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#1

My code is not submitting.
Neither showing testcases failed or passed.
Here is my code:

#2

i guess time limit exceeding. I have done correct but how to reduce time complexity.

#3

it may be becoz of some network issue

#4

soln exists in o(n)

simplest approach : required next smaller element

in a array store the next smaller element in left and right side respectively.
so at each pnt u know the left side and rt side smaller element index
so u can find the area( area=> left smaller index+1 TO right smaller index-1 )
and compare and store the area

hope this will help u
pls rate my work

#5

i got you to the point how to find area but i didnt get how to store left and right smaller element in one go means in O(n)

#6

i need to use nested loops . How to do it in a single loop

#7

bhaiya help! i need to know before tmorrow 1 pm. Tomorrow is my placement drive and i am getting gut feeling for similar kind of que

#8

sorry for the late reply

#9

have you solved a qn named find next smaller element
stack is used in this qn

#10

or next greater element

#11

suppose we have to find next smaller element (left to right)
e.g
1 6 2 8 4
so put 1 onto the stack
now 6 is not smaller than the peek of stack, so put this value also
now stack - 1 , 6
now 2
so 2 is smaller than peek so
pop peek set the ans that 2 is the next smaller element for the 6 element
but 2 can’t pop 1
[
now stack- 1
res== 6 next smaller elem=2
}
now 8
push onto the stack(can’t pop becoz 8 is grtr than peek)
stack-- 1, 8

now 4
4<peek(8)
pop the stack till the peek is smaller than the curr element and set the ans for the peek element
pop 8
set 8 next smll elem - 4
cant pop 1(1<4)

push 4 onto the stack

finallly

stack – 1,4
res==
6 nxt sm ele-2
8 nxt sm ele-4

the elem which r present in stack
don’t have any smaller elem on the right side

O(n)

hope this will help u
pls rate my work
pls