# Basic Calculator IV LeetCode Solution

Here, We see Basic Calculator IV LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

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″]

## Basic Calculator IV LeetCode Solution C++

``````class Solution {
public:
vector<string> basicCalculatorIV(string expression, vector<string>& evalvars, vector<int>& evalints) {
unordered_map<string, int> mp;
int n = evalvars.size();
for (int i = 0; i < n; ++i) mp[evalvars[i]] = evalints[i];
int pos = 0;
unordered_map<string, int> output = helper(expression, mp, pos);
vector<pair<string, int>> ans(output.begin(), output.end());
sort(ans.begin(), ans.end(), mycompare);
vector<string> res;
for (auto& p: ans) {
if (p.second == 0) continue;
res.push_back(to_string(p.second));
if (p.first != "") res.back() += "*"+p.first;
}
return res;
}
private:
unordered_map<string, int> helper(string& s, unordered_map<string, int>& mp, int& pos) {
vector<unordered_map<string, int>> operands;
vector<char> ops;
ops.push_back('+');
int n = s.size();
while (pos < n && s[pos] != ')') {
if (s[pos] == '(') {
pos++;
operands.push_back(helper(s, mp, pos));
}
else {
int k = pos;
while (pos < n && s[pos] != ' ' && s[pos] != ')') pos++;
string t = s.substr(k, pos-k);
bool isNum = true;
for (char c: t) {
if (!isdigit(c)) isNum = false;
}
unordered_map<string, int> tmp;
if (isNum)
tmp[""] = stoi(t);
else if (mp.count(t))
tmp[""] = mp[t];
else
tmp[t] = 1;
operands.push_back(tmp);
}
if (pos < n && s[pos] == ' ') {
ops.push_back(s[++pos]);
pos += 2;
}
}
pos++;
return calculate(operands, ops);
}
unordered_map<string, int> calculate(vector<unordered_map<string, int>>& operands, vector<char>& ops) {
unordered_map<string, int> ans;
int n = ops.size();
for (int i = n-1; i >= 0; --i) {
unordered_map<string, int> tmp = operands[i];
while (i >= 0 && ops[i] == '*')
tmp = multi(tmp, operands[--i]);
int sign = ops[i] == '+'? 1: -1;
for (auto& p: tmp) ans[p.first] += sign*p.second;
}
return ans;
}
unordered_map<string, int> multi(unordered_map<string, int>& lhs, unordered_map<string, int>& rhs) {
unordered_map<string, int> ans;
int m = lhs.size(), n = rhs.size();
for (auto& p: lhs) {
for (auto& q: rhs) {
string t = combine(p.first, q.first);
ans[t] += p.second*q.second;
}
}
return ans;
}
string combine(const string& a, const string& b) {
if (a == "") return b;
if (b == "") return a;
vector<string> strs = split(a, '*');
for (auto& s: split(b, '*')) strs.push_back(s);
sort(strs.begin(), strs.end());
string s;
for (auto& t: strs) s += t +'*';
s.pop_back();
return s;
}
static vector<string> split(const string& s, char c) {
vector<string> ans;
int i = 0, n = s.size();
while (i < n) {
int j = i;
i = s.find(c, i);
if (i == -1) i = n;
ans.push_back(s.substr(j, i-j));
i++;
}
return ans;
}
static bool mycompare(pair<string, int>& a, pair<string, int>& b) {
string s1 = a.first, s2 = b.first;
vector<string> left = split(s1, '*');
vector<string> right = split(s2, '*');
return left.size() > right.size() || (left.size() == right.size() && left < right);
}
};```Code language: PHP (php)```

## Basic Calculator IV LeetCode Solution Java

``````class Solution {
int n;
String s;
char[] arr;
int[] braces;
HashMap<String, Integer> variables = new HashMap<>();

public List<String> basicCalculatorIV(String expression, String[] evalvars, int[] evalints) {
s = expression;
arr = s.toCharArray();
n = arr.length;
braces = new int[n];
Arrays.fill(braces, -1);
int[] stack = new int[n/2];
int index = -1;
for(int i=0; i<n; ++i) {
if(arr[i] == '(') stack[++index] = i;
else if(arr[i] == ')') {
int last = stack[index--];
braces[last] = i;
braces[i] = last;
}
}
for(int i=0; i<evalvars.length; ++i) variables.put(evalvars[i], evalints[i]);
List<Term> terms = dewIt(0, n-1);
TreeMap<String, Integer> map = new TreeMap<>(new Comparator<>() {
public int compare(String a, String b) {
int ca = countStars(a), cb = countStars(b);
if(ca != cb) return cb - ca;
else return a.compareTo(b);
}
private int countStars(String s) {
int ans = 0;
for(char c: s.toCharArray()) if(c == '*') ++ans;
return ans;
}
});
for(Term term: terms) {
if(term.coeff != 0) {
String key = term.getKey();
if(map.containsKey(key)) {
int oldCoeff = map.get(key);
if(oldCoeff == -term.coeff) map.remove(key);
else map.put(key, oldCoeff + term.coeff);
} else map.put(key, term.coeff);
}
}
for(String k: map.keySet()) ans.add(map.get(k) + "" + k);
return ans;
}
private List<Term> dewIt(int a, int b) {
if(braces[a] == b) return dewIt(a+1, b-1);
for(int i=a; i<=b; ) {
int j = i;
List<Term> curr = null;
if(arr[i] == '(') {
j = braces[i] + 1;
curr = dewIt(i+1, j-2);
}
else {
while(j <= b && arr[j] != ' ') ++j;
String exp = s.substring(i, j);
int val = 1;
if(variables.containsKey(exp)) val = variables.get(exp);
else if (exp.charAt(0) <= '9') val = Integer.valueOf(exp);
}
buffer = multiply(buffer, curr);
if(j > b || arr[j+1] == '+' || arr[j+1] == '-') {
}
if(j < b) {
++j;
j += 2;
}
i = j;
}
return ans;
}
private List<Term> multiply(List<Term> a, List<Term> b) {
for(Term x: a) for(Term y: b) {
Term prod = x.clone();
prod.multiply(y);
}
return ans;
}
}
class Term {
int coeff;
List<String> vars;

public Term(int a, List<String> c) {
this.coeff = a;
}
public String getKey() {
StringBuilder b = new StringBuilder();
Collections.sort(vars);
for(String x: vars) {
b.append('*');
b.append(x);
}
return b.toString();
}
public void multiply(Term that) {
this.coeff *= that.coeff;
if(this.coeff == 0) vars.clear();
}
public Term clone() {
return new Term(coeff, vars);
}
}```Code language: PHP (php)```

## Basic Calculator IV LeetCode Solution JavaScript

``````class Unit {
constructor(count, val = []) {
this.count = count
this.val = val
this.val.sort()
this.valStr = this.val.join('*')
}
product(u1 = new Unit) {
return new Unit(this.count * u1.count, this.val.concat(u1.val))
}
toString() {
return this.count + (this.valStr ? '*' + this.valStr : '')
}
}
class Exp {

constructor(children = []) {
this.children = []
this.dic = new Map()
for (let child of children) this.addUnit(child)
}

let { dic, children } = this
if (dic.has(u.valStr))
children[dic.get(u.valStr)].count += u.count
else {
dic.set(u.valStr, children.length)
children.push(u)
}
}

product(exp1 = new Exp) {
let exp = new Exp
this.children.forEach(x => {
exp1.children.forEach(y => {
})
})
return exp
}

return this
}

toArray() {
let group = new Map()
for (let x of this.children) {
if (!group.has(x.val.length)) group.set(x.val.length, [])
x.count != 0 && group.get(x.val.length).push(x)
}
let groupArr = [...group.entries()].sort((a, b) => b[0] - a[0])
return groupArr.map(([_, x]) => {
x.sort((a, b) => a.valStr.localeCompare(b.valStr))
return x.map(x => x.toString())
}).flat()
}
}

var basicCalculatorIV = function (expression, evalvars, evalints) {
let tokens = toTokens(expression)
.map(x => /^\d+\$/.test(x) ? Number.parseInt(x) : x),
vals = toVarDic(evalvars, evalints), i = 0

let exp = nextExp()
return exp.toArray()

function nextExp() {
if (i >= tokens.length) return undefined
let [exp, op1] = [nextUnit(), nextOperator()]
if (!op1 || op1 == ')') return exp
let exp1 = nextUnit()
while (true) {
let op2 = nextOperator()
if (!op2 || op2 == ')') return dealDefault()
let exp2 = nextUnit()
switch (op2) {
case '+':
case '-':
exp = dealDefault()
op1 = op2
exp1 = exp2
continue
case '*':
exp1 = exp1.product(exp2)
continue
default:
return dealDefault()
}
function dealDefault() {
switch (op1) {
case '+':
case '-':
let expz = exp1.product(new Exp([new Unit(-1, [])]))
case '*':
return exp.product(exp1)
}
debugger
}
}
}

function nextUnit() {
if (i >= tokens.length) return undefined
let rtn
if (tokens[i] == '(') {
i++
return nextExp()
}
else if (Number.isInteger(tokens[i])) rtn = new Exp([
new Unit(tokens[i], [])
])
else if (vals.has(tokens[i])) rtn = new Exp([
new Unit(vals.get(tokens[i]), [])
])
else rtn = new Exp([
new Unit(1, [tokens[i]])
])
i++
return rtn
}
function nextOperator() {
if (i >= tokens.length) return undefined
let op = tokens[i]
i++
return op
}
};

function toTokens(str = '') {
let tokens = [], tmp = '', keyWords = {
'(': true,
')': true,
'+': true,
'*': true,
}
for (let char of str) {
if (keyWords[char]) {
if (tmp) tokens.push(tmp)
tokens.push(char)
tmp = ''
continue
}
if (char == ' ') {
if (tmp) {
tokens.push(tmp)
tmp = ''
}
continue
}
tmp += char
}
if (tmp) tokens.push(tmp)
}

function toVarDic(keys = [], vals = []) {
let dic = new Map
for (let i = 0; i < keys.length; i++) {
dic.set(keys[i], vals[i])
}
return dic
}```Code language: JavaScript (javascript)```

## Basic Calculator IV LeetCode Solution Python

``````from collections import deque, defaultdict

class Polynomial:
def __init__(self, variables):
self.variables = variables

new_variables = defaultdict(int)
for name in self.variables:
new_variables[name] += self.variables[name]
for name in other.variables:
new_variables[name] += other.variables[name]
return Polynomial(new_variables)

def __sub__(self, other):
new_variables = defaultdict(int)
for name in self.variables:
new_variables[name] += self.variables[name]
for name in other.variables:
new_variables[name] -= other.variables[name]
return Polynomial(new_variables)

def __mul__(self, other):
new_variables = defaultdict(int)
for name in sorted(self.variables):
for _name in sorted(other.variables):
new_name = '*'.join(part for part in sorted(name.split('*') + _name.split('*')) if part)
new_variables[new_name] += self.variables[name] * other.variables[_name]
return Polynomial(new_variables)

def to_list(self):
result = []
for name in sorted(self.variables, key=lambda x: (-len(x.split('*')), x)):
if self.variables[name] != 0 and name != '':
result.append(str(self.variables[name]) + '*' + name)
if '' in self.variables and self.variables[''] != 0:
result.append(str(self.variables['']))
return result

class Parser:
def __init__(self, variables):
self.variables = dict(variables)
self.tokens = deque()

def evaluate(self, expression):
self.tokens = deque([c for c in expression if c != ' '])

def _match(self, predicate):
return self.tokens and predicate(self.tokens[0])

def _consume(self, predicate):
if self._match(predicate):
return self.tokens.popleft()
else:
raise ValueError('Unexpected token')

value = self._mult()
while self._match(lambda x: x in {'+', '-'}):
op = (lambda x, y: x+y) if self._consume(lambda x: True) == '+' else (lambda x, y: x-y)
_value = self._mult()
value = op(value, _value)
return value

def _mult(self):
value = self._value()
while self._match(lambda x: x in {'*'}):
self._consume(lambda x: True)
value = value * self._value()
return value

def _value(self):
if self._match(lambda x: x.isalpha()):
buffer = []
while self._match(lambda x: x.isalpha()):
buffer.append(self._consume(lambda x: True))
name = ''.join(buffer)
if name in self.variables:
return Polynomial({'': self.variables[name]})
else:
return Polynomial({name: 1})
elif self._match(lambda x: x.isdigit()):
buffer = [self._consume(lambda x: x.isdigit())]
while self._match(lambda x: x.isdigit()):
buffer.append(self._consume(lambda x: True))
return Polynomial({'': int(''.join(buffer))})
else:
self._consume(lambda x: x == '(')