Neelavan
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14of 14 votes
AnswersWrite a program to check whether it is a valid binary tree or not. Check all test cases (e.g. No left Node case).
- Neelavan in United States| Report Duplicate | Flag | PURGE
Amazon SDE-2 Algorithm
public static void RotateString(string myString,int times)
{
char[] myChar = myString.ToArray();
List<char> lst = new List<char>();
int firstStringCount= myChar.Count() - times;
for (int i = firstStringCount; i < myChar.Count(); i++)
{
lst.Add(myChar[i]);
}
for (int i = 0; i < firstStringCount; i++)
{
lst.Add(myChar[i]);
}
lst.ForEach(s=>Console.Write(s));
}
Use a dictionary of rotate values of each integer.
public static void CheckRotationNumber(int myInt)
{
bool flag=true;
Dictionary<int, int> rotateDict = new Dictionary<int, int>();
rotateDict.Add(1, 1); rotateDict.Add(2, 2); rotateDict.Add(3, 3); rotateDict.Add(4, 4);
rotateDict.Add(5, 5); rotateDict.Add(6, 9); rotateDict.Add(7, 7); rotateDict.Add(8, 8); rotateDict.Add(9, 6); rotateDict.Add(0, 0);
char[] myChar = myInt.ToString().ToArray();
int length = myChar.Length-1;
for (int i = 0; i <= length/2; i++)
{
if(rotateDict[myChar[i] - '0'] != myChar[length-i] - '0')
{
flag=false;
break;
}
}
Console.Write(flag.ToString());
}
First track the level of the node and add parent pointer in your tree structure.
static void LeftMostNodeSameLevel(Node myNode)
{
int level = 0;
while (myNode.Parent!=null)
{
myNode = myNode.Parent;
level += 1;
}
while (myNode!=null && level>0)
{
if (myNode.LeftNode != null)
myNode = myNode.LeftNode;
else
myNode = myNode.RightNode;
level-=1;
}
Console.WriteLine("Left Most Node is {0}",myNode.Value);
}
We can keep tracking of previous jump .
public static bool FrogJump(bool[] array)
{
int lastJump = 0;
bool canCross = true;
for (int i = 0; i < array.Count(); i++)
{
if (array[i] == true)
{
int tempJump = 0;
for (int j = 1; j < array.Count()-i; j++)
{
tempJump = j;
if (tempJump > lastJump + 1)
{
canCross = false;
break;
}
if (array[i + tempJump] == false)
{
i = i + j;
break;
}
}
lastJump = tempJump;
}
}
return canCross;
}
It's impossible to do less than O(mn), here is the solution by using DFS recursion.
public static void PrintDFSMatrix(int[,] myInt, int x, int y)
{
if (x < myInt.GetLength(0) && y < myInt.GetLength(1))
{
PrintDFSMatrix(myInt, x + 1, y + 1);
for (int i = y; i < myInt.GetLength(0); i++)
{
Console.WriteLine(myInt[x, i]);
}
for (int i = x + 1; i < myInt.GetLength(1); i++)
{
Console.WriteLine(myInt[i, y]);
}
}
}
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- Neelavan April 09, 2017