best xplanation

you explain much better then my professor, thank you

let me see where i stop i am gonna start this journey

Veri nice

what is the intuition behind calculating profit 2 in backward direction ?

only solution which explained the state & transition properly, thanks a ton

Elegant

if x != 6, it means secret has exactly x matches with word for next iteration, if we keep the ones that have >= x matches with word, how is that going wrong ?

Wow I actually understood this question after going through your solution. Thank you so much

You can do it better by presenting it more visually, rather than just showing exmaples.

well explained in a very short video.

just adding this if len(combination) > k: return which is missing in the python solution saves a hell lot of time

Thanks, Java code is also would be: Arrays.sort(arr); // O(nLogn) for (int i = 0; i < arr.length - 2; i++) { // O(n) int min = i + 1; int max = arr.length - 1; int target = -arr[i]; while (min < max) { // O(n) int sum = arr[min] + arr[max]; if (sum == target) { System.out.println(arr[i] + ", " + arr[min] + ", " + arr[max]); min++; max--; } else if (sum < target) { min++; } else { max--; } }. this code has O(n^2) time complexity! which can be improved by using Hash based collections

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Great Explanation. Thank you!

solutions bhi batadijiye sir...

Thank you for this video. I could code it up on my own after listening to your detailed explanation :D

b ut what if ther would be +8 in the example. then you have to take -2 right? or instead of -2 it would be -1, then -6,-1,7.

class Solution { public: int numDecodings(string s) { int n = s.size(); if (n == 0 || s[0] == '0') return 0; // If the string is empty or starts with '0', it can't be decoded vector<int> dp(n + 1, 0); // dp[i] represents the number of ways to decode s[0:i] dp[0] = 1; // Base case: one way to decode an empty string dp[1] = 1; // One way to decode the string if the first character is valid (non-zero) for (int i = 2; i <= n; i++) { // Check single-digit decode possibility if (s[i - 1] != '0') { dp[i] += dp[i - 1]; } // Check two-digit decode possibility int twoDigit = stoi(s.substr(i - 2, 2)); if (twoDigit >= 10 && twoDigit <= 26) { dp[i] += dp[i - 2]; } } return dp[n]; // Number of ways to decode the full string } };

Backward edge connects between a Parent's parent , Forward edge connects between a Descendant's descendant . ,Tree edge is a connect between a parent And child ,Cross edge is a connect between two children's

awesome

If someone wants a short summary: The idea is to calculate the area between the lines at the two pointers, update the maximum area if the current area is larger, and then move the pointer pointing to the shorter line. By moving the shorter line, you potentially increase the height of the container, as increasing the width alone won't give a better result if one side is short.

thanks for this wonderful explaination

sooooooo difficult to understand for a beginner ,its basically for beginners and if u cant make it easy and simple for them to understand then whats the point.

why we are adding line -> curr_sum -= arr[start++]; just if we assign the number to the curr_sum that would be fine right, why we are subtracting the curr_sum with array again? did i miss to understand any logic here?

Made it look easy, thanks

Among all the other logic explanations on youtube, this is the best and simple for me. thanks man

As a cs student learning C++ and been doing some leetcode problems, this channel looks amazing!

Not good explanation Actually fast pointer will move first then slow will move.

I went through many videos. any everyone was missing the initial point which you completed. Thanks.

do an o(nlogn) soln

first learn how to speak clearly...next time dont make videos while eating !!!!..

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please can you do in java for these questions

Starting my Leetcode journey today. 22-09-24

This is a really good tutorial

Best Explanation...

Thanks for the video.

Why random number?

You need to explain how u have arrived to 2*res+head... It's like u have mugged up the solution and has run a simulation of the solution in this video

thanks a lot for a very clear explanation.

Really good explanation, thank you!

Does this also work for a 2 by 2 or 4 by 4 matrix.. That is, can we always multiply it by 2 regardless of the type of matrix

Awesome clarity between web n app server

That's a great explanation. Thank you

Amazing explanation, thanks!

class Solution { public: int maxArea(vector<int>& height) { int left=0; int right=height.size()-1; int max_area=0; while(left<right){ int current_area=(right-left)*min(height[left],height[right]); max_area=max(current_area,max_area); if(height[left]<height[right]){ left++; } else{ right--; } } return max_area; } };

where is the python code 😭 y u lying like that bro 😭

Can we get notes of this ?

Useful 🎊 sir