To find the minimum start & end point of a n*n matrix, i.e. distance between (0, 0)  & (n-1, n-1). Value at each block represents the cost.


public class Paths {

 public static void main(String[] args) {
  int[] input2 = {5,7,4,6,2,3,5,1,3,1,9,7,1,6,8,8};
  int[] input1 = {4,4};
  System.out.println("Minimum cost : " + no_of_path(input1, input2));
 }

 public static int no_of_path(int[] input1,int[] input2)
 {
  int[][] arr = new int[input1[0]][input1[1]];
  int k = 0;

  for(int i = 0; i < input1[0]; i++){
   for(int j = 0; j < input1[1]; j++){
    arr[i][j] = input2[k];
    k++;
   }
  }

  int cost = paths(arr, 0, 0, input1[0], input1[1]);
  return cost;
 }

 public static int paths(int[][] arr, int i, int j, int row, int col){
  int c = 0;
  int d = 0;
  int r = 0;
  if(i == row-1 && j == col-1)
   return arr[i][j];
  if(i < row && j+1 < col){
   c = paths(arr, i, j+1, row, col);
   c = c + arr[i][j];
  }
  if(i+1 < row && j+1 < col){
   d = paths(arr, i+1, j+1, row, col);
   d = d + arr[i][j];
  }
  if(i+1 < row && j < col){
   r = paths(arr, i+1, j, row, col);
   r = r + arr[i][j];
  }

  if(c ==0 || d == 0 || r == 0)
   return c+d+r;
  else if(c <= d && c <=r)
   return c;
  else if(d <= c && d <=r)
   return d;
  else if(r <= c && r <=d)
   return r;
  else
   return 0;
 }
}