Add solution #1616

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,422 LeetCode solutions in JavaScript
+# 1,423 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1246,6 +1246,7 @@
 1611|[Minimum One Bit Operations to Make Integers Zero](./solutions/1611-minimum-one-bit-operations-to-make-integers-zero.js)|Hard|
 1614|[Maximum Nesting Depth of the Parentheses](./solutions/1614-maximum-nesting-depth-of-the-parentheses.js)|Easy|
 1615|[Maximal Network Rank](./solutions/1615-maximal-network-rank.js)|Medium|
+1616|[Split Two Strings to Make Palindrome](./solutions/1616-split-two-strings-to-make-palindrome.js)|Medium|
 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|
 1669|[Merge In Between Linked Lists](./solutions/1669-merge-in-between-linked-lists.js)|Medium|

solutions/1616-split-two-strings-to-make-palindrome.js

@@ -0,0 +1,46 @@
+/**
+ * 1616. Split Two Strings to Make Palindrome
+ * https://leetcode.com/problems/split-two-strings-to-make-palindrome/
+ * Difficulty: Medium
+ *
+ * You are given two strings a and b of the same length. Choose an index and split both strings at
+ * the same index, splitting a into two strings: aprefix and asuffix where a = aprefix + asuffix,
+ * and splitting b into two strings: bprefix and bsuffix where b = bprefix + bsuffix. Check if
+ * aprefix + bsuffix or bprefix + asuffix forms a palindrome.
+ *
+ * When you split a string s into sprefix and ssuffix, either ssuffix or sprefix is allowed to be
+ * empty. For example, if s = "abc", then "" + "abc", "a" + "bc", "ab" + "c" , and "abc" + "" are
+ * valid splits.
+ *
+ * Return true if it is possible to form a palindrome string, otherwise return false.
+ *
+ * Notice that x + y denotes the concatenation of strings x and y.
+ */
+
+/**
+ * @param {string} a
+ * @param {string} b
+ * @return {boolean}
+ */
+var checkPalindromeFormation = function(a, b) {
+  return check(a, b) || check(b, a);
+
+  function isPalindrome(str, left, right) {
+    while (left < right) {
+      if (str[left++] !== str[right--]) return false;
+    }
+    return true;
+  }
+
+  function check(str1, str2) {
+    let left = 0;
+    let right = str1.length - 1;
+
+    while (left < right && str1[left] === str2[right]) {
+      left++;
+      right--;
+    }
+
+    return isPalindrome(str1, left, right) || isPalindrome(str2, left, right);
+  }
+};