736. Parse Lisp Expression

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. 

 

Approach #1: String. [Java]

class Solution {
    public int evaluate(String expression) {
        return evaluate(expression, new HashMap<>());
    }
    
    private int evaluate(String e, Map<String, Deque<Integer>> map) {
        char c = e.charAt(1);   // the expression must start with "(add " or "(mult " or "(let ".
        if (c == 'a') return evaluateAdd(e, map);      // "add" expression
        else if (c == 'm') return evaluateMult(e, map);     // "mult" expression
        else if (c == 'l') return evaluateLet(e, map);      // "let" expression
        else return 0;      // illegal expression so return 0
    }
    
    private int evaluateAdd(String e, Map<String, Deque<Integer>> map) {
        int offset = 5;     // the expression starts with "(add ", so offset starts at 5.
        String o1 = getOperand(e, offset);      // first operand
        
        offset += o1.length() + 1;
        String o2 = getOperand(e, offset);         // second operand
        
        return evaluateOperand(o1, map) + evaluateOperand(o2, map);
    }
    
    private int evaluateMult(String e, Map<String, Deque<Integer>> map) {
        int offset = 6;
        String o1 = getOperand(e, offset);
        
        offset += o1.length() + 1;
        String o2 = getOperand(e, offset);
        
        return evaluateOperand(o1, map) * evaluateOperand(o2, map);
    }
    
    private int evaluateLet(String e, Map<String, Deque<Integer>> map) {
        List<String> variables = new ArrayList<>();     // list of variables assigned in this "let" expression
        int res = 0;        // the result of this "let" expression
        int offset = 5;     // the expression starts with "(let ", so offset starts at 5.
        
        while (offset < e.length()) {
            String o1 = getOperand(e, offset);
            offset += o1.length() + 1;
            
            String o2 = getOperand(e, offset);
            
            if (o2 == null) {       // if second operand is null, we reached the last operand
                res = evaluateOperand(o1, map);
                break;
            }
                
            offset += o2.length() + 1;

            variables.add(o1);      // record the variable

            if (!map.containsKey(o1)) map.put(o1, new ArrayDeque<>());

            map.get(o1).offerFirst(evaluateOperand(o2, map));       // do the assignment
        }
        
        // pop out assigned values before returning from this "let" expression
        for (int i = variables.size() - 1; i >= 0; --i) {
            String variable = variables.get(i);
            Deque<Integer> stack = map.get(variable);
            stack.pollFirst();
            if (stack.isEmpty()) map.remove(variable);
        }
        
        return res;
    }
    
    private String getOperand(String e, int offset) {
        if (offset >= e.length()) return null;      // invalid offset
        
        char c = e.charAt(offset);
        int start = offset;
        
        if (c == '-' || Character.isDigit(c)) {     // operand is an integer
            if (c == '-') offset++;
            while (offset < e.length() && Character.isDigit(e.charAt(offset))) offset++;
        } else if (Character.isLowerCase(c)) {      // operand is a variable
            while (offset < e.length() && Character.isLetterOrDigit(e.charAt(offset))) offset++;
        } else {        // operand is another expression enclosed in parenthses
            for (int cnt = 0; offset < e.length(); ) {
                c = e.charAt(offset++);
                if (c == '(') cnt++;
                if (c == ')') cnt--;
                if (cnt == 0) break;
            }
        }
        
        return e.substring(start, offset);
    }
    
    private int evaluateOperand(String e, Map<String, Deque<Integer>> map) {
        char c = e.charAt(0);
        
        if (c == '-' || Character.isDigit(c)) {     // operand is an integer so parse it
            return Integer.parseInt(e);
        } else if (Character.isLowerCase(c)) {      // operand is a variable so look it up
            return map.get(e).peekFirst();
        } else {        // operand is another expression so evaluate it recursively
            return evaluate(e, map);
        }
    }
}

　　

　　

Analysis:

These type of problems are notorious for their complex and intertwined nature. A modularized approach would certainly help to clarify the problem. So here is how I would divide the original problem into difference modules.

 

For an input string expression, we have the function evaluate to resolve its value.

 

The input expression must be one of the following three types of expressions -- add, mult and let. Correspondingly we will have three functions to evaluate each of them --evaluateAdd, evaluateMult and evaluateLet.

 

All the three types of expressions can be thought of as composed by operands, where each operand can be an integer, a variable or another expression of the three types mentioned above (note the expression will be enclosed in parentheses).

 

An add expression contains 2 operands and is evaluated to be the sum of the two operands.

 

A mult expression contains 2 operands and is evaluated to be the product of two operands.

 

A let expression contains 2m + 1 operands and is evaluated to be the value of the last operand. The first m pairs of operands correspond to m assignments. For each pair, the first operand is a variable while the second can be an integer, a variable or another expression. To simulate the assignment operations, we will maintain a HashMap, which maps the variable to the assigned values (this also implies the evaluate function in steep 1 should be delegated to a subroutine with an additional HashMap parameter). To simulate the concept of scope, the assigned values will be placed in a stack. Whenever the let expression returns. all assignments performed within it become invalid and should be popped out of the stack.

 

From the analyses above, given an expression e, we need to identify its constituent operands. We will have two functions serving for this purpose.

getOperand: this function will obtain the string representation of the operand starting from the specified offset into the expression e. It will distingguish the three types of operands -- an integer, a variable or another expression (of type add, mult or let).

 

evaluateOperand: This function will evaluate the operand string obtained above. For an operand of integer type, it will look up its value in the hash map; for an operand of expression type, it will recurively call the evaluate function to resolve its value.

 

Reference:

https://leetcode.com/problems/parse-lisp-expression/discuss/109718/Java-modularized-solution