Implemented LargestProductInGrid problem in Java

Solutions/LargestProductInGrid.java

@@ -0,0 +1,77 @@
+package com.luko.LargestProductInGrid;
+
+/**
+ * This short program looks for the cell, that has the biggest sum, when added to their neighbour cells
+ * and prints their position and sum at the end.
+ * @author Lukas Kohlmaier
+ *
+ */
+public class LargestProductInGrid {
+	static int[][] numGrid = { 
+			{ 1,2, 77 },
+			{ 2, 34, 90 },
+			{ 43, 2, 23 },
+			{ 15, 11, 60 }
+	};
+
+
+	public static void main(String[] args) {
+		int sumVar = 0;
+		int xRes = 0;
+		int yRes = 0;
+
+		for(int x = 0; x < numGrid.length; x++) {
+			for(int y = 0; y < numGrid[0].length; y++) {
+				if(calculateGridSum(x, y) > sumVar) {
+					sumVar = calculateGridSum(x, y);
+					xRes = x;
+					yRes = y;
+				}
+			}
+		}
+		System.out.println("The cell with the biggest grid sum is x= " + xRes + " y= " + yRes + " sum: " + sumVar);
+	}
+
+	/**
+	 * 
+	 * @param x The x coordinate of the position to calculate the sum for.
+	 * @param y The y coordinate of the position to calculate the sum for.
+	 * @return The sum of the given cell position with their direct neighbours.
+	 */
+	private static int calculateGridSum(int x, int y) {
+		int finalSum = 0;
+
+		if(y > 0) { //Adding the item above
+			finalSum += numGrid[x][y - 1];
+
+			if(x > 0) { //Adding the item above left
+				finalSum += numGrid[x - 1][y - 1];
+			}
+
+			if(numGrid.length - x > 1) { //Adding the item above right
+				finalSum += numGrid[x + 1][y - 1];
+			}
+		}
+
+		if(x > 0) { //Adding the item left
+			finalSum += numGrid[x - 1][y];
+		}
+
+		if(numGrid.length - x > 1) { //Adding the item right
+			finalSum += numGrid[x + 1][y];
+		}
+
+		if(numGrid[0].length - y > 1) { //Adding the item below
+			finalSum += numGrid[x][y + 1];
+
+			if(x > 0) { //Adding the item below left
+				finalSum += numGrid[x - 1][y + 1];
+			}
+
+			if(numGrid.length - x > 1) { //Adding the item below right
+				finalSum += numGrid[x + 1][y + 1];
+			}
+		}
+		return finalSum;
+	}
+}