Video

Truly the best and most intuitive explanation of joins that I've seen. Everyone just stops at the venn diagrams...which means they probably don't fully understand it themselves haha

It would be nice if you would have explained your reserve logic a bit better with an example...

bla bla

Sensational.

Elegant resolve!

Excellent explanation, neat and clean code, Subscribed. Thank you so much.

Hello Eric. The logic is good, but what happens when both the pointers are at the same value? Ideally we should be looking down from each side and move to a highest next tower. But that isn’t covered in this logic

// base case if (s == null || s.equals("")) return true; String right = (i == n)? "": s.substring(i,n);

Amazing!

Hash set?

very nice explanation !

not need to add => while (!stack.isEmpty()) { Integer topIndex = stack.pop(); map[topIndex] = 0; } since initial map array are initialized to all zero.

excellent!

Good solution, could be a little cleaner in terms of the if statements and that base case isn't required: class Solution { public void moveZeroes(int[] nums) { int left = 0; int right = 1; while(right < nums.length){ if(nums[left] == 0){ if(nums[right] !=0){ int temp = nums[left]; nums[left] = nums[right]; nums[right] = temp; } else{ right++; } } else{ left++; right++; } } } }

I this code will not work, incase of duplicate as during shrink operation you will never know the element which we are removing is it same max or we have duplicate also in windw. storing index is make sense

Hey Eric, can you please make a video for leet code problem 643.

What is the changed time complexity after using min heap?

at 7th line return new int[ ] { l+1; r+1},,, why l+1. why not just l as we want index, l+1 means 1 index more.

in practice four time 29:42 why it's a/b i think it's O(a-(b+1))((

Thank you, good solution

Excellent. Thank you!!

Best explanation on the planet!

this video is a gem. thank you so much

Time complexity !?

Great explanation

You don't need Math. min here, just use left or right index, that's it. The rest is cool, thanks!

public double findMedian() { if(size%2!=0) return maxheap.size() > minheap.size() ? (double) maxheap.peek() : (double) minheap.peek(); return (maxheap.peek()+ minheap.peek()) / 2.0; }

Thank you so much

func twoSum(numbers []int, target int) []int { var result []int l := 0 r := len(numbers) - 1 for r > l { if numbers[l]+numbers[r] == target { result = append(result, l+1) result = append(result, r+1) return result } else if numbers[l]+numbers[r] > target { r-- } else { l++ } } return result }

is there a playlist for dfs?

thank you!

I think the reason we don't have to decrease the mostFreq is because mostFreq only affects our answer when it is larger than the previously largest mostFreq?

Thank you sir this video made my day

Great solution.

I have to revise this videos bc I've been lacking

damn i had no idea there exist stack.toArray() method...

Set would also be log(n) where n is avg number of timestamps for a key, because inserting a new key into treeMap requires sorting that key ascendingly among existing keys

I counted the word basically, you basically said 'basically', basically 500 times

Thank you so much, bro!

Great work! Thanks!

Why are you starting loop from 1?

what kind of drawer tool do you use? is it for macos?

Great video!

best answer Eric, thank you so much

what if the target is equal to the sum of 3 elements of the array instead of 2 elements

Good solution

the first time I tried this exercise with javascript, I wrote this. Math.pow(nums).sort(); hahaha

Thank you

How do you come to the conclusion that you & U Vis need to use the two pointers technique for this problem?

It has a constraint of not using any global variables.