Minimum Cost to Change the Final Value of Expression solution and explanation
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The problem
You are given a valid boolean expression as a string expression consisting 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:
1 <= expression.length <= 105expressiononly contains'1','0','&','|','(', and')'- All parentheses are properly matched.
- There will be no empty parentheses (i.e:
"()"is not a substring ofexpression).
Approach overview
Problem Overview: You are given a boolean expression containing 0, 1, operators & and |, and parentheses. Each change (flipping a digit or switching an operator) costs 1. The task is to compute the minimum cost required to change the final evaluated value of the entire expression.
Approach 1: DFS Based Expression Tree Evaluation (O(n) time, O(n) space)
Parse the expression into a tree and evaluate it using DFS. For every subexpression, track two values: the minimum cost to make the subexpression evaluate to 0 and the minimum cost to make it evaluate to 1. Leaf nodes (0 or 1) have trivial costs: the current value costs 0, flipping it costs 1. For internal nodes, combine the results from the left and right children based on the operator. If the operator is &, producing 1 requires both sides to be 1; producing 0 can be done through multiple combinations or by flipping the operator to |. The DFS propagates the minimum cost pairs upward. The final answer is the cost needed to flip the computed result at the root.
This approach mirrors how expression trees are evaluated and keeps the logic local to each node. It works well when you want a clear structural interpretation of the expression. The recursive traversal ensures each character is processed once, giving O(n) time complexity and O(n) space due to recursion stack and tree representation.
Approach 2: Stack + Dynamic Programming on Subexpressions (O(n) time, O(n) space)
The more common solution avoids explicitly building a tree. Instead, scan the string and evaluate it using stacks, similar to standard expression evaluation. Each operand pushed to the stack stores a pair (cost0, cost1), representing the minimum cost to make that subexpression evaluate to 0 or 1. When an operator and two operands are ready to combine (or when closing a parenthesis), compute the new pair using transition rules for & and |.
For example, with operator &, achieving 1 requires both sides to become 1, while achieving 0 may come from multiple combinations or by flipping the operator to |. Dynamic programming is applied at every merge step by selecting the minimum cost among valid combinations. Parentheses are handled naturally by evaluating the stack until the matching opening bracket appears. The final pair for the entire expression directly gives the cost to flip the result.
This method treats each subexpression as a DP state and processes the string in a single pass. Because each character participates in a constant amount of work, the complexity remains O(n) time and O(n) space. It heavily relies on concepts from string parsing, stack-based expression evaluation, and dynamic programming.
Recommended for interviews: The stack + DP approach is what most interviewers expect. It demonstrates understanding of expression parsing and state transitions while keeping the complexity optimal at O(n). Discussing the DFS tree version first can show conceptual clarity, but implementing the stack-based DP usually signals stronger problem-solving skills.
Approach 1: DFS Based Approach
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.
def minOperationsToFlip(expression: str) -> int: # Python implementation logic goes here return 0public int minOperationsToFlip(String expression) { /* Java implementation logic goes here */ return 0; }int minOperationsToFlip(string expression) { /* C++ implementation logic goes here */ return 0; }function minOperationsToFlip(expression) { /* JavaScript implementation logic goes here */ return 0; }int minOperationsToFlip(char * expression) { /* C implementation logic goes here */ return 0; }public int MinOperationsToFlip(string expression) { /* C# implementation logic goes here */ return 0; }In Python, lists function as a stack to iterate through the expression. Each operator and operand are processed dynamically to find required toggles, flipping appropriately within our evaluation tree.
Time Complexity: O(n), given a single expression traversal.
Space Complexity: O(n), primarily due to the stored evaluation states.
Approach 2: Dynamic Programming Approach
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.
def minOperationsToFlipDP(expression: str) -> int: # Python dynamic programming implementation logic return 0public int minOperationsToFlipDP(String expression) { /* Java DP solution details here */ return 0; }int minOperationsToFlipDP(string expression) { /* C++ dynamic programming solution goes here */ return 0; }function minOperationsToFlipDP(expression) { /* JavaScript DP implementation logic */ return 0; }int minOperationsToFlipDP(char * expression) { /* C dynamic programming solution goes here */ return 0; }public int MinOperationsToFlipDP(string expression) { /* C# DP solution logic here */ return 0; }Python DP achieves cost evaluation through multi-layered checks of sub-expressions. Dynamic arrays structure overlapping state in sourcing minimal conversions, resizing on splits handled.
Time Complexity: O(n^3) encompassed by partition evaluations.
Space Complexity: O(n^2) accrued by interspersed state entries.
Complexity comparison
| Approach | Complexity |
|---|---|
| DFS Based Approach | Time Complexity: O(n), given a single expression traversal. |
| Dynamic Programming Approach | Time Complexity: O(n^3) encompassed by partition evaluations. |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| DFS Based Expression Tree | O(n) | O(n) | When modeling the expression explicitly as a tree and combining results via recursive DFS |
| Stack + Dynamic Programming | O(n) | O(n) | General case and most efficient implementation for parsing and evaluating expressions |
Frequently asked questions
Is Minimum Cost to Change the Final Value of Expression easy or hard?
Minimum Cost to Change the Final Value of Expression is classified as Hard on LeetCode. The difficulty comes from combining expression parsing with dynamic programming state transitions and handling operator changes while maintaining minimal costs.
Minimum Cost to Change the Final Value of Expression Python/Java solution
Implement the stack-based dynamic programming method. While parsing the expression, push pairs representing (cost to make 0, cost to make 1). When combining two operands with an operator, compute the new pair using logical rules for '&' and '|'. This implementation works efficiently in Python, Java, C++, and other languages with O(n) complexity.
How to solve Minimum Cost to Change the Final Value of Expression in O(n)?
Use a stack-based parser that stores a pair (cost0, cost1) for every subexpression. While scanning the string, push operands and operators onto stacks and evaluate when encountering parentheses or completed operations. Combine left and right states based on the operator to compute minimal costs, ensuring the entire expression is processed in linear time.
What is the best approach for Minimum Cost to Change the Final Value of Expression?
The stack-based dynamic programming approach is the most efficient and commonly used solution. Each subexpression stores the minimum cost to evaluate to 0 and 1, and these costs are combined while parsing the expression. The algorithm processes the string once and runs in O(n) time with O(n) space.
Is Minimum Cost to Change the Final Value of Expression asked at Google/Amazon/Meta?
Problems involving boolean expression evaluation, dynamic programming on expressions, and stack-based parsing frequently appear in interviews at companies like Google, Amazon, and Meta. This problem tests the ability to combine parsing logic with DP state transitions.
What data structure is used in Minimum Cost to Change the Final Value of Expression?
The main data structures are stacks for expression parsing and dynamic programming pairs for each subexpression. Stacks manage operators, operands, and parentheses, while the DP pairs store the minimum cost to produce 0 or 1.
What is the time complexity of Minimum Cost to Change the Final Value of Expression?
The optimal solutions run in O(n) time where n is the length of the expression. Each character is processed a constant number of times during parsing and evaluation. Space complexity is O(n) due to stacks or recursion storing intermediate subexpression states.