Monday, August 8, 2016

Google – Average Length of City Path


Google – Average Length of City Path
给n个城市,和n – 1条edge, 这些edges可以保证所有城市都两两互通,每条edge都有长度,求所有城市两两之间path的长度之和,再除以path的数量得到的平均数。
[Solution]
第一种解法就是brute force,O(n^2) time
第二种解法来自一亩三分地,可以在O(n)时间内解决问题。思路非常tricky,类似于Leetcode 310 Minimum Height Tree的解法。
因为题目明确给出了有n个node和n – 1条edges,看到这两个数字应该很本能的反应出这道题和tree拖不了关系。但是Brute Force的解法并没有利用到这个条件。而O(n)的解法正是利用构建Minimum Height Tree来解决问题。下面是算法思路:
(1)算degree
(2)逐层删leaf node来找root。就是Minimum Height Tree的算法。
(3)在删leaf node的过程中多做两件事:建tree和计算sub tree size(包括该节点本身)。如果在第(2)步有两个root,随便选一个作为root,另一个作为子节点就好。这里的主要目的是建Tree.
(4)如果总共有N个结点,当前结点v的sub tree size为K (包括v本身),那么连接v和其parent的那条edge就会被遍历K * (N-K)次。
这是因为如果以v为分界把这个graph分为两部分,一部分有K个点,另一部分为N-K个点。因为每对Pair都需要走一遍,也就是左边的K个点中的任何一个都要去找右边的N-K个点。又因为树的特性,每对pair之间只可能有一条path,所以v到其parent之间的这条edge必定会被遍历K * (N-K)次。
(5)构建完tree,证明完(4),剩下的只需要对tree做一个dfs traversal,就可以得到总的path sum。再除以n-1就可以得到答案了。
class Solution2 {
  int sum = 0;
  int n;
  public int averagePathLength(int n, int[][] graph) {
    if (n <= 1) {
      return 0;
    }
     
    this.n = n;
     
    // get degree of every node
    Map<Integer, Integer> degree = new HashMap<>();
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if (graph[i][j] != 0) {
          degree.put(i, degree.getOrDefault(i, 0) + 1);
          degree.put(j, degree.getOrDefault(j, 0) + 1);
        }
      }
    }
     
    // Build the tree
    TNode root = buildTree(degree, graph);
     
    // dfs traversal to get the path sum
    dfs(root);
     
    return sum / (n - 1);
  }
   
  private void dfs(TNode root) {
    if (root == null) {
      return null;
    }
     
    for (TNode child : root.children) {
      sum += child.size * (n - child.size) * graph[root.val][child.val];
      dfs(child);
    }
  }
   
  private TNode buildTree(Map<Integer, Integer> degree, int[][] graph) {
     
    Map<Integer, TNode> nodeMap = new HashMap<>();
    for (int i = 0; i < n; i++) {
      nodeMap.put(i, new TNode(i));
    }
     
    Queue<Integer> queue = new LinkedList<>();
    for (int v : degree) {
      if (degree.get(v) == 1) {
        queue.add(v);
        nodeMap.get(v).size += 1;
      }
    }
    for (int v : queue) {
      degree.remove(v);
    }
     
    while (true) {
      int size = queue.size();
      for (int i = 0; i < size; i++) {
        int curr = queue.poll();
        int currSize = nodeMap.get(curr).size;
         
        for (int adj = 0; adj < n; adj++) {
          if (graph[curr][adj] != 0) {
            nodeMap.get(adj).children.add(curr);
            nodeMap.get(adj).size += currSize;
             
            degree.put(adj, degree.get(adj) - 1);
            if (degree.get(adj) == 1) {
              queue.offer(adj);
            }
          }
        }
      }
       
      for (int v : queue) {
        degree.remove(v);
      }
       
      if (degree.isEmpty()) {
        break;
      }
    }
     
     
    if (queue.size() > 1) {
      int root1 = queue.poll();
      int root2 = queue.poll();
      nodeMap.get(root1).children.add(root2);
      nodeMap.get(root1).size += nodeMap.get(root2).size;
      return nodeMap.get(root1);
    }
     
    return nodeMap.get(queue.poll());
  }
  
 
  private class TNode {
    int val;
    List<TNode> children;
    int size;
     
    TNode(int val) {
      this.val = val;
      this.children = new ArrayList<>();
    }
  }
}
http://www.1point3acres.com/bbs/thread-139392-1-1.html
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