Add solution #2182

JoshCrozier · JoshCrozier · commit ba18ec578f8c · 2025-07-26T11:38:17.000-05:00
diff --git a/README.md b/README.md
@@ -1979,6 +1979,7 @@
 2179|[Count Good Triplets in an Array](./solutions/2179-count-good-triplets-in-an-array.js)|Hard|
 2180|[Count Integers With Even Digit Sum](./solutions/2180-count-integers-with-even-digit-sum.js)|Easy|
 2181|[Merge Nodes in Between Zeros](./solutions/2181-merge-nodes-in-between-zeros.js)|Medium|
+2182|[Construct String With Repeat Limit](./solutions/2182-construct-string-with-repeat-limit.js)|Medium|
 2183|[Count Array Pairs Divisible by K](./solutions/2183-count-array-pairs-divisible-by-k.js)|Hard|
 2184|[Number of Ways to Build Sturdy Brick Wall](./solutions/2184-number-of-ways-to-build-sturdy-brick-wall.js)|Medium|
 2185|[Counting Words With a Given Prefix](./solutions/2185-counting-words-with-a-given-prefix.js)|Easy|
diff --git a/solutions/2182-construct-string-with-repeat-limit.js b/solutions/2182-construct-string-with-repeat-limit.js
@@ -0,0 +1,60 @@
+/**
+ * 2182. Construct String With Repeat Limit
+ * https://leetcode.com/problems/construct-string-with-repeat-limit/
+ * Difficulty: Medium
+ *
+ * You are given a string s and an integer repeatLimit. Construct a new string
+ * repeatLimitedString using the characters of s such that no letter appears more than
+ * repeatLimit times in a row. You do not have to use all characters from s.
+ *
+ * Return the lexicographically largest repeatLimitedString possible.
+ *
+ * A string a is lexicographically larger than a string b if in the first position where
+ * a and b differ, string a has a letter that appears later in the alphabet than the
+ * corresponding letter in b. If the first min(a.length, b.length) characters do not differ,
+ * then the longer string is the lexicographically larger one.
+ */
+
+/**
+ * @param {string} s
+ * @param {number} repeatLimit
+ * @return {string}
+ */
+var repeatLimitedString = function(s, repeatLimit) {
+  const frequency = new Array(26).fill(0);
+  for (const char of s) {
+    frequency[char.charCodeAt(0) - 97]++;
+  }
+
+  const result = [];
+  let currentChar = 25;
+
+  while (currentChar >= 0) {
+    if (frequency[currentChar] === 0) {
+      currentChar--;
+      continue;
+    }
+
+    const useCount = Math.min(frequency[currentChar], repeatLimit);
+    for (let i = 0; i < useCount; i++) {
+      result.push(String.fromCharCode(currentChar + 97));
+    }
+    frequency[currentChar] -= useCount;
+
+    if (frequency[currentChar] > 0) {
+      let nextChar = currentChar - 1;
+      while (nextChar >= 0 && frequency[nextChar] === 0) {
+        nextChar--;
+      }
+
+      if (nextChar < 0) break;
+
+      result.push(String.fromCharCode(nextChar + 97));
+      frequency[nextChar]--;
+    } else {
+      currentChar--;
+    }
+  }
+
+  return result.join('');
+};
