VERY COMMON FAANG INTERVIEW QUESTION | Leetcode 27 - Remove Element

dynamic programming, leetcode, coding interview question, data structures, data structures and algorithms, faang

[GregHogg]

Master Data Structures & Algorithms For FREE at AlgoMap.io!

[Nazaref]

I thought I was kinda ready for an interview but I had no idea what he was talking about I need to do more learning 💀

[bssandeep]

Fantastic work, Greg! I'm hooked on what you do and consider myself a fan of your work

[Free_Palestine]

prebuilt functions were invented in 1956
people in 1955:

[hit]

To those who didn’t get it, please study basic sorting algo like insertion sort, bubble sort, exertion sort etc

[rdubb]

Or you can just use two indices from the beginning

left = 0

for right in range(len(nums)):
if nums[right] == val: continue
nums[left] = nums[right]
left += 1

return left

[casper_hansen]

No need to swap the elements, just put the element at j as element at i: nums[i++]=nums[j--]

[_VETTRICHEZHIAN]

We can keep i anf j at the zero th index move i and j by one position

Swap the values at i and j when i's value is not equivalent to the key

If the i's value is equivalent to key we stop incrementing the j untill i's value is equal to the key


Whrn i's value becomes not equivalent to key we then swap i's and j's value and increment i and j

[moe]

what about this

index = 0

for i in range(len(nums)):
if nums[i] != val:
nums[index] = nums[i]
index += 1

return index

[sudeepranganath]

Given the question. Isn’t the best approach is to keep chucking 3 to the end no matter what as that is the item to push to the end. Why do you have to have this 3 before that 3 or the other way.

[simonbystricky]

int removeElement(int* nums, int numsSize, int val) {
int head = 0;
for (int i = 0; i < numsSize; i++)
if (nums[i] != val)
nums[head++] = nums[i];
return head;
}

beats 100% on Leetcode

[ramandeepsingh]

It's just like move zeroes to the end

[zFGiveNLATAM]

This is another javascript solution, if you want better benchmark just write up a swap function to swap elements.


function removeElement(nums: number[], val: number): number {
let left = 0, right = nums.length - 1;

while (left <= right) {
if (nums[left] === val) {
[nums[left], nums[right]] = [nums[right], nums[left]];
right -= 1;
} else {
left += 1;
}
}

return left;
}

[ydno]

// My solution in JavaScript

// no order save
function removeElement(nums, val) {
let l = 0,
n = nums.length,
r = n - 1;
while (l <= r) {
[nums[l], nums[r]] = [nums[r], nums[l]];
while (nums[l] != val && ++l <= r);
while (nums[r] == val && l <= --r);
}
return l;
}

// save order
function removeElement(nums, val) {
let l = 0,
n = nums.length,
r = 1;
while (r < n) {
[nums[l], nums[r]] = [nums[r], nums[l]];
while (nums[l] != val && ++l < n);
while (nums[r] == val && ++r < n);
}
return l;
}

[hyperboliq]

Hmm. Couldn’t you just replace every occurrence of Val with -1, then sort the array in reverse, and return the number of val occurrences you replaced? This would return the non-Val elements in the array, and all the -1’s would be at the end. O(nlogn) time, space complexity is O(1), or O(n) if you consider sorting to use an auxiliary array. I submitted this and it worked. Thoughts?

[zangdaarrmortpartout]

this solution is valid but over-complicated.

[millfreedom]

You got the question wrong and therefore your solution is bloated.
In fact you need to have 2 pointers, but going the same way.
Something like this:
int tail=0;
for(i=0;i<nums.size;i++){
if(nums[i]!=val) {
if(tail!=i) nums[tail]=nums[i];
tail++;
}
}
return tail;

[amogh_bakhle]

list1 = [2, 2, 3, 4, 3, 2, 4]
print(list1)
list1=[x for x in list1 if x!=2]+[x for x in list1 if x==2]
print(list1) is it correct? please tell

[mazthespaz]

ok, he blocked off part of the description with the cartoon animal. so i missed that

[belphegor]

why did he say two indexes instead of 2 pointers?