Number of possible BST with given key count n in Scala
Problem -
Total number of possible Binary Search Trees with n keys.
for n = 0 , only one BST is possible (i.e. null or empty tree)
for n = 1 , only one BST is possible
for n = 2 , two BSTs are possible
for n=3 , five different BSTs are possible.
And so on.
Solution in Scala -
import scala.collection.mutable
object BSTCount extends App {
// count(6) = count(5)*count(0) + count(4)*count(1) + count(3)*count(2)
// count(2).count(3) + count(1)*count(4) + count(0)*count(5)
val lookup= mutable.HashMap[Int,Int]((0,1),(1,1),(2,2))
def BSTCount(n: Int): Int = {
n match {
case 0 => 1
case 1 => 1
case 2 => 2
case _ =>
if (!(lookup contains n)) {
val r = ((0 until n) map { i =>
lookup += i -> BSTCount(i)
lookup += (n-i-1) -> BSTCount(n-i-1)
lookup(i) * lookup(n-i-1)
}).sum
lookup += (n -> r)
}
lookup(n)
}
}
(1 to 10) foreach ( num => println(num + " -> " + BSTCount(num)))
}
It generates following output -
1 -> 1
2 -> 2
3 -> 5
4 -> 14
5 -> 42
6 -> 132
7 -> 429
8 -> 1430
9 -> 4862
10 -> 16796
interview-questions tree functional-programming graph scala
asked Jan 2 at 20:35
vikrant
807
add a comment |
Problem -
Total number of possible Binary Search Trees with n keys.
for n = 0 , only one BST is possible (i.e. null or empty tree)
for n = 1 , only one BST is possible
for n = 2 , two BSTs are possible
for n=3 , five different BSTs are possible.
And so on.
Solution in Scala -
import scala.collection.mutable
object BSTCount extends App {
// count(6) = count(5)*count(0) + count(4)*count(1) + count(3)*count(2)
// count(2).count(3) + count(1)*count(4) + count(0)*count(5)
val lookup= mutable.HashMap[Int,Int]((0,1),(1,1),(2,2))
def BSTCount(n: Int): Int = {
n match {
case 0 => 1
case 1 => 1
case 2 => 2
case _ =>
if (!(lookup contains n)) {
val r = ((0 until n) map { i =>
lookup += i -> BSTCount(i)
lookup += (n-i-1) -> BSTCount(n-i-1)
lookup(i) * lookup(n-i-1)
}).sum
lookup += (n -> r)
}
lookup(n)
}
}
(1 to 10) foreach ( num => println(num + " -> " + BSTCount(num)))
}
It generates following output -
1 -> 1
2 -> 2
3 -> 5
4 -> 14
5 -> 42
6 -> 132
7 -> 429
8 -> 1430
9 -> 4862
10 -> 16796
interview-questions tree functional-programming graph scala
asked Jan 2 at 20:35
vikrant
807
add a comment |
Problem -
Total number of possible Binary Search Trees with n keys.
for n = 0 , only one BST is possible (i.e. null or empty tree)
for n = 1 , only one BST is possible
for n = 2 , two BSTs are possible
for n=3 , five different BSTs are possible.
And so on.
Solution in Scala -
import scala.collection.mutable
object BSTCount extends App {
// count(6) = count(5)*count(0) + count(4)*count(1) + count(3)*count(2)
// count(2).count(3) + count(1)*count(4) + count(0)*count(5)
val lookup= mutable.HashMap[Int,Int]((0,1),(1,1),(2,2))
def BSTCount(n: Int): Int = {
n match {
case 0 => 1
case 1 => 1
case 2 => 2
case _ =>
if (!(lookup contains n)) {
val r = ((0 until n) map { i =>
lookup += i -> BSTCount(i)
lookup += (n-i-1) -> BSTCount(n-i-1)
lookup(i) * lookup(n-i-1)
}).sum
lookup += (n -> r)
}
lookup(n)
}
}
(1 to 10) foreach ( num => println(num + " -> " + BSTCount(num)))
}
It generates following output -
1 -> 1
2 -> 2
3 -> 5
4 -> 14
5 -> 42
6 -> 132
7 -> 429
8 -> 1430
9 -> 4862
10 -> 16796
interview-questions tree functional-programming graph scala
asked Jan 2 at 20:35
vikrant
807
Problem -
Total number of possible Binary Search Trees with n keys.
for n = 0 , only one BST is possible (i.e. null or empty tree)
for n = 1 , only one BST is possible
for n = 2 , two BSTs are possible
for n=3 , five different BSTs are possible.
And so on.
Solution in Scala -
import scala.collection.mutable
object BSTCount extends App {
// count(6) = count(5)*count(0) + count(4)*count(1) + count(3)*count(2)
// count(2).count(3) + count(1)*count(4) + count(0)*count(5)
val lookup= mutable.HashMap[Int,Int]((0,1),(1,1),(2,2))
def BSTCount(n: Int): Int = {
n match {
case 0 => 1
case 1 => 1
case 2 => 2
case _ =>
if (!(lookup contains n)) {
val r = ((0 until n) map { i =>
lookup += i -> BSTCount(i)
lookup += (n-i-1) -> BSTCount(n-i-1)
lookup(i) * lookup(n-i-1)
}).sum
lookup += (n -> r)
}
lookup(n)
}
}
(1 to 10) foreach ( num => println(num + " -> " + BSTCount(num)))
}
It generates following output -
1 -> 1
2 -> 2
3 -> 5
4 -> 14
5 -> 42
6 -> 132
7 -> 429
8 -> 1430
9 -> 4862
10 -> 16796
interview-questions tree functional-programming graph scala
interview-questions tree functional-programming graph scala
asked Jan 2 at 20:35
vikrant
807
asked Jan 2 at 20:35
vikrant
807
asked Jan 2 at 20:35
vikrant
807
asked Jan 2 at 20:35
vikrant
807
asked Jan 2 at 20:35
vikrant
807
807
add a comment |
add a comment |
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