# 439. Ternary Expression Parser

## Description

Given a string expression representing arbitrarily nested ternary expressions, evaluate the expression, and return the result of it.

You can always assume that the given expression is valid and only contains digits, '?', ':', 'T', and 'F' where 'T' is true and 'F' is false. All the numbers in the expression are one-digit numbers (i.e., in the range [0, 9]).

The conditional expressions group right-to-left (as usual in most languages), and the result of the expression will always evaluate to either a digit, 'T' or 'F'.

Example 1:

Input: expression = "T?2:3"
Output: "2"
Explanation: If true, then result is 2; otherwise result is 3.


Example 2:

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


Example 3:

Input: expression = "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 ? F : 3)" --> "F"
"(T ? (T ? F : 5) : 3)" --> "(T ? F : 5)" --> "F"


Constraints:

• 5 <= expression.length <= 104
• expression consists of digits, 'T', 'F', '?', and ':'.
• It is guaranteed that expression is a valid ternary expression and that each number is a one-digit number.

## Solutions

• class Solution {
public String parseTernary(String expression) {
Deque<Character> stk = new ArrayDeque<>();
boolean cond = false;
for (int i = expression.length() - 1; i >= 0; --i) {
char c = expression.charAt(i);
if (c == ':') {
continue;
}
if (c == '?') {
cond = true;
} else {
if (cond) {
if (c == 'T') {
char x = stk.pop();
stk.pop();
stk.push(x);
} else {
stk.pop();
}
cond = false;
} else {
stk.push(c);
}
}
}
return String.valueOf(stk.peek());
}
}

• class Solution {
public:
string parseTernary(string expression) {
string stk;
bool cond = false;
reverse(expression.begin(), expression.end());
for (char& c : expression) {
if (c == ':') {
continue;
}
if (c == '?') {
cond = true;
} else {
if (cond) {
if (c == 'T') {
char x = stk.back();
stk.pop_back();
stk.pop_back();
stk.push_back(x);
} else {
stk.pop_back();
}
cond = false;
} else {
stk.push_back(c);
}
}
}
return {stk[0]};
}
};

• class Solution:
def parseTernary(self, expression: str) -> str:
stk = []
cond = False
for c in expression[::-1]:
if c == ':':
continue
if c == '?':
cond = True
else:
if cond:
if c == 'T':
x = stk.pop()
stk.pop()
stk.append(x)
else:
stk.pop()
cond = False
else:
stk.append(c)
return stk[0]


• func parseTernary(expression string) string {
stk := []byte{}
cond := false
for i := len(expression) - 1; i >= 0; i-- {
c := expression[i]
if c == ':' {
continue
}
if c == '?' {
cond = true
} else {
if cond {
if c == 'T' {
x := stk[len(stk)-1]
stk = stk[:len(stk)-2]
stk = append(stk, x)
} else {
stk = stk[:len(stk)-1]
}
cond = false
} else {
stk = append(stk, c)
}
}
}
return string(stk[0])
}