What if we give values like 1,2,3,3,3,3,5,6,6

very well explained!!

class Solution { public int search(int[] nums, int target) { for(int i=0; i<nums.length; i++){ if(nums[i]==target){ return i; } } return -1; } } this colde is also acceptable