LeetCode 399. Evaluate Division

Description

https://leetcode.com/problems/most-common-word/

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.

Explanation

dfs

Python Solution

class Solution:
    def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
        graph = collections.defaultdict(list)
        
        for equation, value  in zip(equations, values):
            numerator, denominator = equation[0], equation[1]
            graph[numerator].append((denominator, value))
            graph[denominator].append((numerator, 1 / value))
            
        result = []
        
        for numerator, denominator in queries:
            visited = set()
            value = self.helper(graph, numerator, denominator, visited)
            result.append(value)
            
        return result
        
        
    def helper(self, graph, numerator, denominator, visited):
        if numerator not in graph:
            return -1.0
        if numerator == denominator:
            return 1.0
        
        visited.add(numerator)
        
        for neighbor, value in graph[numerator]:
            if neighbor == denominator:
                return value
            if neighbor not in visited:
                quotient = self.helper(graph, neighbor, denominator, visited)
                
                if quotient != -1:
                    return value * quotient
                
        return -1.0
        

• Time Complexity: ~NM
• Space Complexity: ~N

Let N be the number of input equations and M be the number of queries.