Some problems in Data structure and algorithm.

Question

Here, you will know various types of numericals and solutions to it.

Answer 1

Conversion of infix expression to prefix expression

P) (A+B) *C

Solution:

⇒[+AB]*C

⇒*+ABC

P:2) (A-B) *(C+D) 

Solution:

⇒[-AB]*[+CD]

⇒*-AB+CD

P:3) (A+B) /(C+D) -(D*E) 

Solution:

⇒[+AB]/[+CD]-[*DE]

⇒[/+AB+CD]-[*DE]

⇒-/+AB+CD*DE

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2. Convert the expression given below into its equivalent prefix and postfix notation. 

((A+B) ^C) -(D-E) ^(F+G)) 

Prefix notation:

⇒(​​​​[+AB]^C-[-DE]^[+FG]) 

⇒-^ABC ^ -DE+FG

Postfix Notation:

⇒AB+C^DE-FG+^-

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3. Evaluate the following postfix expressions for A=2, B=5, C=3, D=2, E=4

ABC+DE*/-

Show stack at each step. 

Solution:

ABC+DE*/-

Given, A=2, B=5, C=3, D=2, E=4

So, putting values of A, B, C, D, E into the above expression , we get

2 5 3+2 4 * / - 

Symbol Scanned	Stack
(1)                 2	2
(2)                 5	2, 5
(3)                 3	2, 5,3
(4)                 +	2,8
(5)                 2	2, 8,2
(6)                 4	2, 8,2,4
(7)                 *	2, 8,8
(8)                 /	2, 1
(9)                -	1

So, the final number in stack is 1, which is the solution. 

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4. Transform the following expression into its equivalent postfix expression using stack. 

A+(B*C-(D/E↑) *G) *H) 

Solution:

Symbol Scanned	Stack	Expression
(1)         A	(	A
(2)         +	(+	A
(3)        (	(+(	A
(4)        B	(+(	AB
(5)        *	(+(*	AB
(6)        C	(+(*	ABC
(7)       -	(+(-	ABC*
(8)       (	(+(-(	ABC*
(9)        D	(+(-(	ABC*D
(10)      /	(+(-(/	ABC*D
(11)      E	(+(-(/	ABC*DE
(12)      ↑	(+(-(/↑	ABC*DE
(13)      F	(+(-(/↑	ABC*DEF
(14)     )	(+(-	ABC*DEF↑/
(15)      *	(+(-*	ABC*DEF↑/
(16)      G	(+(-*	ABC*DEF↑/G
(17)      )	(+	ABC*DEF↑/G*-
(18)      *	(+*	ABC*DEF↑/G*-
(19)     H	(+*	ABC*DEF↑/G*-H
(20)     )		ABC*DEF↑/G*-H*+

The equivalent postfix expression of the above infix expression is: 

ABC*DEF↑/G*-H*+

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5. Calculate the position of element which is present in 7^th row and 8^th column for a 10×10 matrix . Here base address is 6010.

Solution:

By row major order:

 Address[a_ij] =M+(i-1) ×n+j-1

⇒a₇₈=6010+(7-1) ×10+8-1

⇒a₇₈=6077

By coulmn major order:

Address[a_ij]=M+(j-1) ×m+i+1

⇒a₇₈=6010+(8-1) ×10+7-1

⇒a₇₈=6086

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