Thankyou so much Striver for all you efforts throughout in delivering us so much valuable content. Any student / working professional can now be able to transition their career without paying money for courses. Would also like your insights on the point : While preparing for interviews most of the aspirants are going through the videos solely and solving the question after completely watching the video. And also are feeling lazy trying to solve the question on our own. What is the best way to complete any topic without being lazy and how should an aspirant approach any topic/playlist?
by not watching the video first and try to solve the question beforehand. And you talk about laziness to solve a question that help you land a job bro, you shouldn't be even asking this question if you were motivated enough
Striver, your DSA Sheet is absolutely phenomenal! It's been an invaluable resource for mastering data structures and algorithms. Looking forward to the remaining topics, especially the much-anticipated sections on strings and heaps. Thanks for all your hard work!
This is similar to Buy and sell stock 1 here basically we need to figure out whether we can able to jump to last index or not so let's say as he mention from particular index I he can jump to at max 6 and try to traversing the array by calculating from that index what is the jump possobile at any moment lets at index I he can go to max of x but to reach till I the max possible jump we can take is y if y
Waiting for strings ...
do you come to menace and monk streams
@@KartikeyTT no
@@Josuke217 okay
waiting for String playlist ❤ and till now all the videos of greedy are osm
Thank you
We can go from last index to 1st
If we can't go then false else true
Hey striver,in DSA a to z course there is no video on Java collections
Stack and Queue ki playlist daaldo bro please, eagerly waiting. Mail bhi kia thha poochhne ke lie but you did not reply
awesome
Thankyou so much Striver for all you efforts throughout in delivering us so much valuable content. Any student / working professional can now be able to transition their career without paying money for courses.
Would also like your insights on the point :
While preparing for interviews most of the aspirants are going through the videos solely and solving the question after completely watching the video. And also are feeling lazy trying to solve the question on our own. What is the best way to complete any topic without being lazy and how should an aspirant approach any topic/playlist?
by not watching the video first and try to solve the question beforehand. And you talk about laziness to solve a question that help you land a job bro, you shouldn't be even asking this question if you were motivated enough
Mind Benging brooo
ty sir
prefix coding pattern
Bhai kya padhate ho ap never understood anything in my life what u taught
watch pogo
can anyone explain me how is this greedy
yes, for every step the mindset to jump maximum so its greedy method
❤
waste 3500 on pw java course which is not 1% of your free resource
After watching the DP Approach this greedy code is far very easy .
Striver, your DSA Sheet is absolutely phenomenal! It's been an invaluable resource for mastering data structures and algorithms. Looking forward to the remaining topics, especially the much-anticipated sections on strings and heaps. Thanks for all your hard work!
i thought about recursion approach but this is really easy and optimal
Radhe Radhe bhaiya 💖
Hello guys !!! please pay attention iterate the i or n upto size not size -1 else it will not pass the few test cases :)
please make a solution on the right answer
yes this solution is not passed on the[3,2,1,0,4] this case on leet code only 146 / 172 testcases passed
it does, but try for loop until last element
for(int i=0; i maxIndex)
return false;
maxIndex = max(maxIndex, i+nums[i]);
}
bruh
please,anyone can explain intution behind it, why he is not using dp here
This is similar to Buy and sell stock 1 here basically we need to figure out whether we can able to jump to last index or not so let's say as he mention from particular index I he can jump to at max 6 and try to traversing the array by calculating from that index what is the jump possobile at any moment lets at index I he can go to max of x but to reach till I the max possible jump we can take is y if y
bool canJump(vector& nums) {
int current=0;
int available=0;
for(int i=0;iavailable){
available=current;
}
else if(available==0){
return false;
}
available--;
}
return true;
}
please bring the string video first .A humble request from us
please bring the string video first .A humble request from us
Please add the links of these new videos to the A2Z Dsa sheet
please add link to this video in your a2z sheet
"UNDERSTOOD BHAIYA!!"
waiting for Strings playlist
Thank you very much
understood
thanks for the solution
Good morning striver !
Understood
Understood
Best solution
Understood
"Someone did touch you" sounds so wrong haha
This above explained solution is not working for [3,2,1,0,4] if we are starting from starting index
what's wrong in this it will be false in answer will not be able to reach till the last
what if there are multiple zeroes in the array; than the method doesn't seem to work?
why
paaji tussi great ho taufa
kabul karo
😂😂