Add solution #1621

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,426 LeetCode solutions in JavaScript
+# 1,427 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1250,6 +1250,7 @@
 1617|[Count Subtrees With Max Distance Between Cities](./solutions/1617-count-subtrees-with-max-distance-between-cities.js)|Hard|
 1619|[Mean of Array After Removing Some Elements](./solutions/1619-mean-of-array-after-removing-some-elements.js)|Easy|
 1620|[Coordinate With Maximum Network Quality](./solutions/1620-coordinate-with-maximum-network-quality.js)|Medium|
+1621|[Number of Sets of K Non-Overlapping Line Segments](./solutions/1621-number-of-sets-of-k-non-overlapping-line-segments.js)|Medium|
 1657|[Determine if Two Strings Are Close](./solutions/1657-determine-if-two-strings-are-close.js)|Medium|
 1668|[Maximum Repeating Substring](./solutions/1668-maximum-repeating-substring.js)|Easy|
 1669|[Merge In Between Linked Lists](./solutions/1669-merge-in-between-linked-lists.js)|Medium|

solutions/1621-number-of-sets-of-k-non-overlapping-line-segments.js

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+/**
+ * 1621. Number of Sets of K Non-Overlapping Line Segments
+ * https://leetcode.com/problems/number-of-sets-of-k-non-overlapping-line-segments/
+ * Difficulty: 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.
+ */
+
+/**
+ * @param {number} n
+ * @param {number} k
+ * @return {number}
+ */
+var numberOfSets = function(n, k) {
+  const MOD = 1e9 + 7;
+  const comb = Array.from({ length: n + k }, () => Array(n + k).fill(0));
+
+  for (let i = 0; i < n + k; i++) {
+    comb[i][0] = 1;
+    for (let j = 1; j <= i; j++) {
+      comb[i][j] = (comb[i - 1][j - 1] + comb[i - 1][j]) % MOD;
+    }
+  }
+
+  return comb[n + k - 1][2 * k];
+};