# 2912. Number of Ways to Reach Destination in the Grid

## Description

You are given two integers n and m which represent the size of a 1-indexed grid. You are also given an integer k, a 1-indexed integer array source and a 1-indexed integer array dest, where source and dest are in the form [x, y] representing a cell on the given grid.

You can move through the grid in the following way:

• You can go from cell [x1, y1] to cell [x2, y2] if either x1 == x2 or y1 == y2.
• Note that you can't move to the cell you are already in e.g. x1 == x2 and y1 == y2.

Return the number of ways you can reach dest from source by moving through the grid exactly k times.

Since the answer may be very large, return it modulo 109 + 7.

Example 1:

Input: n = 3, m = 2, k = 2, source = [1,1], dest = [2,2]
Output: 2
Explanation: There are 2 possible sequences of reaching [2,2] from [1,1]:
- [1,1] -> [1,2] -> [2,2]
- [1,1] -> [2,1] -> [2,2]


Example 2:

Input: n = 3, m = 4, k = 3, source = [1,2], dest = [2,3]
Output: 9
Explanation: There are 9 possible sequences of reaching [2,3] from [1,2]:
- [1,2] -> [1,1] -> [1,3] -> [2,3]
- [1,2] -> [1,1] -> [2,1] -> [2,3]
- [1,2] -> [1,3] -> [3,3] -> [2,3]
- [1,2] -> [1,4] -> [1,3] -> [2,3]
- [1,2] -> [1,4] -> [2,4] -> [2,3]
- [1,2] -> [2,2] -> [2,1] -> [2,3]
- [1,2] -> [2,2] -> [2,4] -> [2,3]
- [1,2] -> [3,2] -> [2,2] -> [2,3]
- [1,2] -> [3,2] -> [3,3] -> [2,3]


Constraints:

• 2 <= n, m <= 109
• 1 <= k <= 105
• source.length == dest.length == 2
• 1 <= source[1], dest[1] <= n
• 1 <= source[2], dest[2] <= m

## Solutions

Solution 1: Dynamic Programming

We define the following states:

• $f[0]$ represents the number of ways to move from source to source itself;
• $f[1]$ represents the number of ways to move from source to another row in the same column;
• $f[2]$ represents the number of ways to move from source to another column in the same row;
• $f[3]$ represents the number of ways to move from source to another row and another column.

Initially, $f[0] = 1$, and the other states are all $0$.

For each state, we can calculate the current state based on the previous state, as follows:

\begin{aligned} g[0] &= (n - 1) \times f[1] + (m - 1) \times f[2] \\ g[1] &= f[0] + (n - 2) \times f[1] + (m - 1) \times f[3] \\ g[2] &= f[0] + (m - 2) \times f[2] + (n - 1) \times f[3] \\ g[3] &= f[1] + f[2] + (n - 2) \times f[3] + (m - 2) \times f[3] \end{aligned}

We loop $k$ times, and finally check whether source and dest are in the same row or column, and return the corresponding state.

The time complexity is $O(k)$, where $k$ is the number of moves. The space complexity is $O(1)$.

• class Solution {
public int numberOfWays(int n, int m, int k, int[] source, int[] dest) {
final int mod = 1000000007;
long[] f = new long[4];
f[0] = 1;
while (k-- > 0) {
long[] g = new long[4];
g[0] = ((n - 1) * f[1] + (m - 1) * f[2]) % mod;
g[1] = (f[0] + (n - 2) * f[1] + (m - 1) * f[3]) % mod;
g[2] = (f[0] + (m - 2) * f[2] + (n - 1) * f[3]) % mod;
g[3] = (f[1] + f[2] + (n - 2) * f[3] + (m - 2) * f[3]) % mod;
f = g;
}
if (source[0] == dest[0]) {
return source[1] == dest[1] ? (int) f[0] : (int) f[2];
}
return source[1] == dest[1] ? (int) f[1] : (int) f[3];
}
}

• class Solution {
public:
int numberOfWays(int n, int m, int k, vector<int>& source, vector<int>& dest) {
const int mod = 1e9 + 7;
vector<long long> f(4);
f[0] = 1;
while (k--) {
vector<long long> g(4);
g[0] = ((n - 1) * f[1] + (m - 1) * f[2]) % mod;
g[1] = (f[0] + (n - 2) * f[1] + (m - 1) * f[3]) % mod;
g[2] = (f[0] + (m - 2) * f[2] + (n - 1) * f[3]) % mod;
g[3] = (f[1] + f[2] + (n - 2) * f[3] + (m - 2) * f[3]) % mod;
f = move(g);
}
if (source[0] == dest[0]) {
return source[1] == dest[1] ? f[0] : f[2];
}
return source[1] == dest[1] ? f[1] : f[3];
}
};

• class Solution:
def numberOfWays(
self, n: int, m: int, k: int, source: List[int], dest: List[int]
) -> int:
mod = 10**9 + 7
a, b, c, d = 1, 0, 0, 0
for _ in range(k):
aa = ((n - 1) * b + (m - 1) * c) % mod
bb = (a + (n - 2) * b + (m - 1) * d) % mod
cc = (a + (m - 2) * c + (n - 1) * d) % mod
dd = (b + c + (n - 2) * d + (m - 2) * d) % mod
a, b, c, d = aa, bb, cc, dd
if source[0] == dest[0]:
return a if source[1] == dest[1] else c
return b if source[1] == dest[1] else d


• func numberOfWays(n int, m int, k int, source []int, dest []int) int {
const mod int = 1e9 + 7
f := []int{1, 0, 0, 0}
for i := 0; i < k; i++ {
g := make([]int, 4)
g[0] = ((n-1)*f[1] + (m-1)*f[2]) % mod
g[1] = (f[0] + (n-2)*f[1] + (m-1)*f[3]) % mod
g[2] = (f[0] + (m-2)*f[2] + (n-1)*f[3]) % mod
g[3] = (f[1] + f[2] + (n-2)*f[3] + (m-2)*f[3]) % mod
f = g
}

if source[0] == dest[0] {
if source[1] == dest[1] {
return f[0]
}
return f[2]
}

if source[1] == dest[1] {
return f[1]
}
return f[3]
}