Minimum Cost to Change the Final Value of Expression

Problem Description

Ah, the classic dilemma of wanting to change the final value of an expression without breaking the bank—or in this case, without breaking your brain! The problem at hand is to determine the minimum number of operations required to flip the final value of a boolean expression represented as a string.

Imagine you’re at a party, and you’re trying to convince your friends to switch from soda to water. You know that if you can just flip their opinions (0 for soda, 1 for water), you’ll save yourself from a sugar crash. But how many of them do you need to convince? That’s the essence of this problem!

The expression consists of binary values (0s and 1s) and operators (& for AND, | for OR). Your task is to find out how many flips you need to make the expression evaluate to a different value than it currently does.

Code Solution


class Solution {
 public:
  int minOperationsToFlip(string expression) {
    stack> stack;
    pair lastPair;

    for (const char e : expression) {
      if (e == '(' || e == '&' || e == '|') {
        stack.push({e, 0});
        continue;
      }
      if (e == ')') {
        lastPair = stack.top();
        stack.pop();
        stack.pop();  // Pop '('.
      } else {        // e == '0' || e == '1'
        lastPair = {e, 1};
      }
      if (!stack.empty() &&
          (stack.top().first == '&' || stack.top().first == '|')) {
        const char op = stack.top().first;
        stack.pop();
        const auto [a, costA] = stack.top();
        stack.pop();
        const auto [b, costB] = lastPair;
        if (op == '&') {
          if (a == '0' && b == '0')
            lastPair = {'0', 1 + min(costA, costB)};
          else if (a == '0' && b == '1')
            lastPair = {'0', 1};
          else if (a == '1' && b == '0')
            lastPair = {'0', 1};
          else
            lastPair = {'1', min(costA, costB)};
        } else {
          if (a == '0' && b == '0')
            lastPair = {'0', min(costA, costB)};
          else if (a == '0' && b == '1')
            lastPair = {'1', 1};
          else if (a == '1' && b == '0')
            lastPair = {'1', 1};
          else
            lastPair = {'1', 1 + min(costA, costB)};
        }
      }
      stack.push(lastPair);
    }

    return stack.top().second;
  }
};

Approach Explanation

The code uses a stack to evaluate the expression in a bottom-up manner. It processes each character in the expression, pushing operators and operands onto the stack. When it encounters a closing parenthesis, it pops the stack to evaluate the expression inside the parentheses. The cost of flipping the values is calculated based on the operators and the values of the operands.

Time and Space Complexity

Complexity Type	Complexity
Time Complexity	O(n), where n is the length of the expression. Each character is processed once.
Space Complexity	O(n) in the worst case due to the stack storing the operators and operands.

Real-World Example

Let’s say you’re a manager at a fast-food restaurant, and you want to change the menu from burgers (0) to salads (1). You know that some customers love burgers, while others are health-conscious. You need to figure out how many customers you need to convince to switch their orders. This is akin to flipping the values in our boolean expression!