Add solution #949

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,033 LeetCode solutions in JavaScript
+# 1,034 LeetCode solutions in JavaScript
 
 [https://leetcode.com/](https://leetcode.com/)
 
@@ -758,6 +758,7 @@
 946|[Validate Stack Sequences](./solutions/0946-validate-stack-sequences.js)|Medium|
 947|[Most Stones Removed with Same Row or Column](./solutions/0947-most-stones-removed-with-same-row-or-column.js)|Medium|
 948|[Bag of Tokens](./solutions/0948-bag-of-tokens.js)|Medium|
+949|[Largest Time for Given Digits](./solutions/0949-largest-time-for-given-digits.js)|Medium|
 966|[Vowel Spellchecker](./solutions/0966-vowel-spellchecker.js)|Medium|
 970|[Powerful Integers](./solutions/0970-powerful-integers.js)|Easy|
 976|[Largest Perimeter Triangle](./solutions/0976-largest-perimeter-triangle.js)|Easy|

solutions/0949-largest-time-for-given-digits.js

@@ -0,0 +1,46 @@
+/**
+ * 949. Largest Time for Given Digits
+ * https://leetcode.com/problems/largest-time-for-given-digits/
+ * Difficulty: Medium
+ *
+ * Given an array arr of 4 digits, find the latest 24-hour time that can be made using
+ * each digit exactly once.
+ *
+ * 24-hour times are formatted as "HH:MM", where HH is between 00 and 23, and MM is between
+ * 00 and 59. The earliest 24-hour time is 00:00, and the latest is 23:59.
+ *
+ * Return the latest 24-hour time in "HH:MM" format. If no valid time can be made, return
+ * an empty string.
+ */
+
+/**
+ * @param {number[]} arr
+ * @return {string}
+ */
+var largestTimeFromDigits = function(arr) {
+  let result = '';
+  helper(arr, new Array(4).fill(false), []);
+  return result;
+
+  function helper(digits, used, current) {
+    if (current.length === 4) {
+      const hours = parseInt(current.slice(0, 2).join(''));
+      const minutes = parseInt(current.slice(2).join(''));
+      if (hours <= 23 && minutes <= 59) {
+        const time = `${current[0]}${current[1]}:${current[2]}${current[3]}`;
+        if (time > result) result = time;
+      }
+      return;
+    }
+
+    for (let i = 0; i < digits.length; i++) {
+      if (!used[i]) {
+        used[i] = true;
+        current.push(digits[i]);
+        helper(digits, used, current);
+        current.pop();
+        used[i] = false;
+      }
+    }
+  }
+};