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 <= 105expression only contains '1','0','&','|','(', and ')'"()" is not a substring of expression).Description: Use a DFS approach to evaluate each sub-expression. Traverse the expression, and for each binary operation, calculate the cost of flipping the result by inspecting the preceding and succeeding values, along with the operation. Maintain a stack to manage parentheses properly. Each entry in the stack will comprise the current cost, number of required flips, and the results of sub-expressions.
This C solution initializes the parsing of the expression using a stack. Each stack element includes state information like current result, costs, and flip probabilities. Iteratively traverse each character in the expression, updating the stack based on evaluated conditions and operations.
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Time Complexity: O(n) where n is the length of the expression.
Space Complexity: O(n) due to the stack.
Description: Implement a dynamic programming method to split the expression at each binary operator, computing minimal costs for flipping sub-expression results. Capture intermediate results in a table, indexing by part of the expression under current evaluation.
The approach dynamically parses sections of the input towards minimizing toggle costs. We can maintain a table mimicking a recursive call structure, with costs per evaluated expression sequence calculated iteratively.
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Time Complexity: O(n^3) for the combinatorial splits of expressions.
Space Complexity: O(n^2) to store intermediate calculations.
| Approach | Complexity |
|---|---|
| DFS Based Approach | Time Complexity: O(n) where n is the length of the expression. |
| Dynamic Programming Approach | Time Complexity: O(n^3) for the combinatorial splits of expressions. |
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