LeetCode 439: Ternary Expression Parser Solution

Master LeetCode problem 439 (Ternary Expression Parser), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

439. Ternary Expression Parser

Medium

String Stack Recursion

Problem Explanation

Explanation:

This problem can be solved using a recursive approach. We start by iterating through the expression from right to left. Whenever we encounter a '?', we know that we have a ternary expression. We evaluate the condition before the '?' to decide which expression to choose. If the condition is true, we recursively evaluate the expression before the ':', otherwise we recursively evaluate the expression after the ':'. We continue this process until we reach the end of the expression.

Algorithm:

Start iterating the expression from right to left.
If we encounter a '?', recursively evaluate the expression before the '?' if the condition is true, otherwise recursively evaluate the expression after the ':'.
Keep track of the current index while iterating.
Return the final result.

Time Complexity: O(n), where n is the length of the expression. Space Complexity: O(n), for the recursive call stack.

: :

Solution Code

class Solution {
    public String parseTernary(String expression) {
        if (expression == null || expression.length() == 0) {
            return "";
        }
        
        return evaluate(expression, 0, expression.length() - 1);
    }
    
    private String evaluate(String expression, int start, int end) {
        if (start == end) {
            return expression.substring(start, end + 1);
        }
        
        int count = 0;
        int i = end;
        
        while (i >= start) {
            if (expression.charAt(i) == '?') {
                count++;
            } else if (expression.charAt(i) == ':') {
                count--;
                if (count == 0) {
                    break;
                }
            }
            i--;
        }
        
        return expression.charAt(start) == 'T' ? evaluate(expression, start + 2, i - 1) : evaluate(expression, i + 1, end);
    }
}

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