Minimum Cost to Change the Final Value of Expression

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Dynamic Programming

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 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:

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 <= 10⁵
expression only contains '1','0','&','|','(', and ')'
All parentheses are properly matched.
There will be no empty parentheses (i.e: "()" is not a substring of expression).

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What characters are allowed in the expression string beyond 0, 1, &, and |?
What is the maximum length of the expression string?
If it's impossible to change the final value (e.g., an empty expression), what value should I return?
Are parentheses guaranteed to be balanced in the expression string?
Is the input expression string guaranteed to be a valid boolean expression consisting of only 0, 1, &, |, and parentheses?

Brute Force Solution

Approach

The brute force approach to finding the minimum cost to change the expression's final value involves trying every single possible change to the expression. This means testing every combination of flips and then determining the cost. The best possible outcome, costing the least, is chosen in the end.

Here's how the algorithm would work step-by-step:

Consider that you can flip certain characters in the expression (like changing a 0 to a 1 or an AND to an OR).
Start by flipping nothing, then flipping just one character, then flipping every possible pair of characters, then every possible group of three, and so on.
For each of these possibilities, calculate the new value of the expression after making those flips.
Compare the result you got to the final value we want, to see if this set of flips gives us the correct value.
If the flips gives us the value we want, then calculate how much it cost to make all those flips.
Keep track of the minimum cost among all the combinations of flips that resulted in getting to our final value that we want.
Once we've checked every single possible combination of flips, the lowest cost we've recorded is the answer.

Code Implementation

def solve():
    expression = "1&0|1"
    target_value = 0
    minimum_cost = float('inf')
    expression_length = len(expression)

    # Iterate through all possible combinations of flips
    for i in range(1 << expression_length):
        modified_expression = list(expression)
        current_cost = 0

        # For each bit in the bitmask, flip the corresponding character
        for j in range(expression_length):
            if (i >> j) & 1:
                #Flip operator or operand
                if modified_expression[j] == '0':
                    modified_expression[j] = '1'
                    current_cost += 1
                elif modified_expression[j] == '1':
                    modified_expression[j] = '0'
                    current_cost += 1
                elif modified_expression[j] == '&':
                    modified_expression[j] = '|'
                    current_cost += 1
                elif modified_expression[j] == '|':
                    modified_expression[j] = '&'
                    current_cost += 1

        modified_expression_string = "".join(modified_expression)

        # Evaluate the modified expression
        try:
            current_value = eval(modified_expression_string)

            # Check if the modified expression evaluates to the target value
            if current_value == target_value:

                #Update minimum cost
                minimum_cost = min(minimum_cost, current_cost)
        except:
            pass

    if minimum_cost == float('inf'):
        return -1
    else:
        return minimum_cost

#Try expression with no flips

print(solve())

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach involves trying every possible combination of flips. Given an expression of length n, each character can either be flipped or not flipped, resulting in 2 possibilities for each character. Therefore, there are 2^n possible combinations of flips to explore. For each of these 2^n combinations, we need to evaluate the expression which takes O(n) in the worst case. Thus, the overall time complexity is O(n * 2^n), however, because 2^n dominates n, we approximate the big O to O(2^n).

Space Complexity

O(1) – The brute force approach considers all possible subsets of characters to flip. While it doesn't explicitly mention data structures, implicitly the algorithm needs to keep track of which characters are flipped in each combination being tested. However, the plain English explanation doesn't specify any explicit data structures being created to store these combinations. The cost, final value and minimum cost are stored, but these are a fixed number of variables regardless of the input size, N. Therefore, the auxiliary space complexity is constant, O(1).

Optimal Solution

Approach

The goal is to find the cheapest way to make the entire math problem evaluate to the opposite of what it currently does. We use a technique that breaks the problem down into smaller, manageable pieces, figuring out the cost for each piece before combining them to solve the overall problem.

Here's how the algorithm would work step-by-step:

Start by looking at the smallest parts of the problem: the individual numbers (0 or 1). The cost to flip these is already known.
Then, look at slightly bigger chunks of the problem, like 'number operator number' (e.g., '1 & 0'). Figure out the cheapest way to make that entire mini-expression flip its value. This will involve considering flipping one or both of the numbers.
Keep going, building up the expression piece by piece. For each larger piece, you use the solutions you already figured out for the smaller pieces. This means you're reusing work, instead of recalculating.
When you get to a bigger part, look at all the different ways to change its final value (by flipping the numbers inside) and pick the cheapest way to do it.
Continue building the problem back up, considering operator precedence (like doing parentheses first, then ANDs and ORs), until you reach the whole expression.
The cost of flipping the entire expression's value will be the cheapest way you found at the very end.

Code Implementation

def min_cost_to_flip(expression): 
    values = {}
    costs = {}

    def calculate_min_cost(sub_expression): 
        if sub_expression in values:
            return values[sub_expression], costs[sub_expression]

        if sub_expression.isdigit():
            value = int(sub_expression)
            cost_to_flip = 1
            values[sub_expression] = value
            costs[sub_expression] = cost_to_flip
            return value, cost_to_flip

        # Find the last operator to respect precedence
        operator_index = -1
        parenthesis_balance = 0
        for i in range(len(sub_expression) - 1, -1, -1): 
            if sub_expression[i] == ')':
                parenthesis_balance += 1
            elif sub_expression[i] == '(': 
                parenthesis_balance -= 1
            elif parenthesis_balance == 0 and (sub_expression[i] == '&' or sub_expression[i] == '|'):
                operator_index = i
                break

        if operator_index == -1 and sub_expression[0] == '(' and sub_expression[-1] == ')':
            return calculate_min_cost(sub_expression[1:-1])

        # Split based on the operator
        left_expression = sub_expression[:operator_index]
        operator = sub_expression[operator_index]
        right_expression = sub_expression[operator_index + 1:]

        left_value, left_cost = calculate_min_cost(left_expression)
        right_value, right_cost = calculate_min_cost(right_expression)

        if operator == '&':
            current_value = left_value & right_value
            # Calculate costs to flip the result
            cost_to_flip = float('inf')
            cost_to_flip = min(cost_to_flip, left_cost + right_value) 
            cost_to_flip = min(cost_to_flip, right_cost + left_value) 
            cost_to_flip = min(cost_to_flip, left_cost + right_cost) 
        else:
            current_value = left_value | right_value
            # Calculate costs to flip the result
            cost_to_flip = float('inf')
            cost_to_flip = min(cost_to_flip, left_cost + (1 - right_value))
            cost_to_flip = min(cost_to_flip, right_cost + (1 - left_value))
            cost_to_flip = min(cost_to_flip, left_cost + right_cost)

        values[sub_expression] = current_value
        costs[sub_expression] = cost_to_flip
        return current_value, cost_to_flip

    final_value, min_flip_cost = calculate_min_cost(expression)
    return min_flip_cost

Big(O) Analysis

Time Complexity

O(n^3) – The algorithm processes the expression in a bottom-up manner, considering sub-expressions of increasing length. For each sub-expression, it explores all possible combinations of flipping the operands. Since there are potentially O(n^2) sub-expressions (where n is the length of the input string) and for each of these sub-expressions, we might iterate through all its possible operand flip combinations which could be another factor of O(n), the overall time complexity can be approximated as O(n^3). In particular, it iterates over sub-expressions of length `l` (from 1 to n) and starting position `s` (from 0 to n-l), and computes the minimum cost for each of these which itself can take O(n) to find the minimum cost for flips leading to the overall O(n^3) complexity.

Space Complexity

O(N) – The algorithm builds the solution piece by piece, storing the minimum cost to flip the value and the minimum cost to keep the value for each subexpression. This means for each number, operator, and parenthesis in the input expression, we store two cost values. Therefore, the auxiliary space required grows linearly with the size of the expression, where N is the length of the expression string. We store these costs in a data structure that mirrors the structure of the expression tree, implying a space complexity proportional to the number of nodes in the abstract syntax tree which is O(N).

Edge Cases

Case	How to Handle
Empty expression string	Return 0, as no operations are needed on an empty expression.
Null expression string	Throw an IllegalArgumentException or return an error code indicating invalid input.
Expression with a single digit (e.g., '0' or '1')	If the single digit is not equal to the desired final value, the cost is 1, otherwise 0.
Expression with only operators and no operands	Consider the expression invalid and throw an exception or return an error value.
Expression with unbalanced parentheses	Consider the expression invalid and throw an exception or return an error value.
Expression with consecutive operators (e.g., '1&&&0')	Consider the expression invalid and throw an exception or return an error value.
Expression that evaluates to the desired value without any changes	Return 0, as no changes are necessary.
Expression with extremely deep nesting of parentheses, potentially causing stack overflow in recursive solutions	Use iterative dynamic programming instead of recursion to avoid stack overflow errors.

Empty expression string

How to Handle:

Return 0, as no operations are needed on an empty expression.

Null expression string

How to Handle:

Throw an IllegalArgumentException or return an error code indicating invalid input.

Expression with a single digit (e.g., '0' or '1')

How to Handle:

If the single digit is not equal to the desired final value, the cost is 1, otherwise 0.

Expression with only operators and no operands

How to Handle:

Consider the expression invalid and throw an exception or return an error value.

Expression with unbalanced parentheses

How to Handle:

Consider the expression invalid and throw an exception or return an error value.

Expression with consecutive operators (e.g., '1&&&0')

How to Handle:

Consider the expression invalid and throw an exception or return an error value.

Expression that evaluates to the desired value without any changes

How to Handle:

Return 0, as no changes are necessary.

Expression with extremely deep nesting of parentheses, potentially causing stack overflow in recursive solutions

How to Handle:

Use iterative dynamic programming instead of recursion to avoid stack overflow errors.