Add solution #1400

Author: JoshCrozier

README.md

@@ -311,6 +311,7 @@
 1374|[Generate a String With Characters That Have Odd Counts](./1374-generate-a-string-with-characters-that-have-odd-counts.js)|Easy|
 1380|[Lucky Numbers in a Matrix](./1380-lucky-numbers-in-a-matrix.js)|Easy|
 1389|[Create Target Array in the Given Order](./1389-create-target-array-in-the-given-order.js)|Easy|
+1400|[Construct K Palindrome Strings](./1400-construct-k-palindrome-strings.js)|Medium|
 1402|[Reducing Dishes](./1402-reducing-dishes.js)|Hard|
 1408|[String Matching in an Array](./1408-string-matching-in-an-array.js)|Easy|
 1410|[HTML Entity Parser](./1410-html-entity-parser.js)|Medium|

solutions/1400-construct-k-palindrome-strings.js

@@ -0,0 +1,21 @@
+/**
+ * 1400. Construct K Palindrome Strings
+ * https://leetcode.com/problems/construct-k-palindrome-strings/
+ * Difficulty: Medium
+ *
+ * Given a string s and an integer k, return true if you can use all the characters in s to
+ * construct k palindrome strings or false otherwise.
+ */
+
+/**
+ * @param {string} s
+ * @param {number} k
+ * @return {boolean}
+ */
+var canConstruct = function(s, k) {
+  const occurrences = new Array(26).fill(0);
+  s.split('').forEach(c => occurrences[c.charCodeAt(0) - 'a'.charCodeAt(0)]++);
+
+  const oddCount = occurrences.filter(n => n % 2 !== 0).length;
+  return k <= s.length && oddCount <= k;
+};