Formatted question description: https://leetcode.ca/all/736.html

# 736. Parse Lisp Expression

## Level

Hard

## Description

You are given a string `expression`

representing a Lisp-like expression to return the integer value of.

The syntax for these expressions is given as follows.

- An expression is either an integer, a let-expression, an add-expression, a mult-expression, or an assigned variable. Expressions always evaluate to a single integer.
- (An integer could be positive or negative.)
- A let-expression takes the form
`(let v1 e1 v2 e2 ... vn en expr)`

, where`let`

is always the string`"let"`

, then there are 1 or more pairs of alternating variables and expressions, meaning that the first variable`v1`

is assigned the value of the expression`e1`

, the second variable`v2`

is assigned the value of the expression`e2`

, and so on**sequentially**; and then the value of this let-expression is the value of the expression`expr`

. - An add-expression takes the form
`(add e1 e2)`

where`add`

is always the string`"add"`

, there are always two expressions`e1, e2`

, and this expression evaluates to the addition of the evaluation of`e1`

and the evaluation of`e2`

. - A mult-expression takes the form
`(mult e1 e2)`

where`mult`

is always the string`"mult"`

, there are always two expressions`e1, e2`

, and this expression evaluates to the multiplication of the evaluation of`e1`

and the evaluation of`e2`

. - For the purposes of this question, we will use a smaller subset of variable names. A variable starts with a lowercase letter, then zero or more lowercase letters or digits. Additionally for your convenience, the names “add”, “let”, or “mult” are protected and will never be used as variable names.
- Finally, there is the concept of scope. When an expression of a variable name is evaluated,
**within the context of that evaluation**, the innermost scope (in terms of parentheses) is checked first for the value of that variable, and then outer scopes are checked sequentially. It is guaranteed that every expression is legal. Please see the examples for more details on scope.

**Evaluation Examples:**

```
Input: (add 1 2)
Output: 3
Input: (mult 3 (add 2 3))
Output: 15
Input: (let x 2 (mult x 5))
Output: 10
Input: (let x 2 (mult x (let x 3 y 4 (add x y))))
Output: 14
Explanation: In the expression (add x y), when checking for the value of the variable x,
we check from the innermost scope to the outermost in the context of the variable we are trying to evaluate.
Since x = 3 is found first, the value of x is 3.
Input: (let x 3 x 2 x)
Output: 2
Explanation: Assignment in let statements is processed sequentially.
Input: (let x 1 y 2 x (add x y) (add x y))
Output: 5
Explanation: The first (add x y) evaluates as 3, and is assigned to x.
The second (add x y) evaluates as 3+2 = 5.
Input: (let x 2 (add (let x 3 (let x 4 x)) x))
Output: 6
Explanation: Even though (let x 4 x) has a deeper scope, it is outside the context
of the final x in the add-expression. That final x will equal 2.
Input: (let a1 3 b2 (add a1 1) b2)
Output 4
Explanation: Variable names can contain digits after the first character.
```

**Note:**

- The given string
`expression`

is well formatted: There are no leading or trailing spaces, there is only a single space separating different components of the string, and no space between adjacent parentheses. The expression is guaranteed to be legal and evaluate to an integer. - The length of
`expression`

is at most 2000. (It is also non-empty, as that would not be a legal expression.) - The answer and all intermediate calculations of that answer are guaranteed to fit in a 32-bit integer.

## Solution

This problem can be solved recursively, or iteratively using stacks.

For each character `)`

, insert a space before the character, so that the terms can be split.

Create two stacks that stores expressions and the maps that map each variable to the corresponding value. Initially, add an empty element to each stack.

Split the expression into terms using spaces, and loop over the terms from the left to right.

If a term starting with `(`

is met, then a new expression is met, so push a new expression into the into the first stack. If the new expression is a let-expression, then add a copy of the map at the top of the second stack, and push the copy of the map into the second stack.

If a `)`

is met, then an expression ends, so pop the last expression from the first stack. Also pop the second last expression from the first stack. If the last expression is a let-expression, then pop one element from the second stack as well. After the last expression’s type is determined, do corresponding operations and add one element to the second last expression. If the second last expression is a let-expression, then update the second last map in the second stack as well. Finally, after the second last expression is modified, push it into the first stack.

If a variable is met, then update the last expression and the last step accordingly.

If a number is met, then if the last expression is a let-expression, update the last map accordingly.

Finally, there should be only one element in the first stack, which is the result of the expression, so pop it and return it as an integer.

```
class Solution {
public int evaluate(String expression) {
StringBuffer sb = new StringBuffer(expression);
for (int i = sb.length() - 1; i >= 0; i--) {
char c = sb.charAt(i);
if (c == ')')
sb.insert(i, ' ');
}
String[] array = sb.toString().split(" ");
Stack<List<String>> expressionStack = new Stack<List<String>>();
expressionStack.push(new ArrayList<String>());
Stack<Map<String, String>> variableValuesStack = new Stack<Map<String, String>>();
variableValuesStack.push(new HashMap<String, String>());
int length = array.length;
for (int i = 0; i < length; i++) {
String str = array[i];
if (str.charAt(0) == '(') {
List<String> list = new ArrayList<String>();
list.add(str.substring(1));
expressionStack.push(list);
if (list.get(0).equals("let")) {
Map<String, String> prevMap = variableValuesStack.peek();
Set<String> keySet = prevMap.keySet();
Map<String, String> map = new HashMap<String, String>();
for (String key : keySet)
map.put(key, prevMap.get(key));
variableValuesStack.push(map);
}
} else if (str.equals(")")) {
List<String> lastExpression = expressionStack.pop();
List<String> secondLastExpression = expressionStack.pop();
String lastExpressionType = lastExpression.get(0);
if (lastExpressionType.equals("let")) {
String lastStr = lastExpression.get(lastExpression.size() - 1);
if (Character.isLetter(lastStr.charAt(0))) {
Map<String, String> lastMap = variableValuesStack.peek();
secondLastExpression.add(lastMap.get(lastStr));
} else
secondLastExpression.add(lastStr);
variableValuesStack.pop();
} else if (lastExpressionType.equals("add")) {
int curResult = Integer.parseInt(lastExpression.get(1)) + Integer.parseInt(lastExpression.get(2));
secondLastExpression.add(String.valueOf(curResult));
} else if (lastExpressionType.equals("mult")) {
int curResult = Integer.parseInt(lastExpression.get(1)) * Integer.parseInt(lastExpression.get(2));
secondLastExpression.add(String.valueOf(curResult));
}
if (secondLastExpression.get(0).equals("let")) {
Map<String, String> lastMap = variableValuesStack.pop();
int size = secondLastExpression.size();
lastMap.put(secondLastExpression.get(size - 2), secondLastExpression.get(size - 1));
variableValuesStack.push(lastMap);
}
expressionStack.push(secondLastExpression);
} else if (Character.isLetter(str.charAt(0))) {
List<String> lastExpression = expressionStack.pop();
Map<String, String> lastMap = variableValuesStack.pop();
if (lastExpression.get(0).equals("let")) {
if (lastExpression.size() % 2 == 1)
lastExpression.add(str);
else {
String value = lastMap.get(str);
lastMap.put(lastExpression.get(lastExpression.size() - 1), value);
lastExpression.add(value);
}
} else
lastExpression.add(lastMap.get(str));
expressionStack.push(lastExpression);
variableValuesStack.push(lastMap);
} else {
List<String> lastExpression = expressionStack.pop();
if (lastExpression.get(0).equals("let")) {
Map<String, String> lastMap = variableValuesStack.pop();
lastMap.put(lastExpression.get(lastExpression.size() - 1), str);
variableValuesStack.push(lastMap);
}
lastExpression.add(str);
expressionStack.push(lastExpression);
}
}
int result = Integer.parseInt(expressionStack.pop().get(0));
return result;
}
}
```