[Leetcode] DP--629. K Inverse Pairs Array
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.
Solution:
This problem's dp solution is challenging to construct
1.
reference the tutorial of the solution; https://leetcode.com/articles/k-inverse-pairs-array/
very great explanation for 8 approaches, I only came up with the naive method and the recursive method
(1) Define the subproblem
dp[i][j] as the given integer i and the inverse pairs number j
dp[n][k] is the final answer
(2) Find the recursion
dp[i][j] = dp[i-1][j-p] for k = 0 to min(i-1, j)
(3) Get the base case
dp[0][j] = 0 for j = 0 to n
dp[i][0] = 1 for i = 0 to n
1 M = 10**9 + 7 2 dp = [[0]*(k+1) for i in range(0,n+1)] 3 4 #print ("dp: ", dp) 5 for i in range(1, n+1): 6 for j in range(0, k+1): 7 #print ("dp: ", i, j, dp[i][j]) 8 if j == 0: 9 dp[i][j] = 1 10 #print ("atttt: ", i, j, dp[i][0]) 11 12 else: 13 for p in range(0, min(j, i-1)+1): 14 dp[i][j] = (dp[i][j] + dp[i-1][j-p])%M 15 #print ("daaaap: ", i, j, dp[i][0]) 16 17 return dp[n][k]
It has TLE problem
2.
further to observe that if we use accumulated sum for dp[i][j], we can have
dp[i−1][j]+dp[i−1][j−i] if j−i≥0. Otherwise, we add all the elements of the previous row upto the current column j being considered.
1 M = 10**9 + 7 2 dp = [[0]*(k+1) for i in range(0,n+1)] 3 4 #print ("dp: ", dp) 5 for i in range(1, n+1): 6 for j in range(0, k+1): 7 #print ("dp: ", i, j, dp[i][j]) 8 if j == 0: 9 dp[i][j] = 1 10 #print ("atttt: ", i, j, dp[i][0]) 11 12 else: 13 if j >= i: 14 val = (dp[i-1][j] + M - dp[i-1][j-i]) % M 15 else: 16 val = (dp[i-1][j] + M - 0) % M 17 dp[i][j] = (dp[i][j-1] + val)%M 18 #print ("daaaap: ", i, j, dp[i][0]) 19 20 if k > 0: 21 return (dp[n][k] + M - dp[n][k-1])%M 22 else: 23 return (dp[n][k] + M - 0)%M
