Add solution #696

Author: JoshCrozier

README.md

@@ -290,6 +290,7 @@
 684|[Redundant Connection](./0684-redundant-connection.js)|Medium|
 686|[Repeated String Match](./0686-repeated-string-match.js)|Easy|
 695|[Max Area of Island](./0695-max-area-of-island.js)|Medium|
+696|[Count Binary Substrings](./0696-count-binary-substrings.js)|Easy|
 697|[Degree of an Array](./0697-degree-of-an-array.js)|Easy|
 700|[Search in a Binary Search Tree](./0700-search-in-a-binary-search-tree.js)|Easy|
 701|[Insert into a Binary Search Tree](./0701-insert-into-a-binary-search-tree.js)|Medium|

solutions/0696-count-binary-substrings.js

@@ -0,0 +1,33 @@
+/**
+ * 696. Count Binary Substrings
+ * https://leetcode.com/problems/count-binary-substrings/
+ * Difficulty: Easy
+ *
+ * Given a binary string s, return the number of non-empty substrings that have the same
+ * number of 0's and 1's, and all the 0's and all the 1's in these substrings are grouped
+ * consecutively.
+ *
+ * Substrings that occur multiple times are counted the number of times they occur.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var countBinarySubstrings = function(s) {
+  let result = 0;
+
+  for (let i = 0, group = [], count = 1, prevCount = 0; i < s.length; i++) {
+    if (s[i] === s[i + 1]) {
+      count++;
+    } else {
+      if (prevCount) {
+        result += prevCount <= count ? prevCount : count;
+      }
+      prevCount = count;
+      count = 1;
+    }
+  }
+
+  return result;
+};