#coding exercises
1) Find minimum number of merge operations to make a palindrome.
Ex:  {15, 4, 15} => 0
        {1, 4, 5, 1} => 1 because 4 and 5 are merged.
         {11, 14, 15, 19} => 2 because all elements need to be merged.

int GetMergeCount(List<int> numbers)
{
int start = 0;
int end = numbers.Count;
while (start < end)
{
if (numbers[start] != numbers[end])
{
  count += 1;
}
start++;
end--;
}
return count;
}

2) Print diagonal traversal of binary tree

void PrintDiagonal(Node root, int level. ref Dict<int,List<int>> levelValues)
{
if (root == null) return;
levelValues[level].Add(root.data);
PrintDiagonal(root.left, level+1, ref levelValues);
PrintDiagonal(root.right, level, ref levelValues);
}

void PrintDiagonal(ref Dict<int, List<int>> levelValues)
{
foreach (kvp in levelValues)
{
var level = kvp.Key;
console.WriteLine('Level='+level);
var items = kvp.Value;
for (int i = 0; i < items.Count;  i++)
{
  Console.Write(items[i].toString());
}
console.writeLine();
}
}

3) bool isPower(intN, int X)
{
return IsInteger(Math.Log(N)/Math.Log(x));
}

4) void Boggle(List<char> input, List<string> knownwords, ref List<char>candidate, int start, int level)
{
for (int I =start; I< input.Count; I++)
{
candidate[level] = input[I];
if (knownwords.Contains(candidate.toString()) console.writeline(candidate.toString());
if (i < input.Count)
{
Boggle(input, knownwords, ref candidate, start+1, level+1);
}
candidate[level] = '/0';
}
}

5) To determine if there is a backedge from any of the vertex to its ancestors in  a graph other than the tree it is part of, we can maintain two arrays : a discovery time array which measures the time it takes to discover this vertex from root aka levels and a low value array that for a vertex u is the minimum of the discovery time for the vertex u and that of its ancestor w
low[u] = min(disc[u], disc[w])
if ( low[v] >= disc[u]) then there is a backedge.
Also note that by definition of a tree, every node has a parent p

6) void ConnectNodesAtSameLevel(Node root)
{
List<Node> levels = LevelWiseTraversal(root); // using a queue described earlier, append NULL for delimiters
for (int i = 0;i < levels.Count -1; i++)
   if (levels[i])
     levels[i].side = levels[i+1];
}

7) Print top view of a binary tree ( nodes that are on the periphery)

void PrintTopView(Node root)
{
 // similar to level wise traversal
var ret = List<Node> ();
int left = 0;
int right = 0;
var q = new Queue();
q.enqueue(root);
root.dist = 0; // or use a hashtable
q.enqueue(null);
while (q.empty() == false)
{
var item = q.dequeue();
if (item == null){
q.enqueue(null);
}else
{
if (root.dist == 0 { ret.Add(root);}
if (root.dist < left ) { left = root.dist; ret.Add(root);}
if (root.dist > right) { right = root.dist;. ret.Add(root);}
root.left.dist = root.dist-1;
q.enqueue(root.left);
root.right.dist = root.dist+1;
q.enqueue(root.right);
}
}
ret.ForEach( x => console.write(x.toString()));
}