Leetcode Weekly contest 354 - Medium - Maximum Beauty of an Array After Applying Operation

[@atulbartwal2421]

The best thing I like about you is the way you explain a problem. Half of the time, because of that I kinda get the intuition of how to solve the problem. Hope your channel grows exponentially. Cheers.

[@yashvardhanmalve7252]

Your clear explanations and step-by-step approach made it so much easier. Thanks.

[@i_love_api]

Case study approach from [non-optimal solution to optimal] solution ---> one of the best videos I watched.

[@fgffgf9844]

how can one explain such a complex question with such clarity? hatsoff to u man

[@nkvs1248]

That's why I subscribed it, just after seeing a video from this channel :)

[@prakhar3agrwal]

Thanks 🙏

[@prakhar3agrwal]

Thanks 🙏

[@dhruvsakariya3129]

Thank you finally i found some logical explanation instead of bunch of random solution 🙏🙏

[@prakhar3agrwal]

Glad it helped!

[@i_love_api]

I hope in the future we will get more such content.

[@rsKayiira]

This is a very very tricky question but you explained it quite well. This is my second time watching your video on this solution. I needed to first take a break and relook at this problem because the edge cases are quite tricky but this is the best solution out there.

[@loudcapricorn5513]

Sir I was also able to solve this question with the exact same thinking/logic and my implementation was also same. I have been Solving past contests as virtual contest and after that I see your videos for the question which I was not able to solve. This particular type of question where there are ranges and we have to do operation on multiples ranges had appeared in multiple contests recently thus I was able to solve this too [Scanline algorithm is the name i guess]. Your videos are very helpful and provide great learnings , please continue doing so. Lastly Thank You for your efforts

[@prakhar3agrwal]

Thanks. Totally my pleasure.

[@dipanshugoyal491]

Thank you for your nice explanation. I was close and going in same direction just couldn't think of that pre-sum trick.

[@borse2982]

Great Explanation Dude...

[@viveksuman9600]

Very good explanations. Keep it up 👍

[@prakhar3agrwal]

Glad you liked it

[@rsKayiira]

Excellent explanation thank you

[@Abhishek-fo3fc]

Nicely Explained❤

[@prakhar3agrwal]

Glad you liked it!

[@karapureddyjaswanthreddy6053]

great explanation

[@LogicArena01]

Thank you vaiya , pura clear ho geya ❤

[@saivardhanpallerla9]

Hello sir , Thanks for amazing explanation , i have written in c++ , although it had passed all test cases but it didn't got accepted .[ 619 / 619 test cases passed, but took too long.] => We can optimze further for finding the maximum overlapped number instead of iterating from [ 0 , 300002] , we can reduce the iteration by finding the correct indices [ first , second ] . SOLUTION : C++ class Solution { public: int maximumBeauty(vector& nums, int k) { vector freq( 300005 , 0 ); int shift = 100000; int first = INT_MAX , second = INT_MIN; for( int i = 0 ; i < nums.size() ; i++ ) { int start_idx = nums[i] - k ; int end_idx = nums[i] + k; first = min( first , start_idx + shift ); second = max( second , end_idx + shift + 1 ); freq[start_idx + shift]++; freq[end_idx + shift + 1]--; } int ans = freq[first]; for( int i = first + 1 ; i < second + 2 ; i++ ) { freq[i] += freq[i-1]; ans = max( ans , freq[i] ); } return ans; } };

[@suyashmishra6946]

nice

[@shaamidrees6036]

Upload 4th also

[@HarishSumant]

Crystal clear solution as always!! I had a query: This is the C++ version of your solution. It gives TLE after passing 619/619 test cases (????) . Could you please help me with this? class Solution { public: int maximumBeauty(vector& nums, int k) { vectorarr(300005,0); int offset=100000; for(int i=0;i

[@SohanKale-m4z]

mine C++ code got passed

[@saivardhanpallerla9]

class Solution { public: int maximumBeauty(vector& nums, int k) { vector freq( 300005 , 0 ); int shift = 100000; int first = INT_MAX , second = INT_MIN; for( int i = 0 ; i < nums.size() ; i++ ) { int start_idx = nums[i] - k ; int end_idx = nums[i] + k; first = min( first , start_idx + shift ); second = max( second , end_idx + shift + 1 ); freq[start_idx + shift]++; freq[end_idx + shift + 1]--; } int ans = freq[first]; for( int i = first + 1 ; i < second + 2 ; i++ ) { freq[i] += freq[i-1]; ans = max( ans , freq[i] ); } return ans; } };

[@Rajat_maurya]

this is really crazy mine working ditto same class Solution { public: int maximumBeauty(vector& v, int k) { int n = v.size(); int offset = 100000; vector freq(300005, 0); for(int i=0;i

[@siddhantdwivedi7024]

We can also solve this using sliding window but before that we have to sort the array

[@prakhar3agrwal]

Intresting.

[@niteshkhanna690]

i solved by this during contest

[@HI-wl9el]

Yeah done the same . I think it would be better if constraints would be more tougher to implement

[@shaamidrees6036]

Thanks brother

[@shubhamrohan3136]

thankyou sir.

[@tusharaherwar]

Is this line sweep algo

[@divyanshuagarwal1912]

@@tusharaherwar yes

[@nikethdonthula2123]

Very Well Explained Sir !! I was able to write the code just by the intuition provided : Code : class Solution { public int maximumBeauty(int[] nums, int k) { int min = nums[0]; int max = nums[0]; for(int item : nums) { min = Math.min(min,item); max = Math.max(max,item); } HashMap hmap = new HashMap(); for(int i = min-k;i

[@prakhar3agrwal]

Awesome. Best of luck