399. Evaluate Division

Medium (Trung bình) Python 🔗 Xem trên LeetCode

📋 Đề Bài

You are given an array of variable pairs equations and an array of real numbers values, where equations[i] = [Ai, Bi] and values[i] represent the equation Ai / Bi = values[i]. Each Ai or Bi is a string that represents a single variable.

You are also given some queries, where queries[j] = [Cj, Dj] represents the jth query where you must find the answer for Cj / Dj = ?.

Return the answers to all queries. If a single answer cannot be determined, return -1.0.

Note: The input is always valid. You may assume that evaluating the queries will not result in division by zero and that there is no contradiction.

 

Example 1:

Input: equations = [["a","b"],["b","c"]], values = [2.0,3.0], queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]
Output: [6.00000,0.50000,-1.00000,1.00000,-1.00000]
Explanation: 
Given: a / b = 2.0, b / c = 3.0
queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ?
return: [6.0, 0.5, -1.0, 1.0, -1.0 ]

Example 2:

Input: equations = [["a","b"],["b","c"],["bc","cd"]], values = [1.5,2.5,5.0], queries = [["a","c"],["c","b"],["bc","cd"],["cd","bc"]]
Output: [3.75000,0.40000,5.00000,0.20000]

Example 3:

Input: equations = [["a","b"]], values = [0.5], queries = [["a","b"],["b","a"],["a","c"],["x","y"]]
Output: [0.50000,2.00000,-1.00000,-1.00000]

 

Constraints:

  • 1 <= equations.length <= 20
  • equations[i].length == 2
  • 1 <= Ai.length, Bi.length <= 5
  • values.length == equations.length
  • 0.0 < values[i] <= 20.0
  • 1 <= queries.length <= 20
  • queries[i].length == 2
  • 1 <= Cj.length, Dj.length <= 5
  • Ai, Bi, Cj, Dj consist of lower case English letters and digits.

🧠 Thuật Toán & Kỹ Thuật

DFS (Tìm kiếm theo chiều sâu)Hash Table (Bảng băm)Graph (Đồ thị)
⏱️ Thời gian O(V+E)
💾 Không gian O(V)

💻 Lời Giải

Python 0399-evaluate-division.py 
class Solution:
    def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
        adj = defaultdict(list)
        n = len(equations)
        containV = set()
        
        for i in range(n):
            u = equations[i][0]
            v = equations[i][1]
            adj[u].append((v, values[i]))
            adj[v].append((u, 1 / values[i]))
            containV.add(u)
            containV.add(v)
            
        m = len(queries)
        ans = [-1 for _ in range(m)]
        self.visited = set()
        self.check = False
        self.total = 0
        
        def dfs(start, end, s):
            if start not in containV or end not in containV:
                return
            if start == end:
                self.check = True
                self.total = s
                return
            self.visited.add(start)
            for v, c in adj[start]:
                if v not in self.visited:
                    dfs(v, end, s * c)
        
        for i in range(m):
            start = queries[i][0]
            end = queries[i][1]
            dfs(start, end, 1)
            if self.check:
                ans[i] = self.total
            self.check = False
            self.total = 0
            self.visited = set()
            
        return ans