770. Basic Calculator IV LeetCode Solution

In this guide, you will get 770. Basic Calculator IV LeetCode Solution with the best time and space complexity. The solution to Basic Calculator IV problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Basic Calculator IV solution in C++
4. Basic Calculator IV solution in Java
5. Basic Calculator IV solution in Python
6. Additional Resources

Problem Statement of Basic Calculator IV

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 [-231, 231 – 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

Complexity Analysis

• Time Complexity: O(n)
• Space Complexity: O(n)

770. Basic Calculator IV LeetCode Solution in C++

class Poly {
  friend Poly operator+(const Poly& lhs, const Poly& rhs) {
    Poly res(lhs);
    for (const auto& [term, coef] : rhs.terms)
      res.terms[term] += coef;
    return res;
  }

  friend Poly operator-(const Poly& lhs, const Poly& rhs) {
    Poly res(lhs);
    for (const auto& [term, coef] : rhs.terms)
      res.terms[term] -= coef;
    return res;
  }

  friend Poly operator*(const Poly& lhs, const Poly& rhs) {
    Poly res;
    for (const auto& [a, aCoef] : lhs.terms)
      for (const auto& [b, bCoef] : rhs.terms)
        res.terms[merge(a, b)] += aCoef * bCoef;
    return res;
  }

  // Friend ostream& operator<<(ostream& os, const Poly& poly) {
  //   os << "{";
  //   for (const auto& [term, coef] : poly.terms)
  //     os << term << ": " << coef << ", ";
  //   os << "}";
  //   return os;
  // }

 public:
  vector<string> toList() {
    vector<string> res;
    vector<string> keys;
    for (const auto& [term, _] : terms)
      keys.push_back(term);
    ranges::sort(keys, [&](const string& a, const string& b) {
      // the minimum degree is the last
      if (a == "1")
        return false;
      if (b == "1")
        return true;
      const vector<string> as = split(a, '*');
      const vector<string> bs = split(b, '*');
      // the maximum degree is the first
      // Break ties by their lexicographic orders.
      return as.size() == bs.size() ? a < b : as.size() > bs.size();
    });
    auto concat = [&](const string& term) -> string {
      if (term == "1")
        return to_string(terms[term]);
      return to_string(terms[term]) + '*' + term;
    };
    for (const string& key : keys)
      if (terms[key])
        res.push_back(concat(key));
    return res;
  }

  Poly() = default;
  Poly(const string& term, int coef) {
    terms[term] = coef;
  }

 private:
  unordered_map<string, int> terms;

  // e.g. merge("a*b", "a*c") -> "a*a*b*c"
  static string merge(const string& a, const string& b) {
    if (a == "1")
      return b;
    if (b == "1")
      return a;
    string res;
    vector<string> A = split(a, '*');
    vector<string> B = split(b, '*');
    int i = 0;  // A's index
    int j = 0;  // B's index
    while (i < A.size() && j < B.size())
      if (A[i] < B[j])
        res += '*' + A[i++];
      else
        res += '*' + B[j++];
    while (i < A.size())
      res += '*' + A[i++];
    while (j < B.size())
      res += '*' + B[j++];
    return res.substr(1);
  }

  static vector<string> split(const string& token, char c) {
    vector<string> vars;
    istringstream iss(token);
    for (string var; getline(iss, var, c);)
      vars.push_back(var);
    return vars;
  }
};

class Solution {
 public:
  vector<string> basicCalculatorIV(string expression, vector<string>& evalvars,
                                   vector<int>& evalints) {
    vector<string> tokens = getTokens(expression);
    unordered_map<string, int> evalMap;

    for (int i = 0; i < evalvars.size(); ++i)
      evalMap[evalvars[i]] = evalints[i];

    for (string& token : tokens)
      if (const auto it = evalMap.find(token); it != evalMap.cend())
        token = to_string(it->second);

    const vector<string>& postfix = infixToPostfix(tokens);
    return evaluate(postfix).toList();
  }

 private:
  vector<string> getTokens(const string& s) {
    vector<string> tokens;
    int i = 0;
    for (int j = 0; j < s.length(); ++j)
      if (s[j] == ' ') {
        if (i < j)
          tokens.push_back(s.substr(i, j - i));
        i = j + 1;
      } else if (string("()+-*").find(s[j]) != string::npos) {
        if (i < j)
          tokens.push_back(s.substr(i, j - i));
        tokens.push_back(s.substr(j, 1));
        i = j + 1;
      }
    if (i < s.length())
      tokens.push_back(s.substr(i));
    return tokens;
  }

  bool isOperator(const string& token) {
    return token == "+" || token == "-" || token == "*";
  }

  vector<string> infixToPostfix(const vector<string>& tokens) {
    vector<string> postfix;
    stack<string> ops;

    auto precedes = [](const string& prevOp, const string& currOp) -> bool {
      if (prevOp == "(")
        return false;
      return prevOp == "*" || currOp == "+" || currOp == "-";
    };

    for (const string& token : tokens)
      if (token == "(") {
        ops.push(token);
      } else if (token == ")") {
        while (ops.top() != "(")
          postfix.push_back(ops.top()), ops.pop();
        ops.pop();
      } else if (isOperator(token)) {
        while (!ops.empty() && precedes(ops.top(), token))
          postfix.push_back(ops.top()), ops.pop();
        ops.push(token);
      } else {  // isOperand(token)
        postfix.push_back(token);
      }

    while (!ops.empty())
      postfix.push_back(ops.top()), ops.pop();

    return postfix;
  }

  Poly evaluate(const vector<string>& postfix) {
    vector<Poly> polys;
    for (const string& token : postfix)
      if (isOperator(token)) {
        const Poly b = polys.back();
        polys.pop_back();
        const Poly a = polys.back();
        polys.pop_back();
        if (token == "+")
          polys.push_back(a + b);
        else if (token == "-")
          polys.push_back(a - b);
        else  // token == "*"
          polys.push_back(a * b);
      } else if (token[0] == '-' ||
                 ranges::all_of(token, [](char c) { return isdigit(c); })) {
        polys.push_back(Poly("1", stoi(token)));
      } else {
        polys.push_back(Poly(token, 1));
      }
    return polys[0];
  }
};
/* code provided by PROGIEZ */

770. Basic Calculator IV LeetCode Solution in Java

class Poly {
  public Poly add(Poly o) {
    for (final String term : o.terms.keySet())
      terms.merge(term, o.terms.get(term), Integer::sum);
    return this;
  }

  public Poly minus(Poly o) {
    for (final String term : o.terms.keySet())
      terms.merge(term, -o.terms.get(term), Integer::sum);
    return this;
  }

  public Poly mult(Poly o) {
    Poly res = new Poly();
    for (final String a : terms.keySet())
      for (final String b : o.terms.keySet())
        res.terms.merge(merge(a, b), terms.get(a) * o.terms.get(b), Integer::sum);
    return res;
  }

  // @Override
  // Public String toString() {
  //   StringBuilder sb = new StringBuilder();
  //   sb.append("{");
  //   for (final String term : terms.keySet())
  //     sb.append(term).append(": ").append(terms.get(term)).append(", ");
  //   sb.append("}");
  //   return sb.toString();
  // }

  public List<String> toList() {
    List<String> res = new ArrayList<>();
    List<String> keys = new ArrayList<>(terms.keySet());
    Collections.sort(keys, new Comparator<String>() {
      @Override
      public int compare(final String a, final String b) {
        // the minimum degree is the last
        if (a.equals("1"))
          return 1;
        if (b.equals("1"))
          return -1;
        String[] as = a.split("\\*");
        String[] bs = b.split("\\*");
        // the maximum degree is the first
        // Break ties by their lexicographic orders.
        return as.length == bs.length ? a.compareTo(b) : bs.length - as.length;
      }
    });
    for (final String key : keys)
      if (terms.get(key) != 0)
        res.add(concat(key));
    return res;
  }

  public Poly() {}
  public Poly(final String term, int coef) {
    terms.put(term, coef);
  }

  private Map<String, Integer> terms = new HashMap<>();

  // e.g. merge("a*b", "a*c") -> "a*a*b*c"
  private static String merge(final String a, final String b) {
    if (a.equals("1"))
      return b;
    if (b.equals("1"))
      return a;
    StringBuilder sb = new StringBuilder();
    String[] A = a.split("\\*");
    String[] B = b.split("\\*");
    int i = 0; // A's index
    int j = 0; // B's index
    while (i < A.length && j < B.length)
      if (A[i].compareTo(B[j]) < 0)
        sb.append("*").append(A[i++]);
      else
        sb.append("*").append(B[j++]);
    while (i < A.length)
      sb.append("*").append(A[i++]);
    while (j < B.length)
      sb.append("*").append(B[j++]);
    return sb.substring(1).toString();
  }

  private String concat(final String term) {
    if (term.equals("1"))
      return String.valueOf(terms.get(term));
    return new StringBuilder().append(terms.get(term)).append('*').append(term).toString();
  }
}

class Solution {
  public List<String> basicCalculatorIV(String expression, String[] evalvars, int[] evalints) {
    List<String> tokens = getTokens(expression);
    Map<String, Integer> evalMap = new HashMap<>();

    for (int i = 0; i < evalvars.length; ++i)
      evalMap.put(evalvars[i], evalints[i]);

    for (int i = 0; i < tokens.size(); ++i)
      if (evalMap.containsKey(tokens.get(i)))
        tokens.set(i, String.valueOf(evalMap.get(tokens.get(i))));

    List<String> postfix = infixToPostfix(tokens);
    return evaluate(postfix).toList();
  }

  private List<String> getTokens(final String s) {
    List<String> tokens = new ArrayList<>();
    int i = 0;
    for (int j = 0; j < s.length(); ++j)
      if (s.charAt(j) == ' ') {
        if (i < j)
          tokens.add(s.substring(i, j));
        i = j + 1;
      } else if ("()+-*".contains(s.substring(j, j + 1))) {
        if (i < j)
          tokens.add(s.substring(i, j));
        tokens.add(s.substring(j, j + 1));
        i = j + 1;
      }
    if (i < s.length())
      tokens.add(s.substring(i));
    return tokens;
  }

  private boolean isOperator(final String token) {
    return token.equals("+") || token.equals("-") || token.equals("*");
  }

  private boolean precedes(final String prevOp, final String currOp) {
    if (prevOp.equals("("))
      return false;
    return prevOp.equals("*") || currOp.equals("+") || currOp.equals("-");
  }

  private List<String> infixToPostfix(List<String> tokens) {
    List<String> postfix = new ArrayList<>();
    Deque<String> ops = new ArrayDeque<>();

    for (final String token : tokens)
      if (token.equals("(")) {
        ops.push(token);
      } else if (token.equals(")")) {
        while (!ops.peek().equals("("))
          postfix.add(ops.pop());
        ops.pop();
      } else if (isOperator(token)) {
        while (!ops.isEmpty() && precedes(ops.peek(), token))
          postfix.add(ops.pop());
        ops.push(token);
      } else { // isOperand(token)
        postfix.add(token);
      }

    while (!ops.isEmpty())
      postfix.add(ops.pop());

    return postfix;
  }

  private Poly evaluate(List<String> postfix) {
    LinkedList<Poly> polys = new LinkedList<>();
    for (final String token : postfix)
      if (isOperator(token)) {
        final Poly b = polys.removeLast();
        final Poly a = polys.removeLast();
        if (token.equals("+"))
          polys.add(a.add(b));
        else if (token.equals("-"))
          polys.add(a.minus(b));
        else // token == "*"
          polys.add(a.mult(b));
      } else if (token.charAt(0) == '-' || token.chars().allMatch(c -> Character.isDigit(c))) {
        polys.add(new Poly("1", Integer.parseInt(token)));
      } else {
        polys.add(new Poly(token, 1));
      }
    return polys.getFirst();
  }
}
// code provided by PROGIEZ

770. Basic Calculator IV LeetCode Solution in Python

class Poly:
  def __init__(self, term: str = None, coef: int = None):
    if term and coef:
      self.terms = collections.Counter({term: coef})
    else:
      self.terms = collections.Counter()

  def __add__(self, other):
    for term, coef in other.terms.items():
      self.terms[term] += coef
    return self

  def __sub__(self, other):
    for term, coef in other.terms.items():
      self.terms[term] -= coef
    return self

  def __mul__(self, other):
    res = Poly()
    for a, aCoef in self.terms.items():
      for b, bCoef in other.terms.items():
        res.terms[self._merge(a, b)] += aCoef * bCoef
    return res

  # Def __str__(self):
  #   res = []
  #   for term, coef in self.terms.items():
  #     res.append(term + ': ' + str(coef))
  #   return '{' + ', '.join(res) + '}'

  def toList(self) -> list[str]:
    for term in list(self.terms.keys()):
      if not self.terms[term]:
        del self.terms[term]

    def cmp(term: str) -> tuple:
      # the minimum degree is the last
      if term == '1':
        return (0,)
      var = term.split('*')
      # the maximum degree is the first
      # Break ties by their lexicographic orders.
      return (-len(var), term)

    def concat(term: str) -> str:
      if term == '1':
        return str(self.terms[term])
      return str(self.terms[term]) + '*' + term

    terms = list(self.terms.keys())
    terms.sort(key=cmp)
    return [concat(term) for term in terms]

  def _merge(self, a: str, b: str) -> str:
    if a == '1':
      return b
    if b == '1':
      return a
    res = []
    A = a.split('*')
    B = b.split('*')
    i = 0  # A's index
    j = 0  # B's index
    while i < len(A) and j < len(B):
      if A[i] < B[j]:
        res.append(A[i])
        i += 1
      else:
        res.append(B[j])
        j += 1
    return '*'.join(res + A[i:] + B[j:])


class Solution:
  def basicCalculatorIV(
      self,
      expression: str,
      evalvars: list[str],
      evalints: list[int],
  ) -> list[str]:
    tokens = list(self._getTokens(expression))
    evalMap = {a: b for a, b in zip(evalvars, evalints)}

    for i, token in enumerate(tokens):
      if token in evalMap:
        tokens[i] = str(evalMap[token])

    postfix = self._infixToPostfix(tokens)
    return self._evaluate(postfix).toList()

  def _getTokens(self, s: str) -> Iterator[str]:
    i = 0
    for j, c in enumerate(s):
      if c == ' ':
        if i < j:
          yield s[i:j]
        i = j + 1
      elif c in '()+-*':
        if i < j:
          yield s[i:j]
        yield c
        i = j + 1
    if i < len(s):
      yield s[i:]

  def _infixToPostfix(self, tokens: list[str]) -> list[str]:
    postfix = []
    ops = []

    def precedes(prevOp: str, currOp: str) -> bool:
      if prevOp == '(':
        return False
      return prevOp == '*' or currOp in '+-'

    for token in tokens:
      if token == '(':
        ops.append(token)
      elif token == ')':
        while ops[-1] != '(':
          postfix.append(ops.pop())
        ops.pop()
      elif token in '+-*':  # isOperator(token)
        while ops and precedes(ops[-1], token):
          postfix.append(ops.pop())
        ops.append(token)
      else:  # isOperand(token)
        postfix.append(token)
    return postfix + ops[::-1]

  def _evaluate(self, postfix: list[str]) -> Poly:
    polys: list[Poly] = []
    for token in postfix:
      if token in '+-*':
        b = polys.pop()
        a = polys.pop()
        if token == '+':
          polys.append(a + b)
        elif token == '-':
          polys.append(a - b)
        else:  # token == '*'
          polys.append(a * b)
      elif token.lstrip('-').isnumeric():
        polys.append(Poly("1", int(token)))
      else:
        polys.append(Poly(token, 1))
    return polys[0]
 # code by PROGIEZ

Additional Resources

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• Explore all problems on LeetCode website here

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