399. Evaluate Division

399. Evaluate Division

Tag: Depth-First Search Union-Find

Difficulty: Medium

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

You are also given some queries, where queries[j] = [C_j, D_j] represents the j^th query where you must find the answer for C_j / D_j = ?.

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.

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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.

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Depth-First Search

Time Complexity : O(M⋅N), traverse through the graph of size while traversing through the queries

Space Complexity: O(N), extra memory space to build graph

class Solution:
    def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
        graph = collections.defaultdict(set)
        for (u, v), val in zip(equations, values):
            graph[u].add((v, val))
            graph[v].add((u, 1/val))
            
        def dfs(x, y, visited):
            if not x in graph:
                return -1.0
            elif x == y:
                return 1.0
            visited.add(x)
            for key, val in graph[x]:
                if key == y:
                    return val
                elif not key in visited:
                    output = dfs(key, y, visited)
                    if output != -1:
                        return val * output
            return -1.0
            
        res = list()
        for c, d in queries:
            res.append(dfs(c, d, set()))
        return res

class Solution:
    def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
        # Depth-First Search
        visited = set()
        graph = collections.defaultdict(collections.defaultdict)
        # Build the graph out of the list of input equations
        for (dividend, divisor), value in zip(equations, values):
            # Each equation corresponds to two edges in the graph
            graph[dividend][divisor] = value
            graph[divisor][dividend] = 1 / value

        def dfs(curr_node, target_node, acc):
            visited.add(curr_node)
            ret = -1.0
            neighbors = graph[curr_node]
            if target_node in neighbors:
                ret = acc * neighbors[target_node]
            else:
                for neighbor, value in neighbors.items():
                    if neighbor in visited:
                        continue
                    ret = dfs(neighbor, target_node, acc * value)
                    if ret != -1.0:
                        break
            visited.remove(curr_node)
            return ret

        res = list()
        for dividend, divisor in queries:
            # Check if either of the nodes does not exist in the graph
            if not dividend in graph and not divisor in graph:
                ret = -1.0
            # Check if the origin and the destination are the same nodes
            elif dividend == divisor:
                ret = 1.0
            else:
                # Backtracking searching for path between two variables
                ret = dfs(dividend, divisor, 1)
            res.append(ret)

        return res

Optimized “disjoint set” with Path Compression and Union by Rank

class Solution:
    def calcEquation(self, equations: List[List[str]], values: List[float], queries: List[List[str]]) -> List[float]:
        graph = {}

        def find(x):
            if not x in graph:
                graph[x] = (x, 1)
            group_id, node_weight = graph[x]
            if group_id != x:
                new_group_id, group_weight = find(group_id)
                graph[x] = (new_group_id, node_weight * group_weight)
            return graph[x]
        
        def union(dividend, divisor, value):
            dividend_gid, dividend_weight = find(dividend)
            divisor_gid, divisor_weight = find(divisor)
            if dividend_gid != divisor_gid:
                # Merge the two groups together by attaching the dividend group to the one of divisor            
                graph[dividend_gid] = (divisor_gid, divisor_weight * value / dividend_weight)     
        
        # Build the union group
        for (dividend, divisor), value in zip(equations, values):
            union(dividend, divisor, value)
        
        res = list()
        for (dividend, divisor) in queries:
            if not dividend in graph or not divisor in graph:
                res.append(-1.0)
            else:
                dividend_gid, dividend_weight = find(dividend)
                divisor_gid, divisor_weight = find(divisor)
                if dividend_gid != divisor_gid:
                    res.append(-1.0)
                else:
                    res.append(dividend_weight / divisor_weight)
        return res