Reputation: 703
Given 2D array of positive integers. A "path" is a collection of adjacent cells. Two cells are adjacent only from right/left/top/bottom (no diagonal).
The task is to write a function that receives a 2D array mat
, an integer sum
and a 2D array path
(with same size as mat
-empty array all zeros).
The function should check if path exist where the sum of cells equal to sum
, should return true if exist and false otherwise.
The array path
will mark the path (if exist with 1
).
For example if mat
is:
and sum=4
then path
can be one of these three:
My code:
public static void main(String[] args)
{
int[][] mat={{2,41,3,14},
{2,1,24,7},
{2,15,10,54},
{63,22,2,4}};
int[][] path={{0,0,0,0},
{0,0,0,0},
{0,0,0,0},
{0,0,0,0}};
findSum(mat,4,path);
//print(mat);
print(path);
}
public static boolean findSum (int mat[][], int sum, int path[][])
{
return findSum(mat,sum,path,0,0);
}
public static boolean findSum (int mat[][], int sum, int path[][],int i,int j)
{
if(sum==0)
return true;
if(i==mat[0].length||j==mat[1].length)
return false;
boolean result=findSum(mat,sum-mat[i][j],path,i,j+1)||findSum(mat,sum-mat[i][j],path,i+1,j);
if(result)
path[i][j]=1;
return result;
}
private static void print(int[][] arr)
{
for(int i=0;i<arr[0].length;i++)
{
for(int j=0;j<arr[0].length;j++)
{
System.out.print(arr[i][j]+" ");
}
System.out.println();
}
}
My code works fine only if the path starts at (0,0) but does't work for other pathes, for example it doesn't work(path array is all zero) for
sum=16
even that there is such path.
Note:
Upvotes: 4
Views: 774
Reputation: 1032
Nice question... Here is the answer. It was a fun code challenge ;)
public static void main(String[] args)
{
int[][] mat={{2,41,3,14},
{2,1,24,7},
{2,15,10,54},
{63,22,2,4}};
int[][] path={{0,0,0,0},
{0,0,0,0},
{0,0,0,0},
{0,0,0,0}};
if ( findSum(mat,22,path) ) print(path);
else System.out.println("No path found");
}
public static boolean findSum (int mat[][], int sum, int path[][])
{
return startPath(mat, sum, path, -1, 0);
}
// Recursively check every possible starting point
public static boolean startPath(int mat[][], int sum, int path[][], int y, int x)
{
// Iterate y, goto next column if necessary
if (++y == mat.length) {
y = 0;
++x;
}
if (x == mat[0].length) // Bounds check
return false;
if (findSum(mat, sum, path, y, x)) // We've found a successful start point!
{
System.out.println("A successful path starts at " + x + ", " + y);
return true;
}
return startPath(mat, sum, path, y, x); // We'll have to keep looking
}
public static boolean findSum (int mat[][], int sum, int path[][], int i, int j)
{
if(i==mat[0].length || j==mat[1].length || i<0 || j<0) // Bounds check
return false;
if (path[i][j] == 1) // Backtracking check
return false;
sum -= mat[i][j]; // Decrement sum
if (sum >= 0) { // More to go? look around
path[i][j] = 1;
if (sum == 0) return true; // We made it!
// If any path finds the end, don't try other paths
boolean result = findSum(mat, sum, path, i+1, j);
if (result) return true;
result = findSum(mat, sum, path, i, j+1);
if (result) return true;
result = findSum(mat, sum, path, i-1, j);
if (result) return true;
result = findSum(mat, sum, path, i, j-1);
// There was no successful paths, this is a dead end
if (!result) path[i][j] = 0;
return result;
} else { // We're negative, overshot it
return false;
}
}
private static void print(int[][] arr)
{
for(int i=0;i<arr[0].length;i++)
{
for(int j=0;j<arr[0].length;j++)
{
System.out.print(arr[i][j]+" ");
}
System.out.println();
}
}
By the way when checking multidimensional array dimensions you are meaning to do arr.length and arr[0].length
but instead you are doing arr[0].length and arr[1].length
which would get the same dimension twice and cause an error if the array was not a square.
I've also allowed moving in any direction, by using the intended path to prevent re-checking the same node again. This could probably be a boolean array. Sorry if my x/y recursion is rough looking... I really prefer loops or .forEach or => or .map for that...
The -1, 0 can be cleaned up perhaps with optional parameters and not iterating on the first pass.
Let me know if there are any errors or edge cases you may find.
Upvotes: 1