Add solution #1585

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,385 LeetCode solutions in JavaScript
+# 1,386 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1210,6 +1210,7 @@
 1582|[Special Positions in a Binary Matrix](./solutions/1582-special-positions-in-a-binary-matrix.js)|Easy|
 1583|[Count Unhappy Friends](./solutions/1583-count-unhappy-friends.js)|Medium|
 1584|[Min Cost to Connect All Points](./solutions/1584-min-cost-to-connect-all-points.js)|Medium|
+1585|[Check If String Is Transformable With Substring Sort Operations](./solutions/1585-check-if-string-is-transformable-with-substring-sort-operations.js)|Hard|
 1598|[Crawler Log Folder](./solutions/1598-crawler-log-folder.js)|Easy|
 1657|[Determine if Two Strings Are Close](./solutions/1657-determine-if-two-strings-are-close.js)|Medium|
 1668|[Maximum Repeating Substring](./solutions/1668-maximum-repeating-substring.js)|Easy|

solutions/1585-check-if-string-is-transformable-with-substring-sort-operations.js

@@ -0,0 +1,49 @@
+/**
+ * 1585. Check If String Is Transformable With Substring Sort Operations
+ * https://leetcode.com/problems/check-if-string-is-transformable-with-substring-sort-operations/
+ * Difficulty: Hard
+ *
+ * Given two strings s and t, transform string s into string t using the following operation any
+ * number of times:
+ * - Choose a non-empty substring in s and sort it in place so the characters are in ascending
+ *   order.
+ *   - For example, applying the operation on the underlined substring in "14234" results
+ *     in "12344".
+ *
+ * Return true if it is possible to transform s into t. Otherwise, return false.
+ *
+ * A substring is a contiguous sequence of characters within a string.
+ */
+
+/**
+ * @param {string} s
+ * @param {string} t
+ * @return {boolean}
+ */
+var isTransformable = function(source, target) {
+  const digitPositions = Array.from({ length: 10 }, () => []);
+  for (let index = 0; index < source.length; index++) {
+    digitPositions[source[index]].push(index);
+  }
+
+  const currentIndices = new Array(10).fill(0);
+
+  for (const digit of target) {
+    const digitValue = parseInt(digit, 10);
+    if (currentIndices[digitValue] >= digitPositions[digitValue].length) {
+      return false;
+    }
+
+    const position = digitPositions[digitValue][currentIndices[digitValue]];
+    for (let smallerDigit = 0; smallerDigit < digitValue; smallerDigit++) {
+      if (currentIndices[smallerDigit] < digitPositions[smallerDigit].length
+          && digitPositions[smallerDigit][currentIndices[smallerDigit]] < position) {
+        return false;
+      }
+    }
+
+    currentIndices[digitValue]++;
+  }
+
+  return true;
+};