Remove redundant parentheses from an arithmetic expression

Question

This is an interview question, for which I did not find any satisfactory answers on stackoverflow or outside. Problem statement:

Given an arithmetic expression, remove redundant parentheses. E.g. ((a*b)+c) should become a*b+c

I can think of an obvious way of converting the infix expression to post fix and converting it back to infix - but is there a better way to do this?

Answer 1

A pair of parentheses is necessary if and only if they enclose an unparenthesized expression of the form X % X % ... % X where X are either parenthesized expressions or atoms, and % are binary operators, and if at least one of the operators % has lower precedence than an operator attached directly to the parenthesized expression on either side of it; or if it is the whole expression. So e.g. in

q * (a * b * c * d) + c

the surrounding operators are {+, *} and the lowest precedence operator inside the parentheses is *, so the parentheses are unnecessary. On the other hand, in

q * (a * b + c * d) + c

there is a lower precedence operator + inside the parentheses than the surrounding operator *, so they are necessary. However, in

z * q + (a * b + c * d) + c

the parentheses are not necessary because the outer * is not attached to the parenthesized expression.

Why this is true is that if all the operators inside an expression (X % X % ... % X) have higher priority than a surrounding operator, then the inner operators are anyway calculated out first even if the parentheses are removed.

So, you can check any pair of matching parentheses directly for redundancy by this algorithm:

Let L be operator immediately left of the left parenthesis, or nil
Let R be operator immediately right of the right parenthesis, or nil
If L is nil and R is nil:
  Redundant
Else:
  Scan the unparenthesized operators between the parentheses
  Let X be the lowest priority operator
  If X has lower priority than L or R:
    Not redundant
  Else:
    Redundant

You can iterate this, removing redundant pairs until all remaining pairs are non-redundant.

Example:

((a * b) + c * (e + f))

(Processing pairs from left to right):

((a * b) + c * (e + f))   L = nil R = nil --> Redundant
^                     ^   
 (a * b) + c * (e + f)    L = nil R = nil --> Redundant
 ^     ^                  L = nil R = + X = * --> Redundant
  a * b  + c * (e + f)    L = * R = nil X = + --> Not redundant
               ^     ^

Final result:

a * b + c * (e + f)

Answer 2

I just figured out an answer:

the premises are:

1. the expression has been tokenized
2. no syntax error
3. there are only binary operators

input:

list of the tokens, for example:
   (, (, a, *, b, ), +, c, )

output:

set of the redundant parentheses pairs (the orders of the pairs are not important),
for example,
   0, 8
   1, 5

please be aware of that : the set is not unique, for instance, ((a+b))*c, we can remove outer parentheses or inner one, but the final expression is unique

the data structure:

a stack, each item records information in each parenthese pair
the struct is:
   left_pa: records the position of the left parenthese
   min_op: records the operator in the parentheses with minimum priority
   left_op: records current operator

the algorithm

1.push one empty item in the stack
2.scan the token list
    2.1 if the token is operand, ignore
    2.2 if the token is operator, records the operator in the left_op, 
        if min_op is nil, set the min_op = this operator, if the min_op 
        is not nil, compare the min_op with this operator, set min_op as 
        one of the two operators with less priority
    2.3 if the token is left parenthese, push one item in the stack, 
        with left_pa = position of the parenthese
    2.4 if the token is right parenthese, 
        2.4.1 we have the pair of the parentheses(left_pa and the 
             right parenthese)
        2.4.2 pop the item
        2.4.3 pre-read next token, if it is an operator, set it 
             as right operator
        2.4.4 compare min_op of the item with left_op and right operator
             (if any of them exists), we can easily get to know if the pair 
             of the parentheses is redundant, and output it(if the min_op
             < any of left_op and right operator, the parentheses are necessary,
             if min_op = left_op, the parentheses are necessary, otherwise
             redundant) 
        2.4.5 if there is no left_op and no right operator(which also means 
             min_op = nil) and the stack is not empty, set the min_op of top 
             item as the min_op of the popped-up item

examples

example one

((a*b)+c)

after scanning to b, we have stack:

index left_pa min_op left_op
0
1     0       
2     1       *      *       <-stack top

now we meet the first ')'(at pos 5), we pop the item

left_pa = 1 
min_op = *
left_op = *

and pre-read operator '+', since min_op priority '*' > '+', so the pair(1,5) is redundant, so output it. then scan till we meet last ')', at the moment, we have stack

index left_pa min_op left_op
0
1     0       +      + 

we pop this item(since we meet ')' at pos 8), and pre-read next operator, since there is no operator and at index 0, there is no left_op, so output the pair(0, 8)

example two

a*(b+c)

when we meet the ')', the stack is like:

index  left_pa  min_op left_op
0               *      *
1      2        +      +

now, we pop the item at index = 1, compare the min_op '+' with the left_op '*' at index 0, we can find out the '(',')' are necessary