Container With Most Water - (Meta, Google, Amazon) : Explanation ➕ Live Coding 🧑🏻‍💻👩🏻‍💻

[@Kunalbindra]

He never asked us for like , share and subscribe 🙇‍♂. A True GEM...

[@Kaafirpeado54-6ayesha]

Question is so intimidating i have seen this question and left it right away

[@varunupadhyay2488]

great explanation

[@sonukumarram1880]

Ur explanations , the way of teaching .... big fan sir really feeling blessed

[@Satya-g5t]

nice solution and well explained.

[@surajsidar3280]

I am watching the solution after the interview. I was asked how and why we were moving two pointers. I got the answer. We have to find a minimum of two heights of left and right pointer Since we are decreasing the width, to maximize the water contained we will move the pointer pointing to a smaller height.

[@moinkhan2137]

which Company Bro?

[@tutuimam3381]

Amazing explanation

[@prashupandey1304]

You are the best bro !!!🙏🙏

[@codestorywithMIK]

And so are you Prashu Thank you so much 🙏❤️😇

[@isha25tripathi12]

Thank you so much. When I first read the question I was afraid that its too difficult and when you read the question I was able to do it on my own without listening to the whole explanation.

[@PushpendraKushvaha]

same

[@saketjaiswalSJ]

god level explanation brother

[@ikitayadav3272]

omg why i didn't find this channel earlier

[@cosmicthor7330]

bhaeeya if index 2 pey 6 ki jagah koi bara number hota jaisey 45 to area hamara chut jata na ,kiwki dono height same hai start and end ka if uskey neigher ka height bara ho to compare krna pareyga na kis side move krey i ko ya j ko hope mai smjha paya apna question

[@nitkarshchourasia2406]

class Solution { public: int maxArea(vector& height) { // find length // find height int n = height.size(); int left = 0; int right = n -1; int maximum = INT_MIN; while (left < right) { int l = right - left; int h = std::min(height[left], height[right]); (height[right] > height[left]) ? left++ : right--; int k = l * h; if (k > maximum) { maximum = k; cout

[@harshsingh7713]

Bhai aapke subscriber bhut tezz badd rhe Badhai ho

[@abcd76820]

cant we apply trapping rain water problem logic with modification ?

[@rode_atharva]

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

[@AnkitJain-w3l]

Day 3

[@DeepakSingh-fd2ix]

sir how this question is different from maximum area histogram

[@harchitgulati3065]

same doubt

[@parul15137]

in that we were seeing consecutive heightss to form a rectangle by using the concept of nsl and nsr of stack by taking the current idx height but there isnt any consecutiveness , but here we are storing the water even if the heights in b/w are smaller or bigger we can take them cos water will be stored b/w so we dont care what heights are coming in b/w but in the histogram one if the fluctuation of heights occured our rectangle will be disformed which we dont want

[@Codenow10222]

when the height is same why are we allowed to shift any of them i and j . because in the above question when both i and j if we move i then the maxArea is 24 but if we move from j then it will be 16

[@Lakshya-f4l]

same question

[@varshneyshuchi]

whatever you choose .. will it effect maximum?