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.
\n\nexpression = "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:
\n\n'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.
\n
Example 1:
\n\n\nInput: expression = "1&(0|1)"\nOutput: 1\nExplanation: We can turn "1&(0|1)" into "1&(0&1)" by changing the '|' to a '&' using 1 operation.\nThe new expression evaluates to 0. \n\n\n
Example 2:
\n\n\nInput: expression = "(0&0)&(0&0&0)"\nOutput: 3\nExplanation: We can turn "(0&0)&(0&0&0)" into "(0|1)|(0&0&0)" using 3 operations.\nThe new expression evaluates to 1.\n\n\n
Example 3:
\n\n\nInput: expression = "(0|(1|0&1))"\nOutput: 1\nExplanation: We can turn "(0|(1|0&1))" into "(0|(0|0&1))" using 1 operation.\nThe new expression evaluates to 0.\n\n
\n
Constraints:
\n\n1 <= expression.length <= 105expression only contains '1','0','&','|','(', and ')'"()" is not a substring of expression).Compiled with clang 11 using the latest C++ 20 standard.
Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
Most standard library headers are already included automatically for your convenience.
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