Basic Calculator IV - LeetCode solutions

Problem

Given an expression such as expression = "e + 8 - a + 5" and an evaluation map such as {"e": 1} (given in terms of evalvars = ["e"] and evalints = [1]), return a list of tokens representing the simplified expression, such as ["-1*a","14"]

An expression alternates chunks and symbols, with a space separating each chunk and symbol.
A chunk is either an expression in parentheses, a variable, or a non-negative integer.
A variable is a string of lowercase letters (not including digits.) Note that variables can be multiple letters, and note that variables never have a leading coefficient or unary operator like "2x" or "-x".

Expressions are evaluated in the usual order: brackets first, then multiplication, then addition and subtraction.

For example, expression = "1 + 2 * 3" has an answer of ["7"].

The format of the output is as follows:

For each term of free variables with a non-zero coefficient, we write the free variables within a term in sorted order lexicographically.

For example, we would never write a term like "b*a*c", only "a*b*c".

Terms have degrees equal to the number of free variables being multiplied, counting multiplicity. We write the largest degree terms of our answer first, breaking ties by lexicographic order ignoring the leading coefficient of the term.
For example, "a*a*b*c" has degree 4.
The leading coefficient of the term is placed directly to the left with an asterisk separating it from the variables (if they exist.) A leading coefficient of 1 is still printed.
An example of a well-formatted answer is ["-2*a*a*a", "3*a*a*b", "3*b*b", "4*a", "5*c", "-6"]. Terms (including constant terms) with coefficient 0 are not included.
For example, an expression of "0" has an output of [].

Note: You may assume that the given expression is always valid. All intermediate results will be in the range of [-2^31, 2^31 - 1].

Example 1:

Input: expression = "e + 8 - a + 5", evalvars = ["e"], evalints = [1]
Output: ["-1*a","14"]

Example 2:

Input: expression = "e - 8 + temperature - pressure", evalvars = ["e", "temperature"], evalints = [1, 12]
Output: ["-1*pressure","5"]

Example 3:

Input: expression = "(e + 8) * (e - 8)", evalvars = [], evalints = []
Output: ["1*e*e","-64"]

Constraints:

1 <= expression.length <= 250
expression consists of lowercase English letters, digits, '+', '-', '*', '(', ')', ' '.
expression does not contain any leading or trailing spaces.
All the tokens in expression are separated by a single space.
0 <= evalvars.length <= 100
1 <= evalvars[i].length <= 20
evalvars[i] consists of lowercase English letters.
evalints.length == evalvars.length
-100 <= evalints[i] <= 100

Solution

/**
 * @param {string} expression
 * @param {string[]} evalvars
 * @param {number[]} evalints
 * @return {string[]}
 */
const basicCalculatorIV = function(expression, evalvars, evalints) {
  const record = {}; // { variable: value, ... }
  evalvars.forEach((k, i) => record[k] = evalints[i]);
  const expr = calculator(expression); // expr: { expr: coeff, ... }
  const terms = collectTerms(expr);
  const ans = buildOutput(expr, terms);
  return ans;

  // calculator will return the final expr like: { e*e: 1, #: 64 }
  function calculator(expression) {
    if (!expression.length) return {};

    expression += " +"; // this is to trigger adding `cur` to `result`,
                        // otherwise, it might stop when finish calculating `cur`.
    const n = expression.length;
    let result = {}, cur = {};
    let operator = "+", i;
    for (i = 0; i < n; i += 2) { // i += 2 because there's a space between each expression
      const expr = getExpr();    // recursively to get the nested expr
      // first try an expression with no parentheses and log the expr
      // then try an expression with parentheses and log the expr
      // You'll probably get a feel how the recursion works
      if (["+", "-", "*"].includes(expression[i]) || i == n - 1) {
        // update cur based on operators,
        // for example for expression "e + 8", cur will have values: { e: 1 } and { #: 8 } 
        // result will have values: { e: 1 } and then { e: 1, #: 8 }.
        if (operator == "+") cur = add(cur, expr);
        if (operator == "-") cur = sub(cur, expr);
        if (operator == "*") cur = mul(cur, expr);
        // if meet '+' or '-', we add cur to result.
        // also this is why we added the " +" to expression
        // because otherwise, for example when the expression is "a * b * c", then 
        // cur will have values: { a: 1 }, { a*b: 1 }, and { a*b*c: 1 }.
        // but result will not get populated.
        if (["+", "-"].includes(expression[i]) || i == n - 1) {
          result = add(result, cur);
          cur = {};
        }
        operator = expression[i];
      }
    }
    return result;

    // calculator will return an expr like: { a: 1} for non-nested problem, 
    // and { a: 1, b: 1, #: 8 } for a nested problem
    function getExpr() {
      if (i >= n) return result;

      let expr = {}, start = i;
      if (expression[i] == "(") { // found a subproblem
        const stk = [];
        stk.push(i);
        i += 1;
        // find boundaries of the subproblem, 
        // if reaches the end of expression or parentheses pair closed
        while (i < n && stk.length) {
          if (expression[i] == "(") stk.push(i);
          if (expression[i] == ")") stk.pop();
          i += 1;
        }
        // the subproblem could look like this first: ((a - b) * (b - c) + (c - a))
        // then look like: a - b => { a: 1, b: 1 }
        // then look like: b - c => { b: 1, c: -1 }
        // then look like: c - a => { c: 1, a: -1 }
        const sub = expression.slice(start + 1, i-1);
        // for each of the subproblems above, calculator knows how to caculate it on its own,
        // it won't reach this part again, because the subproblem does not have '(' anymore.
        expr = calculator(sub);
      } else if (isdigit(expression[i])) { // collect digits expr, like { #: 84 }
        while (i < n && isdigit(expression[i])) i += 1;
        const val = expression.slice(start, i);
        expr["#"] = +val;
      } else {
        // collect expression key, like { a: 1 },
        // also it could be a pre-defined variable, so it becomes a constant
        while (i < n && expression[i] != " ") i += 1;
        const key = expression.slice(start, i);
        if (key in record) { // found variable constant
          expr["#"] = record[key];
        } else { // found a expression key
          expr[key] = 1;
        }
      }
      i += 1;
      return expr;
    }
  }

  function collectTerms(tree) {
    const terms = [];
    for (const [key, val] of Object.entries(tree)) {
      if (val && key != "#") {
        // order depends on how many '*' the key has
        const order = key.split("").filter(c => c == "*").length;
        terms.push([key, order]);
      }
    }
    // sort by order desc and key asc
    terms.sort((a, b) => a[1] != b[1] ? b[1] - a[1] : a[0].localeCompare(b[0]))
    return terms.map(([key]) => key);
  }

  function buildOutput(tree, terms) {
    const ans = terms.map(term => tree[term] + "*" + term);
    const constant = tree["#"];
    if (constant) ans.push(constant.toString());
    return ans;
  }
};

function isdigit(char) {
  return char >= "0" && char <= "9";
}

function combineKey(s1, s2) {
  return s1.split("*").concat(s2.split("*")).sort().join("*");
}

function add(a, b) {
  Object.entries(b).forEach(([k, v]) => a[k] = (a[k] || 0) + v);
  return a;
}

function sub(a, b) {
  Object.entries(b).forEach(([k, v]) => a[k] = (a[k] || 0) - v);
  return a;
}

function mul(a, b) {
  const ans = {};
  for (const [ak, av] of Object.entries(a)) {
    for (const [bk, bv] of Object.entries(b)) { 
      if (ak == "#") {
        ans[bk] = (ans[bk] || 0) + av * bv;
      } else if (bk == "#") {
        ans[ak] = (ans[ak] || 0) + av * bv;
      } else {
        const key = combineKey(ak, bk);
        ans[key] = (ans[key] || 0) + av * bv;
      }
    }
  }
  return ans;
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).