You are given a valid boolean expression as a string expression
consisting of the characters '1'
,'0'
,'&'
(bitwise AND operator),'|'
(bitwise OR operator),'('
, and ')'
.
"()1|1"
and "(1)&()"
are not valid while "1"
, "(((1))|(0))"
, and "1|(0&(1))"
are valid expressions.Return the minimum cost to change the final value of the expression.
expression = "1|1|(0&0)&1"
, its value is 1|1|(0&0)&1 = 1|1|0&1 = 1|0&1 = 1&1 = 1
. We want to apply operations so that the new expression evaluates to 0
.The cost of changing the final value of an expression is the number of operations performed on the expression. The types of operations are described as follows:
'1'
into a '0'
.'0'
into a '1'
.'&'
into a '|'
.'|'
into a '&'
.Note: '&'
does not take precedence over '|'
in the order of calculation. Evaluate parentheses first, then in left-to-right order.
Example 1:
Input: expression = "1&(0|1)" Output: 1 Explanation: We can turn "1&(0|1)" into "1&(0&1)" by changing the '|' to a '&' using 1 operation. The new expression evaluates to 0.
Example 2:
Input: expression = "(0&0)&(0&0&0)" Output: 3 Explanation: We can turn "(0&0)&(0&0&0)" into "(0|1)|(0&0&0)" using 3 operations. The new expression evaluates to 1.
Example 3:
Input: expression = "(0|(1|0&1))" Output: 1 Explanation: We can turn "(0|(1|0&1))" into "(0|(0|0&1))" using 1 operation. The new expression evaluates to 0.
Constraints:
1 <= expression.length <= 105
expression
only contains '1'
,'0'
,'&'
,'|'
,'('
, and ')'
"()"
is not a substring of expression
).