2400. Number of Ways to Reach a Position After Exactly k Steps

class Solution:
    def numberOfWays(self, startPos: int, endPos: int, k: int) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if i > j or j < 0:
                return 0
            if j == 0:
                return 1 if i == 0 else 0
            return (dfs(i + 1, j - 1) + dfs(abs(i - 1), j - 1)) % mod

        mod = 10**9 + 7
        return dfs(abs(startPos - endPos), k)