Monday, October 24, 2016

LeetCode 439 - Ternary Expression Parser


Related: Pocket Gems - Ternary Expression to Binary Tree
http://bookshadow.com/weblog/2016/10/23/leetcode-ternary-expression-parser/
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 0-9T 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"
X.
https://segmentfault.com/a/1190000008939830
从右往左看,? 作为决定最小单位的符号,每次遇到一个?, 就拆解离?最近的两个小单位。宏观上看是,从小到大。
    public String parseTernary(String expression) {
        if(expression == null || expression.length() == 0) return "";
        Deque<Character> stk = new LinkedList<>();
        
        for(int i=expression.length()-1; i >= 0; i--){
            char c = expression.charAt(i);
            if(!stk.isEmpty() && stk.peek() == '?'){
                stk.pop();  // pop ?
                char first = stk.pop();
                stk.pop(); // pop :
                char second = stk.pop();
                
                if(c == 'T') stk.push(first);
                else stk.push(second);
            } else {
                stk.push(c);
            }
        }
        
        return String.valueOf(stk.peek());
    }
https://discuss.leetcode.com/topic/64409/very-easy-1-pass-stack-solution-in-java-no-string-concat
Iterate the expression from tail, whenever encounter a character before '?', calculate the right value and push back to stack.
P.S. this code is guaranteed only if "the given expression is valid" base on the requirement.
public String parseTernary(String expression) {
    if (expression == null || expression.length() == 0) return "";
    Deque<Character> stack = new LinkedList<>();

    for (int i = expression.length() - 1; i >= 0; i--) {
        char c = expression.charAt(i);
        if (!stack.isEmpty() && stack.peek() == '?') {

            stack.pop(); //pop '?'
            char first = stack.pop();
            stack.pop(); //pop ':'
            char second = stack.pop();

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

    return String.valueOf(stack.peek());
}
https://discuss.leetcode.com/topic/64512/java-short-clean-code-no-stack

https://discuss.leetcode.com/topic/64524/short-python-solutions-one-o-n
栈(Stack)数据结构
循环直到栈中元素为1并且表达式非空:
取栈顶的5个元素,判断是否为一个可以解析的表达式。若是,则解析后压栈
否则从右向左将expression中的字符压入栈stack
Collect chars from back to front on a stack, evaluate ternary sub-expressions as soon as possible:
def parseTernary(self, expression):
    stack = []
    for c in reversed(expression):
        stack.append(c)
        if stack[-2:-1] == ['?']:
            stack[-5:] = stack[-3 if stack[-1] == 'T' else -5]
    return stack[0]
Originally my check was stack[-4::2] == [':', '?'], but @YJL1228's is right, looking for ? is enough.
https://discuss.leetcode.com/topic/64389/easy-and-concise-5-lines-python-java-solution
The time complexity seems very high(may be O(n^2)). Using lastIndexOf means that u need to scan the whole String each loop, which costs a lot of time.
    public String parseTernary(String expression) {
        while (expression.length() != 1) {
            int i = expression.lastIndexOf("?");    // get the last shown '?'
            char tmp;
            if (expression.charAt(i-1) == 'T') { tmp = expression.charAt(i+1); }
            else { tmp = expression.charAt(i+3); }
            expression = expression.substring(0, i-1) + tmp + expression.substring(i+4);
        }
        return expression;
    }

X. Recursion

? : 所组成的最小单位,可以看作一对括号。 ?类似(, :类似 )。
从左往右看,:作为决定一组完整最小单位的符号。每次找到一对就可以按:分为左右两个子问题递归解决。
宏观上看是从大到小拆开,从小到大递归回去。
https://discuss.leetcode.com/topic/68336/5ms-java-dfs-solution
https://yijiajin.github.io/leetcode/2017/01/11/LC439.-Ternary-Expression-Parser/
The key idea is to find the legitimate : separating two "values".
This : has the property that the number of ? ahead of this : should equals the number of :, similar to the case of ( and ).
Only problem is the time complexity is up to O(n^2) in the worse case.
For example, T?T?T?T?T?1:2:3:4:5:6.
    public String parseTernary(String expression) {
        if(expression == null || expression.length() == 0){
            return expression;
        }
        char[] exp = expression.toCharArray();
        
        return DFS(exp, 0, exp.length - 1) + "";
        
    }
    public char DFS(char[] c, int start, int end){
        if(start == end){
            return c[start];
        }
        int count = 0, i =start;
        for(; i <= end; i++){
            if(c[i] == '?'){
                count ++;
            }else if (c[i] == ':'){
                count --;
                if(count == 0){
                    break;
                }
            }
        }
        return c[start] == 'T'? DFS(c, start + 2, i - 1) : DFS(c, i+1,end);
    }
https://discuss.leetcode.com/topic/64389/easy-and-concise-5-lines-python-java-solution
To avoid use lastIndexOf, we can easily scan through whole string at first. I've added modified version below, however it costs extra 3 lines.
    def parseTernary(self, expression):
        # begin with the last question.
        idx = []
        for i in xrange(len(expression)):
            if expression[i] == "?": idx += i,
        while len(expression) != 1:
            j = idx.pop()
            tmp = expression[j+1] if expression[j-1] == 'T' else expression[j+3]
            expression = expression[:j-1] + tmp + expression[j+4:]

        return expression
http://hongzheng.me/leetcode/leetcode-439-ternary-expression-parser/
一个公式由3部分组成 <bool>?<left>:<right>, bool值只能为T或F, left或right可以为子公式或是单一的值. 所以只要能够正确的划分这三个部分, 就能用递归来处理.
这里我们发现?:是成对出现的, 用一个flag来计数, 遇到?加1, 遇到:减一, 当flag == 0的时候, 说明正好为left和right的分界.
http://www.voidcn.com/blog/zqh_1991/article/p-6244665.html
    string parseTernary(string expression) {
        int len=expression.size();
        if(len==1) return expression;
        int count1=0,count2=0;
        for(int i=0;i<len;i++) {
            if(expression[i]=='?') {
                count1++;
            }
            else if(expression[i]==':') {
                count2++;
                if(count1==count2) {
                    if(expression[0]=='T') {
                        return parseTernary(expression.substr(2,i-2));  //表达式的左边
                    }
                    else return parseTernary(expression.substr(i+1));   //表达式的右边
                }
            }
        }
        return "";
    }

X. Build express tree
https://scottduan.gitbooks.io/leetcode-review/ternary_expression_parser.html
Let's assume the input ternary expression is valid, we can build the tree in preorder manner.
Each time we scan two characters, the first character is either ? or :, the second character holds the value of the tree node. When we see ?, we add the new node to left. When we see :, we need to find out the ancestor node that doesn't have a right node, and make the new node as its right child.
public TreeNode convert(char[] expr) {
  if (expr.length == 0) {
    return null;
  }

  TreeNode root = new TreeNode(expr[0]);

  Stack<TreeNode> stack = new Stack<>();

  for (int i = 1; i < expr.length; i += 2) {
    TreeNode node = new TreeNode(expr[i + 1]);

    if (expr[i] == '?') {
      stack.peek().left = node;
    }

    if (expr[i] == ':') {
      stack.pop();
      while (stack.peek().right != null) {
        stack.pop();
      }
      stack.peek().right = node;
    }

    stack.push(node);
  }

  return root;
}
https://discuss.leetcode.com/topic/64397/java-o-n-using-binary-tree
    public String parseTernary(String expression) {
        if(expression == null || expression.length() == 0) return "";
        Node root = buildTree(expression.toCharArray());
        return evaluate(root) + "";
    }
    static class Node {
        char val;
        Node left;
        Node right;
        Node parent;
        
        public Node(char c) {
            val = c;
            left = null;
            right = null;
            parent = null;
        }
    }
    private static Node buildTree(char[] ch) {
        Node root = new Node(ch[0]);
        Node node = root;
        for(int i = 1; i < ch.length; i++) {
            if(ch[i] == '?') {
                Node left = new Node(ch[i + 1]);
                node.left = left;
                left.parent = node;
                node = node.left;
            }
            else if(ch[i] == ':') {
                node = node.parent;
                while(node.right != null && node.parent != null) {
                    node = node.parent;
                }
                Node right = new Node(ch[i + 1]);
                node.right = right;
                right.parent = node;
                node = node.right;
            }
        }
        return root;
    }
    private static char evaluate(Node root) {
        while(root.val == 'T' || root.val == 'F') {
            if(root.left == null && root.right == null) break;
            if(root.val == 'T') root = root.left;
            else root = root.right;
        }
        return root.val;
    }

TODO http://www.cnblogs.com/grandyang/p/6022498.html
Only problem is the time complexity is up to O(n^2) in the worse case.
For example, T?T?T?T?T?1:2:3:4:5:6.


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