3397. Maximum Number of Distinct Elements After Operations | Greedy

[@ARYANMITTAL]

Q3 & Q4 ❤- th-cam.com/video/58VYapr4wCA/w-d-xo.html

[@imPriyansh77]

Great Explanation. Thanks!!!

[@Abhinav_2060]

The easiest and efficient solution I have seen for this solution . Thanks

[@rdforte]

Thanks for this video tutorial. The way you explained and broke down the question was very helpful. You have yourself a new subscriber!

[@mrmani1700]

your Explanation was Awsome! Thank you so much bhaiya❤

[@Its_Shubham_Negi]

Waiting for next two ❤❤❤

[@vishnu9134]

thankyouu

[@ishitaagrawal3300]

why prevmax

[@jitghosh3923]

Check dry run with test case- arr[]=[4,4,4,4] k=1, check here at the last element when we encounter the previous element then we have already achieved the previousMax as 5 which is the upper limit in our range. Further, if we increment the previousMax it will cross the range, that's why prevmax

[@beinghappy9223]

How do you come up with such solution ?
I coded this and got TLE for obvious k constraint
class Solution {
public:
int maxDistinctElements(vector& nums, int k) {
sort(nums.begin(),nums.end());
int n = nums.size();
setb;
for(int i=0;i

[@ManishKumar-kw7qe]

bro you had almost got there
How do you even come up with this 😂

[@samyaksand6337]

Yes, for the same code, did a few changes for range check. Here's my accepted code -
class Solution {
public:
int maxDistinctElements(vector& nums, int k) {
sort(nums.begin(), nums.end());
int n = nums.size();
set b;
for (int i = 0; i < n; ++i) {
int num = nums[i] - k;
int count = (2 * k);
if (!b.empty()) {
int largeinset = *b.rbegin();
if (num