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## Problem

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"


## Code

class Solution {
public String parseTernary(String expression) {
Stack<Character> stack = new Stack<>();

for(int i = expression.length() - 1; i >= 0; i--){
char c = expression.charAt(i);

if(!stack.isEmpty() && stack.peek() == '?'){
stack.pop(); // ?
char first = stack.pop();

stack.pop(); // :
char second = stack.pop();

if(c == 'T'){
stack.push(first);
} else {
stack.push(second);
}
} else {
stack.push(c);
}
}

return String.valueOf(stack.peek());
}
}