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

please continue your leetcode problems series

vs code me public boolean method write kiya to code run nhi huwa kiyu

public void rotate(int[] nums, int k) { int temp = 0; while (k > 0) { temp = nums[nums.length - 1]; for (int i = nums.length - 1; i > 0; i--) { nums[i] = nums[i-1]; } nums[0] = temp; k--; } }

class Solution { public int removeDuplicates(int[] nums) { int k = 1; for (int i = 1; i < nums.length; i++) { if (nums[i] != nums[k-1]) { nums[k] = nums[i]; k++; } } return k; } }

it is very useful...thank you

please explain this -> nums[count] = nums[i];

how to use this method when user input the array size and values? is that possible? pls rwply

Didi you don't explain why we have to place i, j or k pointers at that position only 😢 . Why can't we start our pointers from somewhere else . You should explain also as it will help with thought process

THANKS BRO!

Why you are dividing by 10 to check overflow condition

what does the k%nums.length; do?

Great explanation. the key point is nums[i] != num[i-2]. But why in the code it becomes n != nums[i-2] ?

Can we also solve the same problem with help of for loop

Very Nice Video! Thanks Ma'am

7:05

u r explained in java what abiut python

When you have already converted the string to lower case there was no need to check for A-Z in regex.