LeetCode: Out of Boundary Paths – Prepare For Coder Interview

Out of Boundary Paths

Similar Problems:

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
Explanation:

Example 2:

Input: m = 1, n = 3, N = 3, i = 0, j = 1
Output: 12
Explanation:

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].

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

Solution:

// https://code.dennyzhang.com/out-of-boundary-paths
// Basic Ideas: dynamic programming
//
//  Notice: what if starting point is out of the grid?
//
//     optimal substructure: pos[i][j] N 
//            <- pos[i-1][j] N-1, pos[i+1][j] N-1, ....
//      
// Complexity: Time O(m*n*N), Space O(m*n*N)
import "math"
func findPaths(m int, n int, N int, i int, j int) int {
    if N == 0 {
        if i>=0 && i<m && j>=0 && j<n {
            return 0
        } else {
            return 1
        }
    }
    mod := int(math.Pow(10, 9)+7)
    l := make([][]int, m)
    for i, _ := range l {
        l[i] = make([]int, n)
    }

    l[i][j] = 1
    res := 0
    // Termination condition: N==1
    for N>0 {
        l2 := make([][]int, m)
        for index_i, _ := range l2 {
            l2[index_i] = make([]int, n)
        }
        for r, row := range l {
            for c, _ := range row {
                for _, offset := range [][]int{[]int{1, 0}, []int{-1, 0}, 
                                               []int{0, 1}, []int{0, -1}} {
                    i2, j2 := r+offset[0], c+offset[1]
                    if !(i2>=0 && i2<m && j2>=0 && j2<n) {
                        res = (res+l[r][c])%mod
                    } else {
                        l2[i2][j2] = (l2[i2][j2] + l[r][c])%mod
                    }
                }
            }
        }
        copy(l, l2)
        N--
    }
    return res
}

// https://code.dennyzhang.com/out-of-boundary-paths
// Basic Ideas: dynamic programming
//
//  Notice: what if starting point is out of the grid?
//
//     optimal substructure: pos[i][j] N 
//            <- pos[i-1][j] N-1, pos[i+1][j] N-1, ....
//      
// Complexity: Time O(m*n*N), Space O(m*n*N)
import "math"
func findPaths(m int, n int, N int, i int, j int) int {
    if N == 0 {
        if i>=0 && i<m && j>=0 && j<n {
            return 0
        } else {
            return 1
        }
    }
    mod := int(math.Pow(10, 9)+7)
    l := make([][]int, m)
    for i, _ := range l {
        l[i] = make([]int, n)
    }

    // Termination condition: N==1
    for N>0 {
        l2 := make([][]int, m)
        for index_i, _ := range l2 {
            l2[index_i] = make([]int, n)
        }
        for index_i, row := range l {
            for index_j, _ := range row {
                for _, offset := range [][]int{[]int{1, 0}, []int{-1, 0}, 
                                               []int{0, 1}, []int{0, -1}} {
                    i2, j2 := index_i+offset[0], index_j+offset[1]
                    if !(i2>=0 && i2<m && j2>=0 && j2<n) {
                        l2[index_i][index_j] = (l2[index_i][index_j] + 1)%mod
                    } else {
                        l2[index_i][index_j] = (l2[index_i][index_j] + l[i2][j2])%mod
                    }
                }
            }
        }
        copy(l, l2)
        N--
    }
    return l[i][j]
}

// https://code.dennyzhang.com/out-of-boundary-paths
// Basic Ideas: dynamic programming
//
//  Notice: what if starting point is out of the grid?
//
//     optimal substructure: pos[i][j] N 
//            <- pos[i-1][j] N-1, pos[i+1][j] N-1, ....
//      
// Complexity: Time O(m*n*N), Space O(m*n*N)
import "math"
func findPaths(m int, n int, N int, i int, j int) int {
    if N == 0 {
        if i>=0 && i<m && j>=0 && j<n {
            return 0
        } else {
            return 1
        }
    }
    mod := int(math.Pow(10, 9)+7)
    l := make([][]int, m)
    // Termination condition: N==1
    for i, _ := range l {
        l[i] = make([]int, n)
        for j, _ := range l[i] {
            if i == 0 || i == m-1 || j == 0 || j == n-1 {
                for _, offset := range [][]int{[]int{1, 0}, []int{-1, 0},
                                               []int{0, 1}, []int{0, -1}} {
                                        i2, j2 := i+offset[0], j+offset[1]
                                        if !(i2>=0 && i2<m && j2>=0 && j2<n) {
                                                l[i][j]++      
                                        }
                                }
            }
        }
    }
        res := l[i][j]
    for k:=2; k<=N; k++ {
                l2 := make([][]int, m)
                for i, _ := range l2 {
                        l2[i] = make([]int, n)
                }
        for i, _ := range l {
            for j, _ := range l[i] {
                for _, offset := range [][]int{[]int{1, 0}, []int{-1, 0}, 
                                               []int{0, 1}, []int{0, -1}} {
                    i2, j2 := i+offset[0], j+offset[1]
                    if i2>=0 && i2<m && j2>=0 && j2<n {
                        l2[i][j] = (l2[i][j] + l[i2][j2])%mod
                    }
                }
            }
        }
        copy(l, l2)
        res = (res + l[i][j])%mod
    }
    return res
}

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