LeetCode in Kotlin
1621. Number of Sets of K Non-Overlapping Line Segments
Medium
Given n points on a 1-D plane, where the ith point (from 0 to n-1) is at x = i, find the number of ways we can draw exactly k non-overlapping line segments such that each segment covers two or more points. The endpoints of each segment must have integral coordinates. The k line segments do not have to cover all n points, and they are allowed to share endpoints.
Return the number of ways we can draw k non-overlapping line segments__. Since this number can be huge, return it modulo 109 + 7.
Example 1:
Input: n = 4, k = 2
Output: 5
Explanation: The two line segments are shown in red and blue. The image above shows the 5 different ways {(0,2),(2,3)}, {(0,1),(1,3)}, {(0,1),(2,3)}, {(1,2),(2,3)}, {(0,1),(1,2)}.
Example 2:
Input: n = 3, k = 1
Output: 3
Explanation: The 3 ways are {(0,1)}, {(0,2)}, {(1,2)}.
Example 3:
Input: n = 30, k = 7
Output: 796297179
Explanation: The total number of possible ways to draw 7 line segments is 3796297200. Taking this number modulo 109 + 7 gives us 796297179.
Constraints:
2 <= n <= 10001 <= k <= n-1
Solution
class Solution {
fun numberOfSets(n: Int, k: Int): Int {
return if (n - 1 >= k) {
val dp = IntArray(k)
val sums = IntArray(k)
val mod = (1e9 + 7).toInt()
for (diff in 1 until n - k + 1) {
dp[0] = (diff + 1) * diff shr 1
sums[0] = (sums[0] + dp[0]) % mod
for (segments in 2..k) {
dp[segments - 1] = (sums[segments - 2] + dp[segments - 1]) % mod
sums[segments - 1] = (sums[segments - 1] + dp[segments - 1]) % mod
}
}
dp[k - 1]
} else {
0
}
}
}