Add solution #757

Author: JoshCrozier

README.md

@@ -574,6 +574,7 @@
 753|[Cracking the Safe](./0753-cracking-the-safe.js)|Hard|
 754|[Reach a Number](./0754-reach-a-number.js)|Medium|
 756|[Pyramid Transition Matrix](./0756-pyramid-transition-matrix.js)|Medium|
+757|[Set Intersection Size At Least Two](./0757-set-intersection-size-at-least-two.js)|Hard|
 762|[Prime Number of Set Bits in Binary Representation](./0762-prime-number-of-set-bits-in-binary-representation.js)|Easy|
 763|[Partition Labels](./0763-partition-labels.js)|Medium|
 783|[Minimum Distance Between BST Nodes](./0783-minimum-distance-between-bst-nodes.js)|Easy|

solutions/0757-set-intersection-size-at-least-two.js

@@ -0,0 +1,45 @@
+/**
+ * 757. Set Intersection Size At Least Two
+ * https://leetcode.com/problems/set-intersection-size-at-least-two/
+ * Difficulty: Hard
+ *
+ * You are given a 2D integer array intervals where intervals[i] = [starti, endi] represents
+ * all the integers from starti to endi inclusively.
+ *
+ * A containing set is an array nums where each interval from intervals has at least two
+ * integers in nums.
+ *
+ * For example, if intervals = [[1,3], [3,7], [8,9]], then [1,2,4,7,8,9] and [2,3,4,8,9]
+ * are containing sets.
+ *
+ * Return the minimum possible size of a containing set.
+ */
+
+/**
+ * @param {number[][]} intervals
+ * @return {number}
+ */
+var intersectionSizeTwo = function(intervals) {
+  intervals.sort((a, b) => a[1] - b[1] || b[0] - a[0]);
+  let size = 0;
+  let smallest = -1;
+  let secondSmallest = -1;
+
+  for (const [start, end] of intervals) {
+    const hasTwo = start <= smallest;
+    const hasOne = start <= secondSmallest;
+
+    if (hasTwo) continue;
+    if (hasOne) {
+      smallest = secondSmallest;
+      secondSmallest = end;
+      size++;
+    } else {
+      smallest = end - 1;
+      secondSmallest = end;
+      size += 2;
+    }
+  }
+
+  return size;
+};