Sunday, September 11, 2016

LeetCode 399 - Evaluate Division


http://bookshadow.com/weblog/2016/09/11/leetcode-evaluate-division/
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). Given some queries, return the answers. If the answer does not exist, return -1.0.
Example:
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 ].
The input is: vector<pair<string, string>> euqations, vector<double>& values, vector<pair<string, string>> query . where equations.size() == values.size(),the values are positive. this represents the equations.return vector<double>. .
The example above: equations = [ ["a", "b"], ["b", "c"] ]. values = [2.0, 3.0]. queries = [ ["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"] ].
The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction.
Floyd算法求解传递闭包
输入等式可以看做一个有向图
例如等式a / b = 2.0,可以转化为两条边:<a, b>,<b, a>,其长度分别为2.0,0.5
遍历equations与values,利用g数组保存有向图中各边的长度,利用vset记录顶点集合
最后调用Floyd算法即可
def calcEquation(self, equations, values, query): """ :type equations: List[List[str]] :type values: List[float] :type query: List[List[str]] :rtype: List[float] """ g = collections.defaultdict(lambda: collections.defaultdict(int)) vset = set() for e, v in zip(equations, values): g[e[0]][e[1]] = v g[e[1]][e[0]] = 1.0 / v vset.add(e[0]) vset.add(e[1]) for k in vset: g[k][k] = 1.0 for s in vset: for t in vset: if g[s][k] and g[k][t]: g[s][t] = g[s][k] * g[k][t] ans = [] for s, t in query: ans.append(g[s][t] if g[s][t] else -1.0) return ans

X. DFS
http://blog.csdn.net/mebiuw/article/details/52512370
这道题首先需要做的工作是,已知a/b=x的情况下,推断出b/a=1/x的情况。将所有a作为被除数的情况下能作为除数的全部集合起来。
然后就有搜索的方式(如DFS),搜索有没有一个这样的链:
a/c = a/x1 x1/x2 x2/x3*…..*xn/c
这样就能推导出最终的结果了
https://discuss.leetcode.com/topic/59146/java-ac-solution-using-graph
If a/b = 2.0 and b/c = 3.0, we can treat a,b,c as vertices.
edge(a,b) weight 2.0, edge(b,c) weight 3.0
backward edge(b,a) weight 1/2.0, edge(c,b) weight 1/3.0
query a,c is a path from a to c, distance (a,c) = weight(a,b) * weight(b,c)
public double[] calcEquation(String[][] equations, double[] values, String[][] queries) {
    double[] res = new double[queries.length];
    if(equations.length == 0) return res;
    Map<String, List<Edge>> adjs = new HashMap();
    for(int i=0;i<equations.length;i++){
        String v = equations[i][0];
        String u = equations[i][1];
        Edge ef = new Edge(u, values[i]);
        Edge eb = new Edge(v, 1.0/values[i]);
        if(adjs.containsKey(v)){
            adjs.get(v).add(ef);
        } else {
            List<Edge> adjsV = new ArrayList();
            adjsV.add(ef);
            adjs.put(v, adjsV);
        }
        if(adjs.containsKey(u)){
            adjs.get(u).add(eb);
        } else {
            List<Edge> adjsU = new ArrayList();
            adjsU.add(eb);
            adjs.put(u, adjsU);
        }
    }
    for(int i=0;i<queries.length;i++){
        String s = queries[i][0];
        String t = queries[i][1];
        Set<String> visited = new HashSet();
        dfs(adjs,visited, s, t, 1.0,i, res);
        if(res[i] == 0 && s != t) res[i] = -1.0;
    }

    return res;
}
private void dfs(Map<String, List<Edge>> adjs,Set<String> visited, String s, String  t, double distance, int index, double[] res){
    if(s.equals(t)) { // start, end, result,
        res[index]  = distance;
    }
    if(visited.contains(s)) return;
    visited.add(s);
    if(!adjs.containsKey(s) || !adjs.containsKey(t)) {
        res[index] = -1.0;
        return;
    }
    List<Edge> adjsV = adjs.get(s);
    Iterator<Edge> iter = adjsV.iterator();
    while(iter.hasNext()){
        Edge e = iter.next();
        dfs(adjs,visited, e.to, t, distance * e.weight,index, res);
    }
class Edge{
String to;
double weight;
Edge(String t, double w){
to = t;
weight = w;
}
}
https://discuss.leetcode.com/topic/58321/java-ac-solution-with-explanation
The logic I have used is to construct a Map of maps, that contains all possible a/b and b/a from the given input and their values.
For the given input
equations = [ ["a", "b"], ["b", "c"] ]. values = [2.0, 3.0]
The map that gets constructed is :
[a: [b:2.0]
b: [a:0.5], [c:3.0]
c: [b:0.333]]
For each key in the outer map, the value represents a map, that denotes all possible denominators for the key and the corresponding key/value.
With this map constructed, the logic for evaluating a query is simple in a dfs style:
To find any m/n, if the map of m contains x1, x2, x3
then
m/n = m/x1 * x1/n if this gives a valid result or m/x2 * x2/n or m/x3 * x3/n
public static double[] calcEquation(String[][] equations, double[] values, String[][] query) {
        Map<String, Map<String, Double>> numMap = new HashMap<>();
        int i = 0;
        for(String[] str : equations) {
            insertPairs(numMap, str[0], str[1], values[i]);
            insertPairs(numMap, str[1], str[0], 1.0/values[i]);
            i++;
        }

        double[] res = new double[query.length];
        i = 0;
        for(String[] q: query) {
            Double resObj = handleQuery(q[0], q[1], numMap, new HashSet<>());
            res[i++] = (resObj != null) ? resObj : -1.0;
        }
        return res;
    }

    public static void insertPairs(Map<String, Map<String, Double>> numMap, String num, String denom, Double value) {
        Map<String, Double> denomMap = numMap.get(num);
        if(denomMap == null) {
            denomMap = new HashMap<>();
            numMap.put(num, denomMap);
        }
        denomMap.put(denom, value);
    }

    public static Double handleQuery(String num, String denom, Map<String, Map<String, Double>> numMap, Set<String> visitedSet) {
        String dupeKey = num+":"+denom;
        if(visitedSet.contains(dupeKey)) return null;
        if(!numMap.containsKey(num) || !numMap.containsKey(denom)) return null;
        if(num.equals(denom)) return 1.0;

        Map<String, Double> denomMap = numMap.get(num);

        for(String key : denomMap.keySet()) {
            visitedSet.add(dupeKey);
            Double res = handleQuery(key, denom, numMap, visitedSet);
            if(res != null) {
                return denomMap.get(key) * res;
            }
            visitedSet.remove(dupeKey);
        }
        return null;
    }
https://discuss.leetcode.com/topic/58445/java-solution-using-hashmap-and-dfs
(1) Build the map, the key is dividend, the value is also a map whose key is divisor and value is its parameter. For example, a / b = 2.0, the map entry is <"a", <"b", 2.0>>. To make searching and calculation easier, we also put b / a = 0.5 into the map.
(2) for each query, use DFS to search divisors recursively
https://discuss.leetcode.com/topic/58321/java-ac-solution-with-explanation
public static double[] calcEquation(String[][] equations, double[] values, String[][] query) {
        Map<String, Map<String, Double>> numMap = new HashMap<>();
        int i = 0;
        for(String[] str : equations) {
            insertPairs(numMap, str[0], str[1], values[i]);
            insertPairs(numMap, str[1], str[0], 1.0/values[i]);
            i++;
        }

        double[] res = new double[query.length];
        i = 0;
        for(String[] q: query) {
            Double resObj = handleQuery(q[0], q[1], numMap, new HashSet<>());
            res[i++] = (resObj != null) ? resObj : -1.0;
        }
        return res;
    }

    public static void insertPairs(Map<String, Map<String, Double>> numMap, String num, String denom, Double value) {
        Map<String, Double> denomMap = numMap.get(num);
        if(denomMap == null) {
            denomMap = new HashMap<>();
            numMap.put(num, denomMap);
        }
        denomMap.put(denom, value);
    }

    public static Double handleQuery(String num, String denom, Map<String, Map<String, Double>> numMap, Set<String> visitedSet) {
        String dupeKey = num+":"+denom;
        if(visitedSet.contains(dupeKey)) return null;
        if(!numMap.containsKey(num) || !numMap.containsKey(denom)) return null;
        if(num.equals(denom)) return 1.0;

        Map<String, Double> denomMap = numMap.get(num);
        visitedSet.add(dupeKey);
        for(String key : denomMap.keySet()) {
            Double res = handleQuery(key, denom, numMap, visitedSet);
            if(res != null) {//\\
                return denomMap.get(key) * res;
            }
        }
        visitedSet.remove(dupeKey);
        return null;
    }
https://discuss.leetcode.com/topic/58445/java-solution-using-hashmap-and-dfs
https://discuss.leetcode.com/topic/58399/share-my-straightforward-java-dfs-solution-very-similar-to-the-332-reconstruct-itinerary
http://blog.csdn.net/yeqiuzs/article/details/52506433
用HashMap 存储图的邻接表,并用创建图节点的visited标记。这里是图节点value是string,用数组表示visited不合适,故用hashmap<string,boolean>
  1. Map<String, Map<String, Double>> map = new HashMap<>();//邻接表  
  2.   
  3. public double[] calcEquation(String[][] equations, double[] values, String[][] query) {  
  4.     Set<String> set = new HashSet<String>();//记录表达式中出现的字符串  
  5.     for (int i = 0; i < equations.length; i++) {//建图  
  6.         set.add(equations[i][0]);  
  7.         set.add(equations[i][1]);  
  8.         Map<String, Double> m;  
  9.         if (map.containsKey(equations[i][0])) {  
  10.             m = map.get(equations[i][0]);  
  11.         } else {  
  12.             m = new HashMap<String, Double>();  
  13.         }  
  14.         m.put(equations[i][1], values[i]);  
  15.         map.put(equations[i][0], m);  
  16.   
  17.         if (map.containsKey(equations[i][1])) {  
  18.             m = map.get(equations[i][1]);  
  19.         } else {  
  20.             m = new HashMap<String, Double>();  
  21.         }  
  22.         m.put(equations[i][0], 1.0 / values[i]);  
  23.         map.put(equations[i][1], m);  
  24.   
  25.     }  
  26.   
  27.     double result[] = new double[query.length];  
  28.     for (int i = 0; i < query.length; i++) {  
  29.   
  30.         //初始化visited标记  
  31.         Iterator<String> it = set.iterator();  
  32.         Map<String, Boolean> visited = new HashMap<String, Boolean>();  
  33.         while (it.hasNext()) {  
  34.             visited.put(it.next(), false);  
  35.         }  
  36.   
  37.         if (query[i][0].equals(query[i][1]) && set.contains(query[i][0])) {  
  38.             result[i] = 1;  
  39.             continue;  
  40.         }  
  41.         //dfs  
  42.         double res = dfs(query[i][0], query[i][1], 1, visited);  
  43.         result[i] = res;  
  44.     }  
  45.     return result;  
  46. }  
  47.   
  48. public double dfs(String s, String t, double res, Map<String, Boolean> visited) {  
  49.     if (map.containsKey(s) && !visited.get(s)) {  
  50.         visited.put(s, true);  
  51.         Map<String, Double> m = map.get(s);  
  52.         if (m.containsKey(t)) {  
  53.             return res * m.get(t);  
  54.         } else {  
  55.             Iterator<String> keys = m.keySet().iterator();  
  56.             while (keys.hasNext()) {  
  57.                 String key = keys.next();  
  58.                 double state = dfs(key, t, res * m.get(key), visited);  
  59.                 if (state != -1.0) {  
  60.                     return state;  
  61.                 }  
  62.             }  
  63.         }  
  64.     } else {  
  65.         return -1.0;  
  66.     }  
  67.     return -1.0;  
  68. }  

X. BFS
http://xiadong.info/2016/09/leetcode-399-evaluate-division/
    // 由于对于string的比较等操作很费时, 所以用一个map把string与int对应起来.
    unordered_map<string, int> nodes;
public:
    vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {
        for(int i = 0; i < equations.size(); i++){
            // 给每一个string分配一个下标
            // 注意这里有个隐藏bug, 假如map/unordered_map对象m中不包含a,
            // 那么在使用m[a]时实际上是已经创建一个a的key和对应的value, 导致size加1
            // 所以如果我们想让第n个加入的元素的value为n-1的话,
            // 需要赋值m.size() - 1而不是m.size()
            if(!nodes.count(equations[i].first)){
                nodes[equations[i].first] = nodes.size() - 1;
            }
            if(!nodes.count(equations[i].second)){
                nodes[equations[i].second] = nodes.size() - 1;
            }
        }
        vector<vector<double>> g(nodes.size(), vector<double>(nodes.size(), -1.0));
        for(int i = 0; i < equations.size(); i++){
            // 构建邻接矩阵
            g[getNode(equations[i].first)][getNode(equations[i].second)] = values[i];
            g[getNode(equations[i].second)][getNode(equations[i].first)] = 1 / values[i];
        }
        vector<double> ret(queries.size());
        for(int i = 0; i < queries.size(); i++){
            string a = queries[i].first, b = queries[i].second;
            if(!nodes.count(a) || !nodes.count(b)){
                // 如果出现了不存在的节点
                ret[i] = -1.0;
            }
            else{
                // 使用BFS来搜索路径
                ret[i] = BFS(g, getNode(a), getNode(b));
            }
        }
        return ret;
    }
    
    int getNode(string s){
        return nodes[s];
    }
    
    double BFS(vector<vector<double>> &g, int a, int b){
        // 如果是同一个节点就直接返回
        if(a == b) return 1.0;
        int n = g.size();
        vector<int> visited(n, 0); // 用于保存是否访问过节点
        queue<int> q; // BFS队列, 保存节点下标
        queue<double> v; // 用于保存从a到BFS队列中相应的节点的路径乘积
        q.push(a);
        visited[a] = 1;
        v.push(1.0);
        while(!q.empty()){
            int node = q.front();
            double value = v.front();
            for(int i = 0; i < n; i++){
                if(visited[i] || g[node][i] == -1.0) continue; // 节点i已经访问过或者没有边到达i
                visited[i] = 1;
                q.push(i);
                double len = value * g[node][i]; // 从a到i的路径权值乘积
                // 添加新的边
                g[a][i] = len;
                g[i][a] = 1 / len;
                if(i == b){ // 抵达b点
                    return len;
                }
                v.push(len);
            }
            q.pop();
            v.pop();
        }
        return -1.0;
    }

X. Floyd–Warshall DP
https://discuss.leetcode.com/topic/58981/java-solution-using-floyd-warshall-algorithm
functional programming in java is very slow
Using a variant of Floyd–Warshall algorithm, to find the distance between each reachable pair:
    public double[] calcEquation(String[][] equations, double[] values, String[][] queries) {
        HashMap<String, HashMap<String, Double>> graph = new HashMap<>();
        Function<String, HashMap<String, Double>> function = s -> new HashMap<>();
        for (int i = 0; i < equations.length; i++) {
            graph.computeIfAbsent(equations[i][0], function).put(equations[i][0], 1.0);
            graph.computeIfAbsent(equations[i][1], function).put(equations[i][1], 1.0);
            graph.get(equations[i][0]).put(equations[i][1], values[i]);
            graph.get(equations[i][1]).put(equations[i][0], 1 / values[i]);
        }
        for (String mid : graph.keySet()) {
            for (String src : graph.get(mid).keySet()) {
                for (String dst : graph.get(mid).keySet()) {
                    double val = graph.get(src).get(mid) * graph.get(mid).get(dst);
                    graph.get(src).put(dst, val);
                }
            }
        }
        double[] result = new double[queries.length];
        for (int i = 0; i < result.length; i++) {
            if (!graph.containsKey(queries[i][0])) {
                result[i] = -1;
            } else {
                result[i] = graph.get(queries[i][0]).getOrDefault(queries[i][1], -1.0);
            }
        }
        return result;
    }
https://discuss.leetcode.com/topic/58482/9-lines-floyd-warshall-in-python
A variation of Floyd–Warshall, computing quotients instead of shortest paths. Submitted once, accepted in 35 ms.
def calcEquation(self, equations, values, queries):
    quot = collections.defaultdict(dict)
    for (num, den), val in zip(equations, values):
        quot[num][num] = quot[den][den] = 1.0
        quot[num][den] = val
        quot[den][num] = 1 / val
    for k, i, j in itertools.permutations(quot, 3):
        if k in quot[i] and j in quot[k]:
            quot[i][j] = quot[i][k] * quot[k][j]
    return [quot[num].get(den, -1.0) for num, den in queries]

Variation without the if (submitted twice, accepted in 68 and 39 ms):
def calcEquation(self, equations, values, queries):
    quot = collections.defaultdict(dict)
    for (num, den), val in zip(equations, values):
        quot[num][num] = quot[den][den] = 1.0
        quot[num][den] = val
        quot[den][num] = 1 / val
    for k in quot:
        for i in quot[k]:
            for j in quot[k]:
                quot[i][j] = quot[i][k] * quot[k][j]
    return [quot[num].get(den, -1.0) for num, den in queries]
Could save a line with for i, j in itertools.permutations(quot[k], 2) but it's longer and I don't like it as much here.


X. union-find
I think the key point to induct your algorithm is to find a common divisor ( the root) for the queries. Clever idea :)
We can avoid traverse the entire map by plug in children information to parent. @iambright
This my java implementation based on your idea. Further, I use weighted UnionFind with path compression in order to speed up union and find operation.
    Map<String, GraphNode> map = null;
    public double[] calcEquation(String[][] equations, double[] values, String[][] queries) {
        double[] res = new double[queries.length];
        map = new HashMap<String, GraphNode>();
        for (int i = 0; i < equations.length; i++){
            String a = equations[i][0];
            String b = equations[i][1];
            if (!map.containsKey(a) && !map.containsKey(b)){
                GraphNode nodeA = new GraphNode(values[i]);
                GraphNode nodeB = new GraphNode(1);
                map.put(a, nodeA);
                map.put(b, nodeB);
                nodeA.parent = nodeB;
                nodeB.children.add(nodeA);
            }else if (!map.containsKey(b)){
                GraphNode nodeA = map.get(a);
                GraphNode nodeB = new GraphNode(nodeA.value / values[i]);
                GraphNode parentA = root(nodeA);
                nodeB.parent = parentA;
                parentA.children.add(nodeB);
                map.put(b, nodeB);
            }else if (!map.containsKey(a)){
                GraphNode nodeB = map.get(b);
                GraphNode nodeA = new GraphNode(nodeB.value * values[i]);
                GraphNode parentB = root(nodeB);
                nodeA.parent = parentB;
                parentB.children.add(nodeB);
                map.put(a, nodeA);
            }else {
                union(map.get(a), map.get(b), values[i]);
            }
        }
        for (int i = 0; i < queries.length; i++){
            String a = queries[i][0];
            String b = queries[i][1];
            if (!map.containsKey(a) || !map.containsKey(b) || root(map.get(a)) != root(map.get(b))){
                res[i] = -1.0;
            }else {
                res[i] = map.get(a).value / map.get(b).value;
            }
        }
        return res;
    }
    private class GraphNode{
        double value;
        GraphNode parent;
        List<GraphNode> children;
        int size;
        GraphNode(double value){
            this.value = value;
            size = 1;
            children = new ArrayList<>(Arrays.asList(this));
            parent = this;
        }
    }
    private GraphNode root(GraphNode node){
        while (node != node.parent){
            node.parent = node.parent.parent;
            node = node.parent;
        }
        return node;
    }
    private void union(GraphNode node1, GraphNode node2, double value){
        GraphNode parent1 = root(node1);
        GraphNode parent2 = root(node2);
        if (parent1 == parent2){
            return ;
        }
        if (parent1.size < parent2.size){
            unionHelper(node1, node2, value);
        }else {
            unionHelper(node2, node1, 1/value);
        }
    }
    private void unionHelper(GraphNode node1, GraphNode node2, double value){
        GraphNode parent1 = root(node1);
        GraphNode parent2 = root(node2);
        double ratio = node2.value * value / node1.value;
        for (GraphNode child : parent1.children){
            child.value *= ratio;
            child.parent = parent2;
            parent2.children.add(child);
            parent2.size++;
        }
    }
 I added a path compression optimization that should bring down the complexity.





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