439. Ternary Expression Parser

• Difficulty: Medium

• Topics: Stack, Depth-first Search

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Problem:

Given a string representing arbitrarily nested ternary expressions, calculate the result of the expression. You can always assume that the given expression is valid and only consists of digits 0-9, ?, :, T and F (T and F represent True and False respectively).

Note:

1. The length of the given string is ≤ 10000.
2. Each number will contain only one digit.
3. The conditional expressions group right-to-left (as usual in most languages).
4. The condition will always be either T or F. That is, the condition will never be a digit.
5. The result of the expression will always evaluate to either a digit 0-9, T or F.

Example 1:

Input: "T?2:3"

Output: "2"

Explanation: If true, then result is 2; otherwise result is 3.

Example 2:

Input: "F?1:T?4:5"

Output: "4"

Explanation: The conditional expressions group right-to-left. Using parenthesis, it is read/evaluated as:

             "(F ? 1 : (T ? 4 : 5))"                   "(F ? 1 : (T ? 4 : 5))"
          -> "(F ? 1 : 4)"                 or       -> "(T ? 4 : 5)"
          -> "4"                                    -> "4"

Example 3:

Input: "T?T?F:5:3"

Output: "F"

Explanation: The conditional expressions group right-to-left. Using parenthesis, it is read/evaluated as:

         "(T ? (T ? F : 5) : 3)"                   "(T ? (T ? F : 5) : 3)"
      -> "(T ? F : 3)"                 or       -> "(T ? F : 5)"
      -> "F"                                    -> "F"

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Solutions:

class Solution {
public:
    string parseTernary(string expression) {
        if (expression.length() == 0)   return "";
        int pos = 0;
        return helper(expression, pos);
    }

    string helper(string& expression, int& pos) {
        char c = expression[pos];
        if ((c == 'T' || c == 'F') && (pos + 1 < expression.length()) && (expression[pos + 1] == '?')) {
            ++pos; // remove 'T' or 'F'
            ++pos; // remove '?'
            string left = helper(expression, pos);
            ++pos; // remove ':'
            string right = helper(expression, pos);
            return c == 'T' ? left : right;
        }

        ++pos; // remove base case 
        return string(1, c);
    }
};