Scala Assignment: Recursion

gistfile1.scala

package recfun

import scala.collection.mutable.ListBuffer
import common._

/** https://class.coursera.org/progfun-2012-001/assignment/view?assignment_id=4 */
object Main {
  def main(args: Array[String]) {
    println("Pascal's Triangle")
    for (row <- 0 to 10) {
      for (col <- 0 to row)
        print(pascal(col, row) + " ")
      println()
    }
  }

  /**
   * Exercise 1: Pascal's Triangle
   */
  def pascal(c: Int, r: Int): Int = {
    if (c == 0 || c == r) 1
    else pascal(c - 1, r - 1) + pascal(c, r - 1)
  }

  /**
   * Exercise 2: Parentheses Balancing
   */
  def balance(chars: List[Char]): Boolean = {
    def f(chars: List[Char], numOpens: Int): Boolean = {
      if (chars.isEmpty) {
        numOpens == 0
      } else {
        val h = chars.head
        val n =
          if (h == '(') numOpens + 1
          else if (h == ')') numOpens - 1
          else numOpens
        if (n >= 0) f(chars.tail, n)
        else false
      }
    }

    f(chars, 0)
  }

  /**
   * Exercise 3: Counting Change
   * Write a recursive function that counts how many different ways you can make
   * change for an amount, given a list of coin denominations. For example,
   * there are 3 ways to give change for 4 if you have coins with denomiation
   * 1 and 2: 1+1+1+1, 1+1+2, 2+2.
   */
  def countChange(money: Int, coins: List[Int]): Int = {
    def f(lastMaxCoin_total_coll: List[(Int, Int)], count: Int): Int = {
      if (lastMaxCoin_total_coll.isEmpty) {
        count
      } else {
        val b = ListBuffer[(Int, Int)]()
        var newCount = count
        for ((lastMaxCoin, total) <- lastMaxCoin_total_coll) {
          if (total < money) {
            for (c <- coins) {
              if (c >= lastMaxCoin) {
                val e = (c, total + c)
                b += e
              }
            }
          } else if (total == money) {
            newCount += 1
          }
        }

        f(b.toList, newCount)
      }
    }

    val b = coins.map { c => (c, c) }
    f(b, 0)
  }
}

this is a function program!!!
so
def balance(chars: List[Char]): Boolean = chars.filter(_ == '(').length == chars.filter(_ == ')').length

BalanceParenthesis using stack

def balance(chars: List[Char]): Boolean = {if (pfunc(chars)==0) return true else (return false)}

def pfunc(chars: List[Char]): Int ={
var st = mutable.StackChar
if(!chars.contains('(')) return 1
for (char <-chars) {
if (char == '(') st.push(char)
else if (char == ')' && !st.isEmpty)
{if ( st.top=='(') st.pop()}
//System.out.println(st.toString(),st.size)
};return st.size}

Balance using fold where int accumulator counts the balance between ( and ) and bool accumulator is used for breaking out of the fold, (https://stackoverflow.com/questions/12892701/abort-early-in-a-fold uses opt for that)

  def balance2(chars: Iterable[Char]): Boolean = {
  	val (bool, int) = chars.foldLeft(true -> 0){
  		case ((bool, int), char) => char match {
  			case '(' => bool -> (int + 1)
  			case ')' => (bool && int != 0) -> (int - 1)
  			case _ => bool -> int
  		}
  	}; bool && (int == 0)
  }                                               //> balance2: (chars: Array[Char])Boolean

or even use return to break out of the fold

  def balance22(chars: Iterable[Char]): Boolean = {
  	chars.foldLeft(0){
  		case (acc, char) => char match {
  			case '(' => acc + 1
  			case ')' => if (acc == 0) return false
  				acc - 1
  			case _ => acc
  		}
  	} == 0
  }                                               //> balance22: (chars: Array[Char])Boolean

  balance22("")                                   //> res11: Boolean = true
  balance22("123")                                //> res12: Boolean = true
  balance22("(df())")                             //> res13: Boolean = true
  balance22("())(")                               //> res14: Boolean = false
  balance22("(((")                                //> res15: Boolean = false

Based on https://github.com/hsleep/Coursera/blob/master/parprog1/reductions/src/main/scala/reductions/ParallelParenthesesBalancing.scala

  def balance(chars: Iterable[Char]): Boolean = {
    val code = Map('(' -> 1, ')' -> -1).withDefaultValue(0)

    def loop(chLs: List[Char], acc: Int = 0): Int = chLs match {
      case head::tail if acc >= 0 => loop(tail, acc + code(head))
      case _ => acc
    }
    loop(chars.toList) == 0
  }                                               //> balance3: (chars: Array[Char])Boolean

I'm still trying to grasp functional programming concepts and honestly struggling with it. However, I came up with an imperative solution.

  def balance(chars: List[Char]): Boolean = {
    val stack = new mutable.Stack[Char]()
    for (c <- chars) {
      if (c == '(')
        stack.push(c)
      else {
        if (c == ')') {
          if (stack.nonEmpty) {
            stack.pop()
          } else {
            return false
          }
        }
      }
    }
    if (stack.isEmpty) {
      true
    } else
      false
  }

My solutions for Exercises 1, 2, and 3:

  def pascal(c: Int, r: Int): Int = {
    if (c == 0 || c == r) 1 else pascal(c - 1, r - 1) + pascal(c, r - 1)
  }

  def balance(chars: List[Char]): Boolean = {
    def balanced(chars: List[Char], count: Int): Boolean =
      if (chars.isEmpty) count == 0
      else if (count < 0) false
      else if (chars.head == '(') balanced(chars.tail, count + 1)
      else if (chars.head == ')') balanced(chars.tail, count - 1)
      else balanced(chars.tail, count)

    balanced(chars, 0)
  }

  def countChange(money: Int, coins: List[Int]): Int = {
    if (money == 0) 1
    else if (money < 0) 0
    else if (coins.isEmpty) 0
    else countChange(money - coins.head, coins) + countChange(money, coins.tail)
  }

I tweeked @vigneshwaranr's solution switching the if/elses to a match. Is the same implementation, don't know if better or not, judge yourself:

def balance(chars: List[Char]): Boolean = {
  def balanced(chars: List[Char], open: Int): Boolean =
    if (chars.isEmpty) open == 0
    else chars.head match {
      case '(' => balanced(chars.tail, open + 1)
      case ')' => open > 0 && balanced(chars.tail, open - 1)
      case _ => balanced(chars.tail, open)
    }

  balanced(chars, 0)
}

Can anybody please explain how in earth do you get solution for number 1 in just 2 lines of code? I mean, not how it works but how do you notice the pattern in the first place!??? I have a working solution but it takes 10 lines of code. I would have never notice for the life of me that it was that simple. Or is everybody just copying/pasting from the internet?

@PyAntony it helps to break recursive problems down into a base case and a recursive step. If you find yourself stuck, try figuring out your base case(s) first: the simplest possible input (often 0 or 1) where you can just return a value. Then for the recursive step figure out how you'd get from that to the next case. The base case is usually just a statement (or a couple of statements) and the recursive step is then a (tail) recursive function call.

So in the pascal case, the base case is that you are in the first column or the last column, in which case return 1. (I actually prefer to add another base case of an illegal input where c < 0, r < 0 or c > r, in which case I return 0). Then the recursive step is that assuming you already have everything computed up to the current step, you can get the correct value by adding the adjacent values from the row above, i.e. pascal(c-1, r-1) and pascal(c-1, r).

@jeffreyyun's solutions above are the best. But here are mine.

  def pascal(c: Int, r: Int): Int = {
    if (r < 0 || c < 0 || c > r) 0
    else if (c == 0 || c == r) 1
    else pascal(c - 1, r - 1) + pascal(c, r - 1)
  }

  def balance(chars: List[Char]): Boolean = {
    val OPEN = '('
    val CLOSE = ')'
    def balanceRec(chars: List[Char], numOpen: Int): Boolean = {
      (chars) match {
        case (Nil) => numOpen == 0
        case (c :: tail) =>
          if (numOpen < 0) false
          else if (c == OPEN) balanceRec(tail, numOpen + 1)
          else if (c == CLOSE) balanceRec(tail, numOpen - 1)
          else balanceRec(tail, numOpen)
      }
    }
    balanceRec(chars, 0)
  }

  def countChange(money: Int, coins: List[Int]): Int = {
    (coins) match {
      case (Nil) => 0
      case (c :: tail) =>
        if (money < 0) 0
        else if (money == 0) 1
        else countChange(money - c, coins) + countChange(money, tail)
    }
  }

def balance(chars: List[Char]): Boolean = {
  def balanced(chars: List[Char], open: Int): Boolean = {
    if (open < 0) false else
    chars match {
      case Nil => open == 0
      case _ => {
      chars.head match {
        case '(' => balanced(chars.tail, open + 1)
        case ')' => balanced(chars.tail, open - 1)
        case _ => balanced(chars.tail, open)
       }
     }
   }
 }

  balanced(chars, 0)
}

def balance(chars: List[Char]): Boolean = {
  def balanced(chars: List[Char], open: Int): Boolean = {
    if (open < 0) false else
    chars match {
      case Nil => open == 0
      case _ => {
      chars.head match {
        case '(' => balanced(chars.tail, open + 1)
        case ')' => balanced(chars.tail, open - 1)
        case _ => balanced(chars.tail, open)
       }
     }
   }
 }

  balanced(chars, 0)
}

Hi, I have a much more simplified version of the balance code:

def balance(chars : List[Char]) : Boolean = {
val equal : Int = 0
@tailrec
def equality (chars : List[Char], equal : Int) : Boolean = {
if (chars.isEmpty) equal == 0 else {
chars.head match{
case '(' => equality(chars.tail, equal + 1)
case ')' => equality(chars.tail, equal - 1)
case _ => equality(chars.tail, equal)
}
}
}
equality(chars, 0)
}

:) The answer is really not smart. Code just like in C but not functional language. Use recursive function.

Indeed... the change function, I'm sure works but it uses two if not three, more advanced elements not lectured in the course yet... that's not the point.
maps,
for
and listbuffer