Class Solution
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public final class Solution1896 - Minimum Cost to Change the Final Value of Expression.
Hard
You are given a valid boolean expression as a string
expressionconsisting of the characters'1','0','&'(bitwise AND operator),'|'(bitwise OR operator),'(', and')'.For example,
"()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.
For example, if
expression = "1|1|(0&0)&1", its value is1|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 to0.
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:
Turn a
'1'into a'0'.Turn a
'0'into a'1'.Turn a
'&'into a'|'.Turn 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:
<code>1 <= expression.length <= 10<sup>5</sup></code>
expressiononly contains'1','0','&','|','(', and')'All parentheses are properly matched.
There will be no empty parentheses (i.e:
"()"is not a substring ofexpression).
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Constructor Summary
Constructors Constructor Description Solution()
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Method Summary
Modifier and Type Method Description final IntegerminOperationsToFlip(String expression)-
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Method Detail
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minOperationsToFlip
final Integer minOperationsToFlip(String expression)
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