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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
            f = [1, 0, 0, 0]
            for _ in range(k):
                g = [0] * 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 f[0] if source[1] == dest[1] else f[2]
            return f[1] if source[1] == dest[1] else f[3]
    
    
  • 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]
    }
    

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