Coin Change 2 - Day 7

Author: suriyakrishna

src/main/scala/com/kishan/scala/leetcode/juneChallenges/CoinChange2.scala

@@ -0,0 +1,81 @@
+package com.kishan.scala.leetcode.juneChallenges
+
+
+/*
+* You are given coins of different denominations and a total amount of money.
+* Write a function to compute the number of combinations that make up that amount.
+* You may assume that you have infinite number of each kind of coin.
+*/
+
+
+/*
+* Input: amount = 5, coins = [1, 2, 5]
+  Output: 4
+  Explanation: there are four ways to make up the amount:
+  5=5
+  5=2+2+1
+  5=2+1+1+1
+  5=1+1+1+1+1
+* */
+
+/*
+* Approach: Dynamic Programming
+*
+* Assume above inputs. Creating Empty MultiDimensional Array filled with 0(Space complexity -> 0(n+1*m+1))
+* n -> Length coins array
+* m -> value of the amount
+*
+* 0,1,2,3,4,5
+* [
+* 0       -> [1, 0, 0, 0, 0, 0]
+* 1       -> [1, 1, 1, 1, 1, 1]
+* [1,2]   -> [1, 1, 1+1, 1+1, 1+2, 1+2]
+* [1,2,5] -> [1, 1, 2, 2, 3, 3+1]
+* ]
+*
+* In order to get the probability we have use the below formula
+* element[i][j] = element[i-1][j]
+* when we have difference of j and coin is greater than 0 we have to get the value from the same row for the corresponding index and we have to add it to current index
+* element[i][j] = element[i-1][j] + element[i][j-coin]
+*/
+
+
+object CoinChange2 {
+  def main(args: Array[String]): Unit = {
+    val amount = 5
+    val coins = Array(1, 2, 5)
+    val result = changeSingleDimesionArray(amount, coins)
+    println(result, result == 4)
+  }
+
+  // Memory Efficient
+  def changeSingleDimesionArray(amount: Int, coins: Array[Int]): Int = {
+    val length = coins.length
+    var dp: Array[Int] = Array.ofDim[Int](amount + 1)
+    dp(0) = 1
+    for (i <- 1 until length + 1) {
+      for (j <- 1 until amount + 1) {
+        if (j - coins(i - 1) >= 0) {
+          dp(j) += dp(j - coins(i - 1))
+        }
+      }
+    }
+    return dp(amount)
+  }
+
+  // Multi dimesional Array
+  def change(amount: Int, coins: Array[Int]): Int = {
+    val length = coins.length
+    var dp: Array[Array[Int]] = Array.ofDim[Int](length + 1, amount + 1)
+    dp.foreach(a => a(0) = 1)
+    for (i <- 1 until length + 1) {
+      for (j <- 1 until amount + 1) {
+        dp(i)(j) = dp(i - 1)(j)
+        if (j - coins(i - 1) >= 0) {
+          dp(i)(j) += dp(i)(j - coins(i - 1))
+        }
+      }
+    }
+    return dp(length)(amount)
+  }
+}