Add solution #30

Author: JoshCrozier

README.md

@@ -35,6 +35,7 @@
 27|[Remove Element](./0027-remove-element.js)|Easy|
 28|[Implement strStr()](./0028-implement-strstr.js)|Easy|
 29|[Divide Two Integers](./0029-divide-two-integers.js)|Medium|
+30|[Substring with Concatenation of All Words](./0030-substring-with-concatenation-of-all-words.js)|Hard|
 31|[Next Permutation](./0031-next-permutation.js)|Medium|
 33|[Search in Rotated Sorted Array](./0033-search-in-rotated-sorted-array.js)|Medium|
 34|[Find First and Last Position of Element in Sorted Array](./0034-find-first-and-last-position-of-element-in-sorted-array.js)|Medium|

solutions/0030-substring-with-concatenation-of-all-words.js

@@ -0,0 +1,54 @@
+/**
+ * 30. Substring with Concatenation of All Words
+ * https://leetcode.com/problems/substring-with-concatenation-of-all-words/
+ * Difficulty: Hard
+ *
+ * You are given a string s and an array of strings words. All the strings of words are of the same
+ * length.
+ *
+ * A concatenated substring in s is a substring that contains all the strings of any permutation of
+ * words concatenated.
+ *
+ * - For example, if words = ["ab","cd","ef"], then "abcdef", "abefcd", "cdabef", "cdefab", "efabcd",
+ * and "efcdab" are all concatenated strings. "acdbef" is not a concatenated substring because it is
+ * not the concatenation of any permutation of words.
+ *
+ * Return the starting indices of all the concatenated substrings in s. You can return the answer in
+ * any order.
+ */
+
+/**
+ * @param {string} s
+ * @param {string[]} words
+ * @return {number[]}
+ */
+var findSubstring = function(s, words) {
+  if (!words || words.length === 0) {
+    return [];
+  }
+
+  const result = [];
+  const map = {};
+  const count = words[0].length;
+
+  words.forEach(word => map[word] = ~~map[word] + 1);
+
+  for (let i = 0; i < s.length - words.length * count + 1; i++) {
+    const temp = Object.assign({}, map);
+
+    for (let j = i; j < i + words.length * count; j += count) {
+      const key = s.slice(j, j + count);
+      if (!temp[key]) {
+        break;
+      } else if (--temp[key] === 0) {
+        delete temp[key];
+      }
+    }
+
+    if (Object.keys(temp).length === 0) {
+      result.push(i);
+    }
+  }
+
+  return result;
+};