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

# Given two integers n and k, find how many different arrays consist of numbers
# from 1 to n such that there are exactly k inverse pairs.
#
# We define an inverse pair as following: For ith and jth element in the array,
#     if i < j and a[i] > a[j] then it's an inverse pair; Otherwise, it's not.
#
# Since the answer may very large, the answer should be modulo 109 + 7.
#
# Example 1:
# Input: n = 3, k = 0
# Output: 1
# Explanation: 
# Only the array [1,2,3] which consists of numbers from 1 to 3 has exactly 0 inverse pair.
# Example 2:
# Input: n = 3, k = 1
# Output: 2
# Explanation: 
# The array [1,3,2] and [2,1,3] have exactly 1 inverse pair.
# Note:
# The integer n is in the range [1, 1000] and k is in the range [0, 1000].

class Solution(object):
    def kInversePairs(self, n, k):
        """
        :type n: int
        :type k: int
        :rtype: int
        """
        M = 1000000007
        dp = [[0] * (k+1) for _ in xrange(2)]
        dp[0][0] = 1
        for i in xrange(1, n+1):
            dp[i%2] = [0] * (k+1)
            dp[i%2][0] = 1
            for j in xrange(1, k+1):
                dp[i%2][j] = (dp[i%2][j-1] + dp[(i-1)%2][j]) % M
                if j-i >= 0:
                    dp[i%2][j] = (dp[i%2][j] - dp[(i-1)%2][j-i]) % M
        return dp[n % 2][k]

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

# Given two integers n and k, find how many different arrays consist of numbers
# from 1 to n such that there are exactly k inverse pairs.
#
# We define an inverse pair as following: For ith and jth element in the array,
#     if i < j and a[i] > a[j] then it's an inverse pair; Otherwise, it's not.
#
# Since the answer may very large, the answer should be modulo 109 + 7.
#
# Example 1:
# Input: n = 3, k = 0
# Output: 1
# Explanation: 
# Only the array [1,2,3] which consists of numbers from 1 to 3 has exactly 0 inverse pair.
# Example 2:
# Input: n = 3, k = 1
# Output: 2
# Explanation: 
# The array [1,3,2] and [2,1,3] have exactly 1 inverse pair.
# Note:
# The integer n is in the range [1, 1000] and k is in the range [0, 1000].

class Solution(object):
    def kInversePairs(self, n, k):
        """
        :type n: int
        :type k: int
        :rtype: int
        """
        M = 1000000007
        dp = [[0]*(k+1) for _ in xrange(2)]
        dp[0][0] = 1
        for i in xrange(1, n+1):
            dp[i%2] = [0]*(k+1)
            dp[i%2][0] = 1
            for j in xrange(1, k+1):
                dp[i%2][j] = (dp[i%2][j-1] + dp[(i-1)%2][j]) % M
                if j-i >= 0:
                    dp[i%2][j] = (dp[i%2][j] - dp[(i-1)%2][j-i]) % M
        return dp[n%2][k]

Update k-inverse-pairs-array.py