LeetCode in Kotlin
629. K Inverse Pairs Array
Hard
For an integer array nums, an inverse pair is a pair of integers [i, j] where 0 <= i < j < nums.length and nums[i] > nums[j].
Given two integers n and k, return the number of different arrays consist of numbers from 1 to n such that there are exactly k inverse pairs. Since the answer can be huge, return it 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 pairs.
Example 2:
Input: n = 3, k = 1
Output: 2
Explanation: The array [1,3,2] and [2,1,3] have exactly 1 inverse pair.
Constraints:
1 <= n <= 10000 <= k <= 1000
Solution
@Suppress("NAME_SHADOWING")
class Solution {
fun kInversePairs(n: Int, k: Int): Int {
var k = k
k = Math.min(k, n * (n - 1) / 2 - k)
if (k < 0) {
return 0
}
var dp = IntArray(k + 1)
var dp1 = IntArray(k + 1)
dp[0] = 1
dp1[0] = 1
val mod = 1000000007
for (i in 1..n) {
val temp = dp
dp = dp1
dp1 = temp
var j = 1
val m = Math.min(k, i * (i - 1) / 2)
while (j <= m) {
dp[j] = (dp1[j] + dp[j - 1] - if (j >= i) dp1[j - i] else 0) % mod
if (dp[j] < 0) {
dp[j] += mod
}
j++
}
}
return dp[k]
}
}