in DS edited by
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in DS edited by
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considering the stack : 1,0,2,4,5,7 with top = 7

queue 1 : 3,6,7,8,9,0 with front as 3 and rear as 0.

now as we know that   :-> stack : lifo    queue : fifo

1st case : x=7 , y = 3 ==> x=x+y =10

so now queue1=6,7,8,9,0,10 ans queue 2 = 3,

similarly we will execute the while loop since stack is empty.

which gives out..

queue 1 : 10,11,11,10,9,1

queue 2 : 3,6,7,8,9,0

so answer is A

 

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