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Evaluate Division
Sar Champagne Bielert edited this page Apr 8, 2024
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- 🔗 Leetcode Link: https://leetcode.com/problems/evaluate-division
- 💡 Problem Difficulty: Medium
- ⏰ Time to complete: __ mins
- 🛠️ Topics: Graphs, Breadth-First Search, Depth-First Search
- 🗒️ Similar Questions: Check for Contradictions in Equations
Understand what the interviewer is asking for by using test cases and questions about the problem.
- Established a set (2-3) of test cases to verify their own solution later.
- Established a set (1-2) of edge cases to verify their solution handles complexities.
- Have fully understood the problem and have no clarifying questions.
- Have you verified any Time/Space Constraints for this problem?
-
How many equations can we have?
- We can have up to 20 questions.
-
Can there be no questions?
- No, there must be a minimum of one question.
-
How long can variable names be?
- There can be up to 5 characters in a variable name.
-
Can variable names be empty?
- No, variable names must have at least one alphabetical character
-
How many queries are performed?
- There can be up to 20 queries.
HAPPY CASE
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]
EDGE CASE
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]Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.
For graph problems, some things we want to consider are:
- BFS: We can use BFS to traverse the equation graph, and accumulate the products of the edges to find the answer.
- DFS: We can use DFS to traverse the equation graph, and accumulate the products of the edges to find the answer, while ignoring invalid paths.
- Adjacency List: We can use an adjacency list to store the graph, especially when the graph is sparse.
- Adjacency Matrix: We can use an adjacency matrix to store the graph, but a sparse graph will cause an unneeded worst-case runtime.
- Topological Sort: We can use topological sort when a directed graph is used and returns an array of the nodes where each node appears before all the nodes it points to. In order to have a topological sorting, the graph must not contain any cycles.
- Map: We can use a map to store the edges in the graph to lookup equations by name, and store all neighbors by name.
- Union Find: Are there find and union operations here? Can you perform a find operation where you can determine which subset a particular element is in? This can be used for determining if two elements are in the same subset. Can you perform a union operation where you join two subsets into a single subset? Can you check if the two subsets belong to same set? If no, then we cannot perform union.
- BONUS – We can use floyd warshall to perform all node minimum distance computation, allowing O(1) query.
Plan the solution with appropriate visualizations and pseudocode.
General Idea: We will build a graph of the equation where the nodes are variables, edges are the equations, and edge weights are the equation values. Then, we will perform a topological sort to return the weight contribution of the children to the parent node.
- Create a map as our graph.
- For each equation, a) Add edge from A to B with V[i] b) Add edge from B to A with 1/V[i]
- For each query a) Perform topological sort from start node i) Keep track of product weights for each neighbor, computed by neighbor * topoSort(neighbor) ii) If all neighbor product weights were -1, return -1 iii) Return tracked product weights b) Add result of topological sort call from start to end variable to list
- Return result list
- Make sure you really clarify all inputs. When treating this problem as a graph problem, they may assume information that can significantly affect runtime complexity. For instance, assuming the lengths of the variable names or number of equations greatly affects which algorithms can be used and how they are used. In this case, students should cover most, if not all, clarifications to develop an efficient algorithm.
Implement the code to solve the algorithm.
public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
// build graph
Map<String, Map<String, Double>> graph = buildGraph(equations, values);
double[] result = new double[queries.size()];
for (int i = 0; i < queries.size(); i++) {
result[i] = getPathWeight(queries.get(i).get(0), queries.get(i).get(1), new HashSet<>(), graph);
}
return result;
}
private double getPathWeight(String start, String end, Set<String> visited, Map<String, Map<String, Double>> graph) {
// rejection case
if (!graph.containsKey(start))
return -1.0;
// accepting case
if (graph.get(start).containsKey(end))
return graph.get(start).get(end);
visited.add(start);
for (Map.Entry<String, Double> neighbour : graph.get(start).entrySet()) {
if (!visited.contains(neighbour.getKey())) {
double productWeight = getPathWeight(neighbour.getKey(), end, visited, graph);
if (productWeight != -1.0)
return neighbour.getValue() * productWeight;
}
}
return -1.0;
}
private Map<String, Map<String, Double>> buildGraph(List<List<String>> equations, double[] values) {
Map<String, Map<String, Double>> graph = new HashMap<>();
String u, v;
for (int i = 0; i < equations.size(); i++) {
u = equations.get(i).get(0);
v = equations.get(i).get(1);
graph.putIfAbsent(u, new HashMap<>());
graph.get(u).put(v, values[i]);
graph.putIfAbsent(v, new HashMap<>());
graph.get(v).put(u, 1 / values[i]);
}class Solution:
def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
graph = defaultdict(defaultdict)
def backtrack_evaluate(curr_node, target_node, acc_product, visited):
visited.add(curr_node)
ret = -1.0
neighbors = graph[curr_node]
if target_node in neighbors:
ret = acc_product * neighbors[target_node]
else:
for neighbor, value in neighbors.items():
if neighbor in visited:
continue
ret = backtrack_evaluate(
neighbor, target_node, acc_product * value, visited)
if ret != -1.0:
break
visited.remove(curr_node)
return ret
# build the graph from the equations
for (dividend, divisor), value in zip(equations, values):
# add nodes and two edges into the graph
graph[dividend][divisor] = value
graph[divisor][dividend] = 1 / value
# evaluate each query via backtracking (DFS)
# by verifying if there exists a path from dividend to divisor
results = []
for dividend, divisor in queries:
if dividend not in graph or divisor not in graph:
# case 1): either node does not exist
ret = -1.0
elif dividend == divisor:
# case 2): origin and destination are the same node
ret = 1.0
else:
visited = set()
ret = backtrack_evaluate(dividend, divisor, 1, visited)
results.append(ret)
return resultsReview the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
- Trace through your code with an input to check for the expected output
- Catch possible edge cases and off-by-one errors and verify the code works for the happy and edge cases you created in the “Understand” section
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
Let N be the number of input equations and M be the number of queries.
Time Complexity: O(N*M)
Space Complexity: O(N)