Count chars in subtree mapping to string.

Author: deepakhb2

Interview-Questions/count_nodes_in_subtree.rb

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+=begin
+
+Nodes in a Subtree
+You are given a tree that contains N nodes, each containing an integer u which corresponds to a lowercase character c in the string s using 1-based indexing.
+You are required to answer Q queries of type [u, c], where u is an integer and c is a lowercase letter. The query result is the number of nodes in the subtree of node u containing c.
+Signature
+int[] countOfNodes(Node root, ArrayList<Query> queries, String s)
+Input
+A pointer to the root node, an array list containing Q queries of type [u, c], and a string s
+Constraints
+N and Q are the integers between 1 and 1,000,000
+u is an integer between 1 and N
+s is of the length of N, containing only lowercase letters
+c is a lowercase letter contained in string s
+Node 1 is the root of the tree
+Output
+An integer array containing the response to each query
+Example
+        1(a)
+        /   \
+      2(b)  3(a)
+s = "aba"
+RootNode = 1
+query = [[1, 'a']]
+Note: Node 1 corresponds to first letter 'a', Node 2 corresponds to second letter of the string 'b', Node 3 corresponds to third letter of the string 'a'.
+output = [2]
+Both Node 1 and Node 3 contain 'a', so the number of nodes within the subtree of Node 1 containing 'a' is 2.
+
+=end
+
+class Node
+  attr_reader :value
+  attr_accessor :children 
+  
+  def initialize(value, children)
+    @value = value
+    @children = children
+  end
+end
+
+# Should use the substring to count.
+def count_of_nodes(root, queries, s)
+  output = []
+  queries.each do |query|
+    r_node = nil
+    queue = [root]
+    while(queue.size > 0)
+      temp_node = queue.shift
+      if(temp_node.value == query[0])
+        r_node = temp_node
+        break
+      else
+        queue = queue + temp_node.children
+      end
+    end
+    queue = [r_node]
+    count = 0
+    while(queue.size > 0)
+      temp_node = queue.shift
+      if(s[temp_node.value-1] == query[1])
+        count+=1
+      end
+      queue = queue + temp_node.children
+    end
+    output << count
+  end
+  output
+end
+
+node4 = Node.new(4, [])
+node1 = Node.new(2, [node4])
+node2 = Node.new(3, [])
+node3 = Node.new(1, [node1, node2])
+
+p count_of_nodes(node3, [[1, 'a']], 'abaa')