Why we are calling optimal solution as greedy algorithm ? My perspective : If I see it, we have optimized our Iterative DP (Tabulation) by going to every index from last to first and asking if i can reach target or not.
Sir I am big fan of your leetcode playlist, Regularly folllowing it ,Please continue doing more videos on leetcode ,Waiting for more Leetcode problems ❤
I watched so many videos n i could not understand the problem , after watching your video i finally understood it🥺🔥 the visualisation helped alot to understand
amazing explanation, love the video. this is my algorithm before watching ur video, it only passed 120/170 test cases when i tried to submit it. So i just wanted to know if my approach to this question is definitely incorrect. class Solution { public boolean canJump(int[] nums) { int size = nums.length-1; int sum = 0; for (int i=0; i < nums.length-1; i++){ sum += nums[i]; } if (sum-(nums.length-2)>= size){ return true; } else if (nums[0]>= nums.length-1){ return true; } else{ return false; } } } again, thx for the video
You're basically looking for the last reachable index at each iteration. That is not a greedy approach. Can you explain what do you mean by greedy approach?
Hi.. Thanks for the vide, your explanation is really good and helpful. But I do have doubt here and a request while explaining , pls consider the code also . I feel like the explanation and the dry run code somewhere I am unable to understabnd(may be I need more practice but still..) Example-> while explaining you said to go back step from 1 , that is 0, you cant reach the destination => agree but in dry run code-> idx+nums[idx] , how are you bringing these terms, like how did you think its should be in this way , its like idx=7,nums[7]=0 and you are adding both 7+0=7, so i am not getting how your idea is to add idx+nums[idex].
If you have understood the explanation, try to write the code on your own. That is the only way you will learn. If everything else fails, only then refer to someone else’s code.
i really hate kind of videos which doesnt tells the intuition why we are doing so.....there are many videos avl for this pblm and many of them are just doing the dry run of the code without telling the intuition behind their though process.... But i am really thanks to you sir that you focused more on the intuition behind the code and have not just done the dry run 😌😌
i actually tried this code and come across a wrong ans for [1] as it is reachable at any cost so i run the loop from nums.length -1 to 0 and that worked.... and thank you for this amazing solution i stuck on this for 3 hrs straight...
What if the second last element is zero? Let's dry run the provided array [3, 2, 1, 0, 4] through the given canJump method: Dry run: Initial State: lastElement = 4 (index of the last element). Iteration 1 (i = 3): i + nums[i] = 3 + 0 >= 4, which is less than lastElement. No update. Iteration 2 (i = 2): i + nums[i] = 2 + 1 >= 4, which is less than lastElement. No update Iteration 3 (i = 1): i + nums[i] = 1 + 2 >= 4, which is less than lastElement. No update Iteration 4 (i = 0): i + nums[i] = 0 + 3 >= 4, which is less than lastElement. No update Return: lastElement == 0, which is true. So, for the array [3, 2, 1, 0, 4], the canJump method returns true, indicating that it is possible to jump from the first element to the last element. Please explain I'm not able to understand the false case?
Why we are calling optimal solution as greedy algorithm ?
My perspective :
If I see it, we have optimized our Iterative DP (Tabulation) by going to every index from last to first and asking if i can reach target or not.
This is actually a very good explanation. I was able to understand because of the dry-run of the code. Thanks a lot.
your explanation is great. I have tried many dsa channels to follow. Then I find this channel. It is so great and underrated.
Your solutions and explanations are great!! thank you
watching your video for the first time, really liked your explanation. Would watch more of your videos :) thanks!
Glad you like them!
I'm fan of ur way of teaching
I learnt trees because of u
Hope you start dp playlist like trees
please ♥️😇
He is great.
So nice of you
Sir I am big fan of your leetcode playlist, Regularly folllowing it ,Please continue doing more videos on leetcode ,Waiting for more Leetcode problems ❤
i am adding more and more problems when I get time. Trying to cover important problems first :)
@@nikoo28 🤍
For this problem i seen many videos, but this one was crystal clear and i never forget.
Very Good Job Sir.
So nice of you
class Solution {
public:
bool canJump(vector& nums) {
int n=nums.size();
int maxJump=0;
for(int i=0;imaxJump)
return false;
maxJump = max(maxJump,i+nums[i]);
if(i>=n-1)
return true;
}
return false;
}
};
I watched so many videos n i could not understand the problem , after watching your video i finally understood it🥺🔥 the visualisation helped alot to understand
your explanation was very simple. made me understand the problem.
underrated channel
amazing explanation, love the video.
this is my algorithm before watching ur video, it only passed 120/170 test cases when i tried to submit it. So i just wanted to know if my approach to this question is definitely incorrect.
class Solution {
public boolean canJump(int[] nums) {
int size = nums.length-1;
int sum = 0;
for (int i=0; i < nums.length-1; i++){
sum += nums[i];
}
if (sum-(nums.length-2)>= size){
return true;
} else if (nums[0]>= nums.length-1){
return true;
} else{
return false;
}
}
}
again, thx for the video
Hi Sir, You are gem. I am learning from you a lot. Thanks, Sir for this Help.
It's my pleasure
Incredible presentation as always. Would love to have you do a problem solving mindset tips and tricks.
that is a really great idea, I will add it to my pipeline of upcoming videos
Excellent solution
It was really a clean and clear explanation ❤
Just Wow...
I understand after watching first time this video.
fantastic brilliant,explanation sir,you deserve a lot
perfect explantion
Best Explaination 👍
Cool explanation bhai...and an advice...keep content concise and outro subtle
i try my best, but everyone has their own learning pace. for quick learners, i have chapter markers for faster navigation 😄
thank u sir ... for such a great explanation❣❣
Please add your chair also in your Recording Gear? Did you buy it from amazon ?
Links in the description :)
@@nikoo28 i didn't find it.
Chair is from Autonomous.
You're basically looking for the last reachable index at each iteration. That is not a greedy approach. Can you explain what do you mean by greedy approach?
My greed is that I want to reach the last pointer from where I am standing
really helpful. Thanks a lot!
this is a gem of a video.
Nice explaination, Thank you
You are welcome
man every question like this just feels so exhausting. It's either some kind of trick or hack to solve the problem.
Thanks
Dry run really helped! thanks a tonne!
Great to hear!
i like how you explain with Animation
great!
Very good explaination!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
your explanation is super
thats the video i was searching exactly
Great Video
Best explanation ever
sir your explaination is awesome... keep uploading more videos.
thanks for your feedback, keep watching :)
You could have used your Jump Game 2 solution here. Both problems are almost same.
Very nice explanation, thanks, keep it up :)
Hi.. Thanks for the vide, your explanation is really good and helpful.
But I do have doubt here and a request while explaining , pls consider the code also .
I feel like the explanation and the dry run code somewhere I am unable to understabnd(may be I need more practice but still..)
Example-> while explaining you said to go back step from 1 , that is 0, you cant reach the destination => agree
but in dry run code-> idx+nums[idx] , how are you bringing these terms, like how did you think its should be in this way , its like idx=7,nums[7]=0 and you are adding both 7+0=7, so i am not getting how your idea is to add idx+nums[idex].
If you have understood the explanation, try to write the code on your own. That is the only way you will learn. If everything else fails, only then refer to someone else’s code.
@@nikoo28 ok sure, Thank you, I will take your advice and implement the same.
Thanks a ton
i really hate kind of videos which doesnt tells the intuition why we are doing so.....there are many videos avl for this pblm and many of them are just doing the dry run of the code without telling the intuition behind their though process....
But i am really thanks to you sir that you focused more on the intuition behind the code and have not just done the dry run 😌😌
glad you liked it
very nice and clear explanation thanks !!!
Glad it was helpful!
Thank you for your work!
My pleasure!
outstanding explination plz try to do playlist for DP ur explination is 🥳
I have a playlist on DP. Constantly adding more and more problems to it: ua-cam.com/play/PLFdAYMIVJQHPXtFM_9mpwwQtIdzP6kxHS.html
Thank you, you helped me so much!
You're very welcome!
Awesome explanation 🔥
With this approach, we never stop on the 0’s right? We are checking if somehow we are able to bypass
Yes
Understood
Nice content
Super useful.💯
Take a value and show it by dry run so we understand a bit more Thanks
Okay so, I don't usually comment but yeah this video was great.
Thank you so much
could you please create a video for leetcode 2483?
Best one
thankyou so much sir its too good
brilliant
Bhai quality explaintaion h apka great baki UA-cam channel toh bs code padh dete h intuition toh batate v nhi h
i like to focus on the problem solving, rather than the language. Languages will come and go. 😅, logic will stay
i understand from u
i actually tried this code and come across a wrong ans for [1] as it is reachable at any cost so i run the loop from nums.length -1 to 0 and that worked.... and thank you for this amazing solution i stuck on this for 3 hrs straight...
Got itt👍
Its O(N**2) ?? Can anyone explain in case of DP
why do you want a solution with a poor time complexity?
What if the second last element is zero?
Let's dry run the provided array [3, 2, 1, 0, 4] through the given canJump method:
Dry run:
Initial State: lastElement = 4 (index of the last element).
Iteration 1 (i = 3):
i + nums[i] = 3 + 0 >= 4, which is less than lastElement. No update.
Iteration 2 (i = 2):
i + nums[i] = 2 + 1 >= 4, which is less than lastElement. No update
Iteration 3 (i = 1):
i + nums[i] = 1 + 2 >= 4, which is less than lastElement. No update
Iteration 4 (i = 0):
i + nums[i] = 0 + 3 >= 4, which is less than lastElement. No update
Return: lastElement == 0, which is true.
So, for the array [3, 2, 1, 0, 4], the canJump method returns true, indicating that it is possible to jump from the first element to the last element.
Please explain I'm not able to understand the false case?
you need to start from the last element, not the first one. watch the explanation that starts at 9:17