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OutOfBoundaryPaths.java
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OutOfBoundaryPaths.java
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package Leetcode;
/**
* @author kalpak
*
* There is an m by n grid with a ball.
* Given the start coordinate (i,j) of the ball, you can move the ball to adjacent cell or cross the grid boundary in four directions (up, down, left, right).
* However, you can at most move N times. Find out the number of paths to move the ball out of grid boundary.
*
* The answer may be very large, return it after mod 109 + 7.
*
*
* Example 1:
* Input: m = 2, n = 2, N = 2, i = 0, j = 0
* Output: 6
*
* Example 2:
* Input: m = 1, n = 3, N = 3, i = 0, j = 1
* Output: 12
*
*
* Note:
*
* Once you move the ball out of boundary, you cannot move it back.
* The length and height of the grid is in range [1,50].
* N is in range [0,50].
*
*/
public class OutOfBoundaryPaths {
private static final int MOD = 1000000007;
public static int findPathsRecursive(int m, int n, int N, int i, int j) {
if(i == m || j == n || i < 0 || j < 0)
return 1; // path available to take the ball out
if(N == 0)
return 0;
return findPathsRecursive(m, n, N - 1, i - 1, j) +
findPathsRecursive(m, n, N - 1, i + 1, j) +
findPathsRecursive(m, n, N - 1, i, j - 1) +
findPathsRecursive(m, n, N - 1, i, j + 1);
}
public static int findPaths(int m, int n, int N, int i, int j) {
if (m < 0 || n < 0 || N < 0 || i < 0 || i >= m || j < 0 || j >= n)
return 0;
Long[][][] dp = new Long[m][n][N + 1];
return (int)findPathsDFS(dp, m, n, N, i, j);
}
private static long findPathsDFS(Long[][][] dp, int m, int n, int N, int i, int j) {
if(i < 0 || j < 0 || i >= m || j >= n)
return 1;
if(N < 1)
return 0;
if (dp[i][j][N] != null)
return dp[i][j][N];
long down = findPathsDFS(dp, m, n, N - 1,i + 1, j);
long up = findPathsDFS(dp, m, n, N - 1, i - 1, j);
long left = findPathsDFS(dp, m, n, N - 1, i, j - 1);
long right = findPathsDFS(dp, m, n, N - 1, i, j + 1);
dp[i][j][N] = (down + up + left + right) % MOD;
return dp[i][j][N];
}
public static void main(String[] args) {
System.out.println(findPaths(1, 3, 3, 0, 1));
}
}