Remove Element | Leetcode problem 27

Remove Element
Leetcode problem number 27
Solution in JAVA

JAVA interview programming playlist:

#java #interviewquestions #leetcode

[varshapriyadarshini]

Very nice explanation, after watching 5 videos finally I understood here.

[_NITESHMOTIVATION]

thanks mam !that was a excellent solution....i am happy to learn this approach as it was small and effecient

[rinkuchoudhary]

simplest way to explain, love your voice , thanks for this amazing content

[umeshnaik]

Many people solve this problem by swapping

Something like this

public int removeElement(int[] nums, int val)
{
int l = 0;
int r = nums.length - 1;
while (l <= r) {

if (nums[l] == val) {
int t = nums[l];
nums[l] = nums[r];
nums[r] = t;
r--;

} else {
l++;
}

}

return l;
}


But it is crisp and short

[NikCoder]

But in this question saying that we do not have to use any extra space right?

[AmadYT]

this is only the first part of the quetion the other is missing

[sravanokumar]

How count value returning new array as output i dont understand explain

[Saurabh-febg]

Well this solution is good in its place, what if the question considers to be having an impact with moving the other digits to the end ...that would be more interesting

[mohanant]

From the question, I can see that we should not use another array that very much increases the space complexity and it defeats the whole purpose of two pointer concept

[prakashradhakrishnan]

Hi mam! you using for loop .by using that time complexity o(n)if there (N) no of arr, what is the best method to use under the time complexity

[thelazycat]

but the question says you have to do in-place.

[brahmithegod]

It will not work because extra space is allocted