Add solution #2048

JoshCrozier · JoshCrozier · commit 356288021f4c · 2025-05-08T13:17:02.000-04:00
diff --git a/README.md b/README.md
@@ -1,4 +1,4 @@
-# 1,708 LeetCode solutions in JavaScript
+# 1,709 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1570,6 +1570,7 @@
 2044|[Count Number of Maximum Bitwise-OR Subsets](./solutions/2044-count-number-of-maximum-bitwise-or-subsets.js)|Medium|
 2045|[Second Minimum Time to Reach Destination](./solutions/2045-second-minimum-time-to-reach-destination.js)|Hard|
 2047|[Number of Valid Words in a Sentence](./solutions/2047-number-of-valid-words-in-a-sentence.js)|Easy|
+2048|[Next Greater Numerically Balanced Number](./solutions/2048-next-greater-numerically-balanced-number.js)|Medium|
 2053|[Kth Distinct String in an Array](./solutions/2053-kth-distinct-string-in-an-array.js)|Medium|
 2071|[Maximum Number of Tasks You Can Assign](./solutions/2071-maximum-number-of-tasks-you-can-assign.js)|Hard|
 2085|[Count Common Words With One Occurrence](./solutions/2085-count-common-words-with-one-occurrence.js)|Easy|
diff --git a/solutions/2048-next-greater-numerically-balanced-number.js b/solutions/2048-next-greater-numerically-balanced-number.js
@@ -0,0 +1,39 @@
+/**
+ * 2048. Next Greater Numerically Balanced Number
+ * https://leetcode.com/problems/next-greater-numerically-balanced-number/
+ * Difficulty: Medium
+ *
+ * An integer x is numerically balanced if for every digit d in the number x, there are exactly
+ * d occurrences of that digit in x.
+ *
+ * Given an integer n, return the smallest numerically balanced number strictly greater than n.
+ */
+
+/**
+ * @param {number} n
+ * @return {number}
+ */
+var nextBeautifulNumber = function(n) {
+  let candidate = n + 1;
+  while (candidate <= 10000000) {
+    if (isBalanced(candidate)) return candidate;
+    candidate++;
+  }
+
+  return -1;
+
+  function isBalanced(num) {
+    const freq = new Array(10).fill(0);
+    const str = num.toString();
+
+    for (const digit of str) {
+      freq[digit]++;
+    }
+
+    for (const digit of str) {
+      if (freq[digit] !== parseInt(digit)) return false;
+    }
+
+    return true;
+  }
+};
