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You are given equations in the form of a / b = k, where a and b are variables represented as strings, and k is a floating-point number. Some equations are given to you as input, along with some queries, where each query is in the form of c / d. Return the answers to all queries. If such a division cannot be determined, return -1.0.

Example:

equations = [["a", "b"], ["b", "c"]]
values = [2.0, 3.0]
queries = [["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"]]

Output:

[6.0, 0.5, -1.0, 1.0, -1.0]

Clarifying Questions

  1. Are all given variables and equations guaranteed to form valid divisions (i.e., no zero denominators)?
  2. How many variables and equations can we assume at maximum (to understand expected input size and optimize accordingly)?
  3. Are variables case-sensitive?

Strategy

  1. Graph Representation:
    • Use a graph where each variable is a node, and an edge represents the division relationship with the given value.
  2. Building the Graph:
    • For each equation a / b = k, create an edge from a to b with weight k, and an edge from b to a with weight 1/k.
  3. Query Resolution:
    • Use Depth-First Search (DFS) or Breadth-First Search (BFS) to find the path from the numerator to the denominator within the graph and to compute the product of edge weights along this path.
  4. Handling Unreachable Nodes:
    • If there’s no path between the nodes in the graph for a given query, return -1.0.
  5. Edge Cases:
    • Self-division (e.g., ["a", "a"] should return 1.0).
    • Variables not present in any equation should yield -1.0.

Code

from collections import defaultdict, deque

def calcEquation(equations, values, queries):
    # Graph initialization
    graph = defaultdict(dict)
    
    # Building the graph
    for (dividend, divisor), value in zip(equations, values):
        graph[dividend][divisor] = value
        graph[divisor][dividend] = 1 / value
    
    # Function to perform BFS to find the result of a single query
    def bfs(start, end):
        if start not in graph or end not in graph:
            return -1.0
        queue = deque([(start, 1.0)])
        visited = set()
        
        while queue:
            current, product = queue.popleft()
            if current == end:
                return product
            visited.add(current)
            for neighbor in graph[current]:
                if neighbor not in visited:
                    queue.append((neighbor, product * graph[current][neighbor]))
        return -1.0
    
    # Process each query using the BFS function
    results = []
    for dividend, divisor in queries:
        result = bfs(dividend, divisor)
        results.append(result)
    
    return results

# Example usage
equations = [["a", "b"], ["b", "c"]]
values = [2.0, 3.0]
queries = [["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"]]
print(calcEquation(equations, values, queries))  # Output: [6.0, 0.5, -1.0, 1.0, -1.0]

Time Complexity

The overall time complexity for Q queries would thus be O(Q * (V + E)). This should be efficient for typical constraints given by coding interview problems.

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