DSCPPAS270 - Editorial

editorial
,
1

Problem Statement:

You are given a string representing a prefix expression (also known as Polish notation), where operators precede their operands. The task is to evaluate the expression and return the result. The expression will contain single-digit integers and the operators +, -, *, and /.

Approach:

To evaluate a prefix expression, you typically process the expression from right to left. This is because in prefix notation, the operator appears before its operands. Using a stack is a convenient way to handle this as you can push operands onto the stack and, when an operator is encountered, you can pop the necessary operands, perform the operation, and push the result back onto the stack.

Here’s the step-by-step approach:

Step 1: Parse the Expression

  • First, split the input expression into tokens. Each token will either be an operator or an operand.

Step 2: Use a Stack for Evaluation

  • Iterate over the tokens from right to left.
  • If a token is an operand, push it onto the stack.
  • If a token is an operator, pop the top two elements from the stack, apply the operator, and push the result back onto the stack.

Step 3: Handle Modular Arithmetic

  • In some cases, particularly in competitive programming, results might need to be computed under a modulo. If needed, the addition, subtraction, multiplication, and division operations should be performed modulo a specific value (e.g., 10^9 + 7).
  • Division in modular arithmetic requires computing the modular inverse using Fermat’s Little Theorem.

Example:

For the prefix expression * + 2 3 4:

  1. Start by pushing 2 and 3 to the stack.
  2. When encountering +, pop 2 and 3, compute 2 + 3 = 5, and push 5 back onto the stack.
  3. Push 4 onto the stack.
  4. Finally, when encountering *, pop 5 and 4, compute 5 * 4 = 20, and push 20 back onto the stack.
  5. The result is 20.

Time Complexity:

The time complexity of this solution is O(N), where N is the number of tokens in the expression, as each token is processed exactly once.

Space Complexity:

The space complexity is O(N) due to the stack used to store intermediate results.