Add solution #1433

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,300 LeetCode solutions in JavaScript
+# 1,301 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1093,6 +1093,7 @@
 1425|[Constrained Subsequence Sum](./solutions/1425-constrained-subsequence-sum.js)|Hard|
 1431|[Kids With the Greatest Number of Candies](./solutions/1431-kids-with-the-greatest-number-of-candies.js)|Easy|
 1432|[Max Difference You Can Get From Changing an Integer](./solutions/1432-max-difference-you-can-get-from-changing-an-integer.js)|Medium|
+1433|[Check If a String Can Break Another String](./solutions/1433-check-if-a-string-can-break-another-string.js)|Medium|
 1436|[Destination City](./solutions/1436-destination-city.js)|Easy|
 1437|[Check If All 1's Are at Least Length K Places Away](./solutions/1437-check-if-all-1s-are-at-least-length-k-places-away.js)|Easy|
 1443|[Minimum Time to Collect All Apples in a Tree](./solutions/1443-minimum-time-to-collect-all-apples-in-a-tree.js)|Medium|

solutions/1433-check-if-a-string-can-break-another-string.js

@@ -0,0 +1,35 @@
+/**
+ * 1433. Check If a String Can Break Another String
+ * https://leetcode.com/problems/check-if-a-string-can-break-another-string/
+ * Difficulty: Medium
+ *
+ * Given two strings: s1 and s2 with the same size, check if some permutation of string s1 can
+ * break some permutation of string s2 or vice-versa. In other words s2 can break s1 or vice-versa.
+ *
+ * A string x can break string y (both of size n) if x[i] >= y[i] (in alphabetical order) for all
+ * i between 0 and n-1.
+ */
+
+/**
+ * @param {string} s1
+ * @param {string} s2
+ * @return {boolean}
+ */
+var checkIfCanBreak = function(s1, s2) {
+  const sorted1 = s1.split('').sort();
+  const sorted2 = s2.split('').sort();
+
+  let canBreak1 = true;
+  let canBreak2 = true;
+
+  for (let i = 0; i < s1.length; i++) {
+    if (sorted1[i] < sorted2[i]) {
+      canBreak1 = false;
+    }
+    if (sorted2[i] < sorted1[i]) {
+      canBreak2 = false;
+    }
+  }
+
+  return canBreak1 || canBreak2;
+};