Introduction

In this blog post, we will explore how to solve the LeetCode problem "1896" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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).

Explanation

Here's a breakdown of the solution:

• Dynamic Programming with Recursion: The core idea is to use dynamic programming (DP) combined with recursion to efficiently compute the minimum cost to achieve a target value (0 or 1) for each subexpression. The DP memoization avoids redundant calculations.
• Bottom-Up Evaluation: The recursion essentially implements a bottom-up approach, starting from the innermost parenthesized expressions and working outwards.

• Cost Calculation: The DP table stores the minimum cost to make a subexpression evaluate to 0 and the minimum cost to make it evaluate to 1. The costs are updated based on the operator ('&' or '|') and the costs of the subexpressions it connects.

• Time Complexity: O(n), where n is the length of the expression. Space Complexity: O(n)

Code

    def min_operations_to_target(expression):
    """
    Calculates the minimum cost to change the final value of a boolean expression.

    Args:
        expression: A valid boolean expression string.

    Returns:
        The minimum cost to change the final value of the expression.
    """

    def calculate_cost(s, memo):
        """
        Recursively calculates the minimum cost to achieve 0 or 1 for a subexpression.

        Args:
            s: The subexpression string.
            memo: A dictionary for memoization.

        Returns:
            A tuple (cost_to_0, cost_to_1) representing the minimum cost to make the subexpression
            evaluate to 0 and 1, respectively.
        """

        if s in memo:
            return memo[s]

        if len(s) == 1:
            if s == '0':
                memo[s] = (0, 1)
            else:
                memo[s] = (1, 0)
            return memo[s]

        if s[0] == '(':
            balance = 0
            for i in range(len(s)):
                if s[i] == '(':
                    balance += 1
                elif s[i] == ')':
                    balance -= 1
                if balance == 0:
                    if i == len(s) - 1:
                        return calculate_cost(s[1:-1], memo)
                    break

        min_cost_0 = float('inf')
        min_cost_1 = float('inf')

        for i in range(1, len(s)):
            if s[i] == '&' or s[i] == '|':
                left = s[:i]
                right = s[i + 1:]
                left_cost_0, left_cost_1 = calculate_cost(left, memo)
                right_cost_0, right_cost_1 = calculate_cost(right, memo)

                if s[i] == '&':
                    # Original operation is AND
                    min_cost_0 = min(min_cost_0, min(left_cost_0 + right_cost_1, left_cost_1 + right_cost_0, left_cost_0 + right_cost_0) + 1)
                    min_cost_1 = min(min_cost_1, left_cost_1 + right_cost_1)
                else:
                    # Original operation is OR
                    min_cost_0 = min(min_cost_0, left_cost_0 + right_cost_0)
                    min_cost_1 = min(min_cost_1, min(left_cost_0 + right_cost_1, left_cost_1 + right_cost_0, left_cost_1 + right_cost_1) + 1)

        memo[s] = (min_cost_0, min_cost_1)
        return memo[s]

    cost_to_0, cost_to_1 = calculate_cost(expression, {})
    return cost_to_0