Find the Number of K-Even Arrays — Python

# Time:  O(n)
# Space: O(n)

import collections


# stars and bars, combinatorics
class Solution(object):
    def countOfArrays(self, n, m, k):
        """
        :type n: int
        :type m: int
        :type k: int
        :rtype: int
        """
        MOD = 10**9+7
        fact, inv, inv_fact = [[1]*2 for _ in xrange(3)]
        def nCr(n, k):
            while len(inv) <= n:  # lazy initialization
                fact.append(fact[-1]*len(inv) % MOD)
                inv.append(inv[MOD%len(inv)]*(MOD-MOD//len(inv)) % MOD)  # https://cp-algorithms.com/algebra/module-inverse.html
                inv_fact.append(inv_fact[-1]*inv[-1] % MOD)
            return (fact[n]*inv_fact[n-k] % MOD) * inv_fact[k] % MOD

        def nHr(n, k):
            return nCr(n+k-1, k)

        lookup = collections.defaultdict(list)
        def pow(a, b):
            while len(lookup[a]) <= b:
                lookup[a].append((lookup[a][-1]*a)%MOD if lookup[a] else 1)
            return lookup[a][b]

        MOD = 10**9+7
        even, odd = m//2, (m+1)//2
        result = 0
        if k == 0:
            result = (result+pow(odd, n))%MOD
        for x in xrange(1, (n+1-k)//2+1):  # since (n-(k+x))-((x+1)-2) >= 0, so x <= (n+1-k)/2
            result = (result+(nHr(x, (k+x)-x)*nHr(x+1, (n-(k+x))-((x+1)-2))*pow(even, k+x)*pow(odd, n-(k+x))%MOD))%MOD
        return result


# Time:  O(n * k)
# Space: O(k)
# dp
class Solution2(object):
    def countOfArrays(self, n, m, k):
        """
        :type n: int
        :type m: int
        :type k: int
        :rtype: int
        """
        MOD = 10**9+7
        even, odd = m//2, (m+1)//2
        dp = [[0]*(k+1) for _ in xrange(2)]
        dp[0][0], dp[1][0] = even, odd
        for _ in xrange(n-1):
            for i in reversed(xrange(k+1)):
                dp[0][i], dp[1][i] = (((dp[0][i-1] if i-1 >= 0 else 0)+dp[1][i])*even)%MOD, ((dp[0][i]+dp[1][i])*odd)%MOD
        return (dp[0][k]+dp[1][k])%MOD