I guess you all know the "strawberry" problem that some give you in job interviews, where you need to calculate the path between 2 corners of a 2D array that you can only move up or to the right and you have the calculate the maximum valued path. I have a perfectly working code that does it in Recursion, but it's complexity is to high. i also solved the problem in the "for loop" solution that does it in O(n^2) complexity. but in this solution i just couldn't figure out a way to print the route like i did in the recursion solution. This is my code (it is quite long to read here so i guess you should copy,compile and run). look at the results of the recursion solution, BTW - The path needs to be from the left bottom corner to the right upper corner I want to print the route the same way in the better solution:
public class Alg
{
public static void main(String args[])
{
String[] route = new String[100];
int[][]array = {{4,-2,3,6}
,{9,10,-4,1}
,{-1,2,1,4}
,{0,3,7,-3}};
String[][] route2 = new String[array.length][array[0].length];
int max = recursionAlg(array,array.length-1,0,route);
int max2 = loopAlg(array,array.length-1,0,route2);
System.out.println("The max food in the recursion solution is: "+max);
System.out.println("and the route is: ");
printRouteArray(route);
System.out.println("The max food in the loop solution: "+max2);
System.out.println("The route is: ");
//SHOULD PRINT HERE THE ROUTE
}
public static int loopAlg(int [][] arr,int x, int y, String[][] route)
{
int n=0;
int[][]count = new int[arr.length][arr[0].length];
for(int i = x; i>=0 ; i--)
{
for(int j = 0; j<arr[0].length; j++)
{
if (i==x && j==0) {count[i][j]=arr[i][j];}
else if (i == x) { count[i][j]=count[i][j-1]+arr[i][j];}
else if (j == 0) { count[i][j]=count[i+1][j]+arr[i][j]; }
else{
if (count[i][j-1]>count[i+1][j]) {count[i][j]=count[i][j-1]+arr[i][j];}
else { count[i][j]= count[i+1][j]+arr[i][j];}
}
}
}
return count[0][arr[0].length-1];
}
public static int recursionAlg(int [][] arr, int x, int y,String[] route)
{
return recursionAlg(arr,0,x,y,arr[0].length-1,route,0);
}
public static int recursionAlg(int[][]arr,int count,int x, int y, int max_y, String[] route, int i)
{
if (x == 0 && y == max_y) {return count;}
else if (x == 0) {
route[i]="Right";
return recursionAlg(arr,count+arr[x][y+1],x,y+1,max_y,route,i+1);
}
else if (y==max_y){
route[i]="Up";
return recursionAlg(arr,count+arr[x-1][y],x-1,y,max_y,route,i+1);
}
else if (recursionAlg(arr,count+arr[x-1][y],x-1,y,max_y,route,i+1)>recursionAlg(arr,count+arr[x][y+1],x,y+1,max_y,route,i+1))
{
route[i]="Up";
return recursionAlg(arr,count+arr[x-1][y],x-1,y,max_y,route,i+1);
}
else
{
route[i]="Right";
return recursionAlg(arr,count+arr[x][y+1],x,y+1,max_y,route,i+1);
}
}
public static void printRouteArray(String[] arr)
{
int i=0;
while (i<arr.length && (arr[i]=="Up" || arr[i]=="Right"))
{
System.out.print(arr[i]+"-->");
i++;
}
System.out.println("End");
}
}
Hope you can help, thanks!
You need another 2-dimensional array inside
loopAlg
that memorizes which step to take to come to this next entry for every entry in your initial 2-dim array. See the following code and https://ideone.com/kM8BAZ for a demo: