stream solution for n sales

src/main/java/Solution.java

@@ -1,20 +1,57 @@
 import java.util.*;
-import java.io.*;
+import java.util.stream.IntStream;
-import java.math.*;
 
-/**
- * Template code to help you parse the standard input 
- * according to the problem statement.
- **/
 class Solution {
 	
 	public static void main( String[] args ) {
-		Scanner in = new Scanner( System.in );
 		// read values with in.next...() methods
-		
+		var scanner = new Scanner( System.in );
-		// code your solution here
+		var sellCount = scanner.nextInt(); 
+		var prices = scanner.tokens()
+				.mapToInt( Integer::valueOf )
+				.toArray();
 		
 		// Write result with System.out.println()
-		System.out.println( "value" );
+		System.out.println( maxProfits( sellCount, prices ));
+	}
+	
+	// code your solution here
+	// 0 - 1 | ... | x-sellCount - x  -->  0 - x-1 | ... | x-1 - x
+	private static int maxProfits( int sellCount, int[] prices ) {
+		return IntStream.range( 0, sellCount )
+				.parallel()
+				.map( sale -> {
+					// recursion end condition
+					if ( sellCount == 1 ) {  
+						return maxProfit( prices );
+					}
+					else {
+						return IntStream.range( 2, prices.length )
+								.parallel()
+								.map( index -> {
+									var firstRange = Arrays.copyOfRange( prices, 0, index );
+									final int max1 = maxProfit( firstRange );
+									
+									var secondRange = Arrays.copyOfRange( prices, index - 1, prices.length );
+									// reduce range sizes recursively
+									final int max2 = maxProfits( sellCount - 1, secondRange );
+									
+									return max1 + max2;
+								}).max().orElse( 0 );
+					}
+				})
+				.max().orElse( 0 );
+	}
+	
+	private static int maxProfit( int[] prices ){
+		return IntStream.range( 0, prices.length )
+				.map( index -> {
+					var range = Arrays.copyOfRange( prices, Math.min( index + 1, prices.length - 1 ), prices.length );
+					int max = IntStream.of( range ).max().orElse( 0 );
+					// difference between current price and future maximum
+					return max - prices[index];
+				})
+				.max().orElse( 0 );
 	}
+	
 }