Description
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
Solutions
Dynamic programming + combinatorics, deadly combination.
Language: javascript(2026-09-16 08:34)DONE
CPU Performance33.03%
Memory Performance82.51%
/** By Leetcode */
var numberOfSets = function (n, k) {
const MOD = 1_000_000_007;
const dp = Array(n).fill(1);
const prefixSums = Array(n + 1).fill(0);
for (let j = 0; j < n; j++) {
prefixSums[j + 1] = (prefixSums[j] + dp[j]) % MOD;
}
for (let i = 1; i <= k; i++) {
dp[0] = 0;
for (let j = 1; j < n; j++) {
dp[j] = (dp[j - 1] + prefixSums[j]) % MOD;
}
for (let j = 0; j < n; j++) {
prefixSums[j + 1] = (prefixSums[j] + dp[j]) % MOD;
}
}
return dp[n - 1];
};