Add solution #3029

JoshCrozier · JoshCrozier · commit e64a4159d399 · 2025-06-06T22:59:13.000-05:00
diff --git a/README.md b/README.md
@@ -2138,6 +2138,7 @@
 3025|[Find the Number of Ways to Place People I](./solutions/3025-find-the-number-of-ways-to-place-people-i.js)|Medium|
 3027|[Find the Number of Ways to Place People II](./solutions/3027-find-the-number-of-ways-to-place-people-ii.js)|Hard|
 3028|[Ant on the Boundary](./solutions/3028-ant-on-the-boundary.js)|Easy|
+3029|[Minimum Time to Revert Word to Initial State I](./solutions/3029-minimum-time-to-revert-word-to-initial-state-i.js)|Medium|
 3042|[Count Prefix and Suffix Pairs I](./solutions/3042-count-prefix-and-suffix-pairs-i.js)|Easy|
 3066|[Minimum Operations to Exceed Threshold Value II](./solutions/3066-minimum-operations-to-exceed-threshold-value-ii.js)|Medium|
 3068|[Find the Maximum Sum of Node Values](./solutions/3068-find-the-maximum-sum-of-node-values.js)|Hard|
diff --git a/solutions/3029-minimum-time-to-revert-word-to-initial-state-i.js b/solutions/3029-minimum-time-to-revert-word-to-initial-state-i.js
@@ -0,0 +1,33 @@
+/**
+ * 3029. Minimum Time to Revert Word to Initial State I
+ * https://leetcode.com/problems/minimum-time-to-revert-word-to-initial-state-i/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed string word and an integer k.
+ *
+ * At every second, you must perform the following operations:
+ * - Remove the first k characters of word.
+ * - Add any k characters to the end of word.
+ *
+ * Note that you do not necessarily need to add the same characters that you removed.
+ * However, you must perform both operations at every second.
+ *
+ * Return the minimum time greater than zero required for word to revert to its initial state.
+ */
+
+/**
+ * @param {string} word
+ * @param {number} k
+ * @return {number}
+ */
+var minimumTimeToInitialState = function(word, k) {
+  const n = word.length;
+
+  for (let i = 1; i * k <= n; i++) {
+    if (word.slice(i * k) === word.slice(0, n - i * k)) {
+      return i;
+    }
+  }
+
+  return Math.ceil(n / k);
+};
