One Pager Cheat Sheet

• Let's program a very basic calculator, handling positive, negative symbols, parentheses, and spaces, without using eval function!
• In most languages, you can evaluate code strings quickly and easily using eval.
• We can break down the problem into parsing and evaluating symbols such as integers, signs/operations, parenthesis and spaces, and iterate through an expression string to transform it into a form understandable by the machine, with some caveats in the processing order.
• Adding a negative number to our running result is the key to solving the problem of subtraction.
• We can keep track of our sign while converting the signs of numbers by using 1 or -1 to represent the + or - operations respectively.
• Leveraging two stacks to keep track of the result and operations within the parentheses is the key to solving this challenging problem.
• We skip that iteration if the current character is a space.
• We can iterate through the input of length n in O(n) time and space complexity.
• We can extend our previous solution to handle Multiplication and Division by using a stack to store multipliers and dividers and multiplying or dividing the result of each for loop iteration by this stack as it pops values off.
• Byrecreatingthe parentheses expression on a stack, we can still use the left-to-right approach to parse"4*(3+2)", instead of being "drawn to the parentheses".

This is our final solution.

To visualize the solution and step through the below code, click Visualize the Solution on the right-side menu or the VISUALIZE button in Interactive Mode.

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}

var assert = require('assert');

​

function calculator(str) {

  let result = 0,

    sign = 1;

  const stack = [],

    operationStack = [];

​

  for (let i = 0; i < str.length; i++) {

    const curr = str.charAt(i);

    if (curr === ' ') {

      continue;

    } else if (curr === '+') {

      sign = 1;

    } else if (curr === '-') {

      sign = -1;

    } else if (curr >= '0' && curr <= '9') {

      let num = curr;

      while (

        i + 1 < str.length &&

        str.charAt(i + 1) >= '0' &&

        str.charAt(i + 1) <= '9'

      ) {

        num += str.charAt(i + 1);

        i++;

      }

      result += sign * parseInt(num);

    } else if (curr === '(') {

      stack.push(result);

      result = 0;

      operationStack.push(sign);

      sign = 1;

That's all we've got! Let's move on to the next tutorial.

If you had any problems with this tutorial, check out the main forum thread here.