@allwell8570
You are just awesome! Here is what I learnt --> In general, suppose to calculate dp[x], if we need dp[y], then dp[y] must be calculated before evaluating dp[x]. In this problem, as he has explained, to calculate dp[x][y][z], we need dp[x-1][y+1][z], dp[x][y-1][z+1] and dp[x][y][z-1]. Suppose if your outermost loop is y, then for dp[...][y][...] you need dp[...][y+1][...], dp[...][y-1][...] and dp[...][y][...]. For particular y, you need both y-1 and y+1. This can not be done in an iterative version.
@pleasesirmorevideos4684
one of the best videos on order of evaluation in case of iterative dp
@Adityag0243
Great video bhaiya❤
@adityachawla2163
Why did you iterate over ones first, then twos and then threes? Why not the other way around? I mean, how do you know that states required for transition have already been calculated before?
@tuhinmukherjee8141
What's the maths behind the reason you added a one to dp[a][b][c]?
@anubhavmishra318
Once again amazing explanation.
@tusharsinha1772
Very insightful video.
@PravinShankhapal1
How are you so op?
@gopal7338
Nice!
@ardhendureja6037
bhaiya thora zoom karliya karo sublime ko
@fatakful
Demoralizer sir I am your fan how can I get your autograph
@saurabhsharma7123
bhai tera editor N * N * N * N per runtime error kese show kr ra hai? Which plugin😅??
@muditjain4499
Sir when are you doing a face reveal??
@Acodedaily