Add solution #2523

Author: JoshCrozier

README.md

@@ -745,6 +745,7 @@
 2482|[Difference Between Ones and Zeros in Row and Column](./2482-difference-between-ones-and-zeros-in-row-and-column.js)|Medium|
 2490|[Circular Sentence](./2490-circular-sentence.js)|Easy|
 2493|[Divide Nodes Into the Maximum Number of Groups](./2493-divide-nodes-into-the-maximum-number-of-groups.js)|Hard|
+2523|[Closest Prime Numbers in Range](./2523-closest-prime-numbers-in-range.js)|Medium|
 2529|[Maximum Count of Positive Integer and Negative Integer](./2529-maximum-count-of-positive-integer-and-negative-integer.js)|Easy|
 2535|[Difference Between Element Sum and Digit Sum of an Array](./2535-difference-between-element-sum-and-digit-sum-of-an-array.js)|Easy|
 2542|[Maximum Subsequence Score](./2542-maximum-subsequence-score.js)|Medium|

solutions/2523-closest-prime-numbers-in-range.js

@@ -0,0 +1,44 @@
+/**
+ * 2523. Closest Prime Numbers in Range
+ * https://leetcode.com/problems/closest-prime-numbers-in-range/
+ * Difficulty: Medium
+ *
+ * Given two positive integers left and right, find the two integers num1 and num2 such that:
+ * - left <= num1 < num2 <= right
+ * - Both num1 and num2 are prime numbers.
+ * - num2 - num1 is the minimum amongst all other pairs satisfying the above conditions.
+ *
+ * Return the positive integer array ans = [num1, num2]. If there are multiple pairs satisfying
+ * these conditions, return the one with the smallest num1 value. If no such numbers exist,
+ * return [-1, -1].
+ */
+
+/**
+ * @param {number} left
+ * @param {number} right
+ * @return {number[]}
+ */
+var closestPrimes = function(left, right) {
+  const group = new Uint8Array(right + 1).fill(1);
+  let result = [-1, -1];
+
+  group[0] = group[1] = 0;
+  for (let i = 2; i * i <= right; i++) {
+    if (group[i]) {
+      for (let j = i * i; j <= right; j += i) {
+        group[j] = 0;
+      }
+    }
+  }
+  for (let i = Math.max(2, left), prev = -1, min = Infinity; i <= right; i++) {
+    if (group[i]) {
+      if (prev !== -1 && i - prev < min) {
+        min = i - prev;
+        result = [prev, i];
+      }
+      prev = i;
+    }
+  }
+
+  return result;
+};