Maximum element deque STL

#1

INPUT= 123456 (I’m taking sample case of my own)

So, in this question we first process k elements separately
by using ’ i ’ as our iteration variable so our window is (123) for k=3. I completely understood this part but when we process next window we are simply skipping initialization of ’ i ’ in loop (line number 29). And, now our window will be from (345) right?? because we processed first k elements separately but don’t you think we should first process (234) because that way we are incrementing our window by 1. I hope I made myself clear!

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#2

hello @div_yanshu07

here u went wrong. value of i after processing first k elements will be k (becuase 0 based indexing) .
so the first window that we will consider in second for loop will be [2,3,4] -> [ i-k+1…i].

#3

i will become k right i=3
Now in second loop we are not initializing i so our i will be at 3rd index only right??
How i is becoming 1?

#4

ok let me clear this first.
i (in second loop) -> it is index of right most element of current window.
starting index is i-k+1 .
so when i is 3 (ending index). then i-k+1 -> 3-3+1 = 1 wiil be starting index of the considered window

#5

Okay, I did a dry run. Now, I understood we are handling that case inside the loop. Thank you!

#6

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