Add solution #1540

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,354 LeetCode solutions in JavaScript
+# 1,355 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1176,6 +1176,7 @@
 1536|[Minimum Swaps to Arrange a Binary Grid](./solutions/1536-minimum-swaps-to-arrange-a-binary-grid.js)|Medium|
 1537|[Get the Maximum Score](./solutions/1537-get-the-maximum-score.js)|Hard|
 1539|[Kth Missing Positive Number](./solutions/1539-kth-missing-positive-number.js)|Easy|
+1540|[Can Convert String in K Moves](./solutions/1540-can-convert-string-in-k-moves.js)|Medium|
 1550|[Three Consecutive Odds](./solutions/1550-three-consecutive-odds.js)|Easy|
 1551|[Minimum Operations to Make Array Equal](./solutions/1551-minimum-operations-to-make-array-equal.js)|Medium|
 1566|[Detect Pattern of Length M Repeated K or More Times](./solutions/1566-detect-pattern-of-length-m-repeated-k-or-more-times.js)|Easy|

solutions/1540-can-convert-string-in-k-moves.js

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+/**
+ * 1540. Can Convert String in K Moves
+ * https://leetcode.com/problems/can-convert-string-in-k-moves/
+ * Difficulty: Medium
+ *
+ * Given two strings s and t, your goal is to convert s into t in k moves or less.
+ *
+ * During the ith (1 <= i <= k) move you can:
+ * - Choose any index j (1-indexed) from s, such that 1 <= j <= s.length and j has not been chosen
+ *   in any previous move, and shift the character at that index i times.
+ * - Do nothing.
+ *
+ * Shifting a character means replacing it by the next letter in the alphabet (wrapping around so
+ * that 'z' becomes 'a'). Shifting a character by i means applying the shift operations i times.
+ *
+ * Remember that any index j can be picked at most once.
+ *
+ * Return true if it's possible to convert s into t in no more than k moves, otherwise return false.
+ */
+
+/**
+ * @param {string} s
+ * @param {string} t
+ * @param {number} k
+ * @return {boolean}
+ */
+var canConvertString = function(source, target, maxMoves) {
+  if (source.length !== target.length) return false;
+  const shiftCounts = new Array(26).fill(0);
+
+  for (let i = 0; i < source.length; i++) {
+    const shiftNeeded = (target.charCodeAt(i) - source.charCodeAt(i) + 26) % 26;
+    if (shiftNeeded > 0) {
+      shiftCounts[shiftNeeded]++;
+    }
+  }
+
+  for (let shift = 1; shift < 26; shift++) {
+    if (shiftCounts[shift] === 0) continue;
+    const totalMoves = shift + 26 * (shiftCounts[shift] - 1);
+    if (totalMoves > maxMoves) return false;
+  }
+
+  return true;
+};