Given a valid boolean expression string consisting of characters "1", "0", "&", "|", "(", and ")", determine the minimum number of operations needed so that after making changes the final value of the expression is flipped. An operation can flip a digit (changing "1" to "0" or vice versa) or flip an operator (changing "&" to "|" or "|" to "&"). Note that evaluation follows parentheses first and then proceeds left-to-right (i.e. no precedence among "&" and "|").
Key Insights
The expression can be broken down into subexpressions, each of which has two associated values: the minimum cost to force a result of 0 and the minimum cost to force a result of 1.
For a literal digit, the cost to obtain its inherent value is zero while flipping it incurs a cost of one.
For an operator combining two subexpressions, you can either keep the operator or flip it (incurring an extra cost) and compute the result based on the DP values of the left and right subexpressions.
A stack-based evaluation (similar to evaluating an expression) can be used to merge subresults in left-to-right order while taking parentheses into account.
At each binary operation the optimal cost is the minimum between keeping the operator (and computing the cost based on its semantics) and flipping the operator (adding one extra operation plus computing cost using the flipped semantics).
Space and Time Complexity
Time Complexity: O(n), where n is the length of the expression. Each character is processed once and every operator and operand is combined in constant time.
Space Complexity: O(n) due to the use of stacks to hold operands and operators.
Solution
The solution uses dynamic programming on subexpressions. Every digit or subexpression is represented by a pair (cost0, cost1), meaning the minimum cost required for that part of the expression to evaluate to 0 and 1, respectively.
For a literal "0", the pair is (0, 1) because it costs 0 operations to be 0 and 1 operation to flip to 1.
For a literal "1", the pair is (1, 0).
When combining two subexpressions with an operator, we compute:
For the "&" operator, the result is 1 only if both subexpressions are 1; any other combination gives 0.
For the "|" operator, the result is 0 only if both subexpressions are 0; any other combination gives 1.
For each operator, we also consider flipping it (at an additional cost of 1) and take the minimum cost among both possibilities.
A stack-based approach is used to simulate the evaluation order (taking into account parentheses and left-to-right evaluation). Two stacks are maintained: one for operands (the dp pairs) and one for operators. As you parse the expression, operands are pushed, and when you identify an operator (or when a closing parenthesis indicates the end of a subexpression), the last two operands are combined using the operator at the top of the operator stack.
Trick: When an operator is encountered and the operator stack is not empty (and the top is not a parenthesis), perform an evaluation immediately because all operators have equal precedence (left-to-right evaluation).
Code Solutions
# Define a very large number as infinity since we need to take minimumsINF =10**9defminCostToFlip(expression):# Helper function to combine two dp pairs using opdefcombine(left, op, right):# left and right are tuples (cost_to_get_0, cost_to_get_1)# For op: if we keep it:if op =='&':# Without flipping: cost_keep_1 = left[1]+ right[1]# Only possibility is both 1 cost_keep_0 =min(left[0]+ right[0],# 0 & 0 = 0 left[0]+ right[1],# 0 & 1 = 0 left[1]+ right[0])# 1 & 0 = 0# If we flip '&' to '|' (cost + 1): cost_flip_0 = left[0]+ right[0]# 0 | 0 = 0 cost_flip_1 =min(left[1]+ right[0],# 1 | 0 = 1 left[0]+ right[1],# 0 | 1 = 1 left[1]+ right[1])# 1 | 1 = 1else:# op == '|'# Without flipping: cost_keep_0 = left[0]+ right[0]# Only possibility is both 0 cost_keep_1 =min(left[1]+ right[0],# 1 | 0 = 1 left[0]+ right[1],# 0 | 1 = 1 left[1]+ right[1])# 1 | 1 = 1# If we flip '|' to '&' (cost + 1): cost_flip_1 = left[1]+ right[1]# 1 & 1 = 1 cost_flip_0 =min(left[0]+ right[0],# 0 & 0 = 0 left[0]+ right[1],# 0 & 1 = 0 left[1]+ right[0])# 1 & 0 = 0# Total cost = min(cost with keeping operator, cost with an extra flip cost) final0 =min(cost_keep_0,1+ cost_flip_0) final1 =min(cost_keep_1,1+ cost_flip_1)return(final0, final1)# Stacks for operands (dp pairs) and operators operand_stack =[] operator_stack =[]# Process the expression character by characterfor ch in expression:if ch ==' ':continue# skip spaces if anyif ch =='(': operator_stack.append(ch)elif ch =='0'or ch =='1':# For a literal digit: cost is 0 to get its value and 1 to flip.if ch =='0': operand_stack.append((0,1))else: operand_stack.append((1,0))elif ch in'&|':# Process pending operators (all have same precedence and are left associative)while operator_stack and operator_stack[-1]in"&|": op = operator_stack.pop() right = operand_stack.pop() left = operand_stack.pop() operand_stack.append(combine(left, op, right)) operator_stack.append(ch)elif ch ==')':# Process until matching '(' is encountered.while operator_stack and operator_stack[-1]!='(': op = operator_stack.pop() right = operand_stack.pop() left = operand_stack.pop() operand_stack.append(combine(left, op, right)) operator_stack.pop()# pop the '('# Process any remaining operators.while operator_stack: op = operator_stack.pop() right = operand_stack.pop() left = operand_stack.pop() operand_stack.append(combine(left, op, right))# Final dp pair for the whole expression# We need to flip the final value so return the cost for the opposite value. final_dp = operand_stack[-1]# If the expression originally evaluates to 0, we want to output cost for 1; else cost for 0.# However, problem asks for "minimum cost to change the final value",# meaning flip the computed final result.# Determine the current evaluation by taking the minimum cost option for each value: final_value =0if final_dp[0]< final_dp[1]else1return final_dp[1]if final_value ==0else final_dp[0]# Example usage:# expression = "1&(0|1)"# Expected output is 1.#print(minCostToFlip("1&(0|1)"))