Trapping Rain Water | without stack | Made Super Easy | Leetcode 42 | codestorywithMIK

[@abhayshukla322]

Thank you so much bhai mai yeh video dekhne se pehle khud bhi try kiya tha to solve but I am not able to figure out that. Although aapka video dekhne se ekdum clear ho gya hai jo aapne small example lekr samjhaya hai na wo toh jbardast tha

[@codestorywithMIK]

Thank you so much Abhay 😇❤️🙏

[@anubhavbhutani8512]

Currently at 4:26 , Stopped the video, tried it by myself and boom done! Thank you so much Mik!

[@EB-ot8uu]

I completed your arrays playlist 2 days ago binge watched all of them. today solved this on my own. thanks a lot

[@Aniket_0314]

which is his arrays playlist? I only found this one? Can you show me the playlist you're talking about?
Thanks

[@FanIQQuiz]

This channel is a blessing. What a fine explanation.

[@aggarwalsachin4854]

bro, you just killed it

[@raushankumargupta7018]

Aapka explanation boht hi easy & top quality hota h hmesa. Thank you very much bhai. Happy Eid🎉

[@wearevacationuncoverers]

Congratulations in advance for 41K. your channel will blast and will reach soon to 1M

[@aws_handles]

the legend is here

[@mohdmehraj4546]

I was Expecting O(1) space By seeing the thumbnail !!

[@gui-codes]

mind blowing. i now comment on every video of yours because you are too good man

[@opsahadat9567]

OMG, ITS SOMTING HERD FOR ME, BUT FINALLY DONE 💜💜

[@MohdDanish-j7p]

what a fabulous explanation..

[@codestorywithMIK]

Thank you 🙏😇

[@vivekgupta_3085]

Good approach brother thank you

[@ankitajha3962]

I was able to code on my own (: Great explanation

[@anmolverma075]

Sir what an amazing and simple explaination!
Thanks a lot!

[@priyanshpal5070]

just earned a sub bro, keep up the good work

[@SamriddhiSharma-xc6iw]

Bhaiya please make a video for leetcode problem 394 , "Decode String". It will be a great help. And thanks a lot for such easy and explanatory videos.

[@saanchit1]

@codestorywithMIK I love the work you are doing, for this question I think you could have added the optimised solution (2 pointer) as it's space complexity is o(n)

[@codestorywithMIK]

Appreciate your kind words Saanchit.
I agree, I should have added. Actually this us an old video, I will try to create a separate video covering the 2 pointer solution ❤️❤️

[@akbar55555shaikh]

Now I feel, this problem should be marked as easy on leetcode 😊.

[@thefinalfit]

If you have time this weekend, can you also upload stack approach. thanks

[@ujjawaltyagi8540]

Please make video on other mentioned approaches as well!!

[@bhuppidhamii]

i already solved it but here for intuition

[@intromaker400]

Sir attach the java code on every video and also try to explain the code in java, it helps me a lot

[@codestorywithMIK]

Yes yes.
Now all videos have JAVA code also ❤️🙏

[@advitkothari7302]

gr8 ! job very helpful

[@ManojKrVerma-vw4dx]

Please make a video for this problem using the two pointers method

[@udaytiwari2855]

Thanks a lot.👍👍

[@gauravbanerjee2898]

Thanks a lot bhaiya ❤️

[@manishv3992]

With stack bhaiya plz ❤

[@purushottamkumar3140]

int trap(vector& height) {
int n = height.size();
int left_max[n], right_max[n];

int leftMax = height[0], rightMax = height[n-1];
for(int i=0; ileftMax) leftMax = height[i];
if(height[n-i-1]>rightMax) rightMax = height[n-i-1];
}

int sum = 0;

for(int i=0; i

[@ajinkyakhete1151]

please upload monotonic stack

[@lahirishreemanto19]

Good video, but I was expecting O(1) solution please!!

[@factacewar]

Day 2 done ✅

[@bhaveshjhawar9532]

Sir please Tapping Rain Water 2 bhi solve krdijiyeeee PLEASEEEEEEEE

[@AkOp-bf9vm]

1.can be done with stack also?? how??
2. i think we can do it with O(N) space only for this approach and also it can be done in O(1) space.

[@abhishekprasad5527]

bhaiya you should consider making a-z dsa course on yt, pls bhaiya

[@dhruvp1777]

youre a god

[@youknowsuraj69.]

Hello Mik ,
You done a silly mistake while writing a previous element ,
in last element of previous element its should be prev[i] = max(arr[i],prev[i-1]) = 5.
so the previous array look like this ;-
{4,4,4,4,4,5}

[@harshtiwari416]

Bhai iski stack or dp approach bhi bana do

[@harchitgulati3065]

bhaiya ye stack se ho sakta hei kya using just greater element approach

[@AnkitJain-w3l]

Day -2

[@sarathkumardunga5240]

Hi bhaiya! I have tried for the approach using just variables instead of arrays but i couldnt figure out.. could u help me with the approach using just the variables

[@codestorywithMIK]

Can you share your code ?

[@sarathkumardunga5240]

@@codestorywithMIK i can maintain the leftMax variable and as i go from (0 to n-1) and can update the leftMax variable if the current height of the building is > present leftMax variable. But for rightMax variable each time for the current height of the building i have to traverse all the buildings to right from the current building to figure the max height to right. Is there any way that i can figure out both leftMax height and rightMaxheight at one go.. OR do i need to change my approach??

[@sarathkumardunga5240]

@@codestorywithMIK
int trap(vector& height) {
//APPROACH 1: (USING 2 POINTERS)
//TC: O(N) SC: O(1)
int n = height.size();

int left = 0;
int right = n-1;
int leftbar = height[0];
int rightbar = height[n-1];
int waterStored = 0;
while(left = limmiting bar)
// || if(currworking bar < limiting bar)
if(height[left] >= leftbar){
waterStored+= 0;
leftbar = height[left];
}
else{
waterStored += (leftbar - height[left]);
}
left++;
}
else{
//same as above
if(height[right] >= rightbar){
waterStored+= 0;
rightbar = height[right];
}
else{
waterStored += (rightbar - height[right]);
}
right--;
}
}

return waterStored;
}
--------------------------------------------------------------------------------------------------------
/* //APPROACH 2
// TC: O(N) SC: O(N)
int n = height.size();

vector rmax(n, 0); //for an element i , rmax[i] represents the highest building to its right
vector lmax(n, 0);
rmax[0] = height[0];
lmax[n-1] = height[n-1];
for(int i = 1 ; i < n; i++){
rmax[i] = max(rmax[i-1], height[i]);
lmax[n-i-1] = max(lmax[n-i], height[n-i-1]);
}
int waterStored = 0;
for(int i = 0 ; i < n; i++){
int waterOnTop = min(rmax[i], lmax[i]) - height[i];
waterStored += waterOnTop;
}
return waterStored;
}
*/

[@aws_handles]

Difficult with just variables

[@Pranav_147]

A SMALL MISTAKE BRO........LEFT MAX AT 5TH INDEX IS 5 NOT 4

[@Nofaltuguy1]

hehe daily challenge guys