Doubt in the second method

#1

In the second method when we map -3 to 3 to an array from 0 to 6, we did r-c+n-1, but isnt it wrong? because when r-c=-3 we add n-1(which means 7-1) to it which maps -3 to 3 and +3 to 6

#2

also, if n=5, shouldnt we add n-2 instead of n-1? because -3+5-2=0 but if we add n-1 we get -3+5-1 which is equal to 1

#3

Hi Vipul. to which question are u referring to?? can u share the link of the same

#4

N-queens question(the second method)

#5

can u tell at what time of video are u referring to??

#6

around the 41 mins time mark

#7

Hi Vipul.
so ur first doubt was this -
In the second method when we map -3 to 3 to an array from 0 to 6, we did r-c+n-1, but isnt it wrong? because when r-c=-3 we add n-1(which means 7-1) to it which maps -3 to 3 and +3 to 6
considering 5*5 array when we add n-1 (ie 5-1 = 4) -3 is mapped to 1 and 3 is mapped to 7.
but as u said if we add n-2 then -3 is mapped to 0 and 3 is mapped to 6.

#8

Hope this clears ur doubts. there is also another way out if u don’t want to preform all this addition and all. u can maintain a hashmap as it can store negative values as well. space complexity remains the same.

#9

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