Add medium problem.

Author: das-jishu

Leetcode/medium/evaluate-division.py

@@ -0,0 +1,81 @@
+"""
+# EVALUATE DIVISION
+
+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.
+"""
+
+class Solution:
+    def calcEquation(self, equations, values, queries):
+        graph = {}
+        for i in range(len(equations)):
+            num, den = equations[i]
+            ans = values[i]
+            if num not in graph:
+                graph[num] = {num: 1}
+            graph[num][den] = ans
+            
+            if den not in graph:
+                graph[den] = {den: 1}
+            graph[den][num] = (1 / ans)
+        
+        results = []
+        for n, d in queries:
+            visited = set()
+            visited.add(n)
+            results.append(self.dfs(n, d, 1, visited, graph))
+            
+        return results
+            
+    def dfs(self, num, den, value, visited, graph):
+        if num not in graph:
+            return -1.0
+        
+        possibleValues = graph[num]
+        if den in possibleValues:
+            return value * possibleValues[den]
+        
+        for item in possibleValues:
+            if item in visited:
+                continue
+            visited.add(item)
+            check = self.dfs(item, den, value * possibleValues[item], visited, graph)
+            if check != -1.0:
+                return check
+            visited.remove(item)
+        return -1.0