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| 1 | +package com.interview.graph; |
| 2 | + |
| 3 | +import java.util.HashMap; |
| 4 | +import java.util.Map; |
| 5 | + |
| 6 | +/** |
| 7 | + * Date 10/31/2016 |
| 8 | + * @author Tushar Roy |
| 9 | + * |
| 10 | + * Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). |
| 11 | + * Given some queries, return the answers. If the answer does not exist, return -1.0. |
| 12 | + * |
| 13 | + * Solution |
| 14 | + * Do Flyod warshall algorithm initialized as values between equations. Do Flyod Warshall to create |
| 15 | + * all possible paths b/w two strings. |
| 16 | + * |
| 17 | + * Time complexity O(n * n * n) + O(m) |
| 18 | + * where n is total number of strings in equations and m is total number of queries |
| 19 | + * |
| 20 | + * Reference |
| 21 | + * https://leetcode.com/problems/evaluate-division/ |
| 22 | + */ |
| 23 | +public class EvaluateDivison { |
| 24 | + public double[] calcEquation(String[][] equations, double[] values, String[][] queries) { |
| 25 | + if (equations.length == 0) { |
| 26 | + return new double[0]; |
| 27 | + } |
| 28 | + Map<String, Integer> index = new HashMap<>(); |
| 29 | + int count = 0; |
| 30 | + for (int i = 0; i < equations.length; i++) { |
| 31 | + String first = equations[i][0]; |
| 32 | + String second = equations[i][1]; |
| 33 | + if (!index.containsKey(first)) { |
| 34 | + index.put(first, count++); |
| 35 | + } |
| 36 | + if (!index.containsKey(second)) { |
| 37 | + index.put(second, count++); |
| 38 | + } |
| 39 | + } |
| 40 | + |
| 41 | + double graph[][] = new double[count][count]; |
| 42 | + for (int i = 0; i < graph.length; i++) { |
| 43 | + for (int j = 0; j < graph[i].length; j++) { |
| 44 | + graph[i][j] = -1; |
| 45 | + } |
| 46 | + } |
| 47 | + |
| 48 | + for (int i = 0; i < equations.length; i++) { |
| 49 | + String first = equations[i][0]; |
| 50 | + String second = equations[i][1]; |
| 51 | + int i1 = index.get(first); |
| 52 | + int i2 = index.get(second); |
| 53 | + graph[i1][i1] = graph[i2][i2] = 1.0; |
| 54 | + graph[i1][i2] = values[i]; |
| 55 | + graph[i2][i1] = 1/values[i]; |
| 56 | + } |
| 57 | + |
| 58 | + for (int i = 0 ; i < graph.length; i++) { |
| 59 | + for (int j = 0; j < graph.length; j++) { |
| 60 | + if (graph[i][j] != -1) { |
| 61 | + continue; |
| 62 | + } |
| 63 | + for (int k = 0; k < graph.length; k++) { |
| 64 | + if (graph[i][k] == -1 || graph[k][j] == -1) { |
| 65 | + continue; |
| 66 | + } |
| 67 | + graph[i][j] = graph[i][k] * graph[k][j]; |
| 68 | + } |
| 69 | + } |
| 70 | + } |
| 71 | + |
| 72 | + double[] result = new double[queries.length]; |
| 73 | + for (int i = 0; i < queries.length; i++) { |
| 74 | + String first = queries[i][0]; |
| 75 | + String second = queries[i][1]; |
| 76 | + if (!index.containsKey(first) || !index.containsKey(second)) { |
| 77 | + result[i] = -1; |
| 78 | + } else { |
| 79 | + int i1 = index.get(first); |
| 80 | + int i2 = index.get(second); |
| 81 | + result[i] = graph[i1][i2]; |
| 82 | + } |
| 83 | + } |
| 84 | + |
| 85 | + return result; |
| 86 | + } |
| 87 | +} |
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