Відео

Really Helpful

Very nice explenation as always

Greatest Explanation ❤

can this be solved using segment trees?

Thanks Sir🙏

public class Solution { public long IncremovableSubarrayCount(int[] nums) { int n=nums.Length; int left=0; for(left=0;left<n-1;left++) { if(nums[left]>=nums[left+1]) { break; } } if(left==n-1) return (long) n*(n+1)/2; int right=n-1; while(nums[right]>nums[right-1]) { right--; } long count=n-right+1; for(int i=0,j=right;i<=left;i++) { while(j<n && nums[j]<=nums[i]) { j++; } count+=n-j+1; } return count; } }

subscribed and liked! Your channel needs more traffic! Cheers bro

Can you provide the google doc link which you show initially in vdo

How on earth these gems are so underrated. This channwl is Goldmine for someone who is trying to learn computational problem solving - kudos

Bro! Crazy solution and explanation Please continue to this great thing of doing Multiple approaches

Loved your approach towards questions !!!!

Thanks so much. Really simplified it.

I HAVENT SEEN ANY GOOD CHANNEL FOR LEETCODE GRIND THAN THIS,THANK YOU SO MUCH FOR YOUR EFFORTS

Bro, what will be the dp approach for this?

Awesome Solution sir...

Thank you, sir

Thank you, so much, sir

I solved just using a single loop: "class Solution { public long maximumValueSum(int[] arr, int k, int[][] edges) { int n = arr.length; int[] brr = new int[n]; for(int i = 0; i < n; i++) brr[i] = (arr[i] ^ k) - arr[i]; Arrays.sort(brr); for(int i = 0; i < n / 2; i++){ int temp = brr[i]; brr[i] = brr[n - i - 1]; brr[n - i - 1] = temp; } // System.out.println("Printing brr: "); // for(int i = 0; i < n; i++) System.out.print(brr[i] + " "); // System.out.println(" ------------------------"); long ans = 0; for(int i = 0; i < n; i++){ if(brr[i] > 0) ans += brr[i]; else { if(i % 2 == 1){ long mySum = brr[i] + brr[i - 1]; if(mySum > 0) ans += brr[i]; else ans -= brr[i - 1]; } else{ break; } } } if(n % 2 == 1 && brr[n - 1] > 0) ans -= brr[n - 1]; long initSum = 0; for(long x: arr) initSum += x; // 263 // System.out.println("initSum: " + initSum); return initSum + ans; } }"

🙌🙌🙌

Thank you, sir

Thank you sir

Did using max segment tree and min segment tree and binary serach

Thank you, sir

Best explainer over yt. Love to watch your videos. Thank you, sir !!!

Best solution video I have seen so far. Sir please do find some time to create playlists for other concepts. Thank you for your hard work.

The DSU solution is so genius

Thank you sir

I did not understand, can you please explain in Hindi.

Nicely Explained

QUALITY>>>

Great explanation,

very impressive explanation. keep doing these videos

Wowwwww

Lol This is straightforward How is this difficult

@codingmohan I am not able to understand about one thing what about the cases where the optimal doesn't have first factory

39:28 Why will there be k/2 ones to left and (k/2)-1 to right?

Top quality content

The code is not handling the test cases updated by leetcode

Great explanation

wow, this makes so much sense now , thank you

Could someone provide me with some problems that can be solved using this technique?

why this depth[st[root -> val]] = height instead of depth[root -> val] = height explain please any one explain further clearly why we do this

HUGE Respect to you

Wow i learned somthing new euler tour thx man

You are not explaing the question very well and not take at least 2 example for better understanding. Take it as a feedback .

Why you are doing aaa aaa aaa in between it's irritating