Add solution #567

Author: JoshCrozier

README.md

@@ -118,6 +118,7 @@
 557|[Reverse Words in a String III](./0557-reverse-words-in-a-string-iii.js)|Easy|
 565|[Array Nesting](./0565-array-nesting.js)|Medium|
 566|[Reshape the Matrix](./0566-reshape-the-matrix.js)|Easy|
+567|[Permutation in String](./0567-permutation-in-string.js)|Medium|
 606|[Construct String from Binary Tree](./0606-construct-string-from-binary-tree.js)|Easy|
 617|[Merge Two Binary Trees](./0617-merge-two-binary-trees.js)|Easy|
 628|[Maximum Product of Three Numbers](./0628-maximum-product-of-three-numbers.js)|Easy|

solutions/0567-permutation-in-string.js

@@ -0,0 +1,38 @@
+/**
+ * 567. Permutation in String
+ * https://leetcode.com/problems/permutation-in-string/
+ * Difficulty: Medium
+ *
+ * Given two strings s1 and s2, return true if s2 contains a permutation of s1, or false otherwise.
+ *
+ * In other words, return true if one of s1's permutations is the substring of s2.
+ */
+
+/**
+ * @param {string} s1
+ * @param {string} s2
+ * @return {boolean}
+ */
+var checkInclusion = function(s1, s2) {
+  const getCharCode = c => c.charCodeAt() - 'a'.charCodeAt();
+  const isMatch = (a1, a2) => a1.every((n, i) => a2[i] === n);
+
+  if (s1.length > s2.length) {
+    return false;
+  }
+
+  const map1 = new Array(26).fill(0);
+  const map2 = new Array(26).fill(0);
+  for (let i = 0; i < s1.length; i++) {
+    map1[getCharCode(s1[i])]++;
+    map2[getCharCode(s2[i])]++;
+  }
+
+  for (let i = 0; i < s2.length - s1.length; i++) {
+    if (isMatch(map1, map2)) return true;
+    map2[getCharCode(s2[i + s1.length])]++;
+    map2[getCharCode(s2[i])]--;
+  }
+
+  return isMatch(map1, map2);
+};