Add solution #1545

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,358 LeetCode solutions in JavaScript
+# 1,359 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1180,6 +1180,7 @@
 1541|[Minimum Insertions to Balance a Parentheses String](./solutions/1541-minimum-insertions-to-balance-a-parentheses-string.js)|Medium|
 1542|[Find Longest Awesome Substring](./solutions/1542-find-longest-awesome-substring.js)|Hard|
 1544|[Make The String Great](./solutions/1544-make-the-string-great.js)|Easy|
+1545|[Find Kth Bit in Nth Binary String](./solutions/1545-find-kth-bit-in-nth-binary-string.js)|Medium|
 1550|[Three Consecutive Odds](./solutions/1550-three-consecutive-odds.js)|Easy|
 1551|[Minimum Operations to Make Array Equal](./solutions/1551-minimum-operations-to-make-array-equal.js)|Medium|
 1566|[Detect Pattern of Length M Repeated K or More Times](./solutions/1566-detect-pattern-of-length-m-repeated-k-or-more-times.js)|Easy|

solutions/1545-find-kth-bit-in-nth-binary-string.js

@@ -0,0 +1,48 @@
+/**
+ * 1545. Find Kth Bit in Nth Binary String
+ * https://leetcode.com/problems/find-kth-bit-in-nth-binary-string/
+ * Difficulty: Medium
+ *
+ * Given two positive integers n and k, the binary string Sn is formed as follows:
+ * - S1 = "0"
+ * - Si = Si - 1 + "1" + reverse(invert(Si - 1)) for i > 1
+ *
+ * Where + denotes the concatenation operation, reverse(x) returns the reversed string x, and
+ * invert(x) inverts all the bits in x (0 changes to 1 and 1 changes to 0).
+ *
+ * For example, the first four strings in the above sequence are:
+ * - S1 = "0"
+ * - S2 = "011"
+ * - S3 = "0111001"
+ * - S4 = "011100110110001"
+ *
+ * Return the kth bit in Sn. It is guaranteed that k is valid for the given n.
+ */
+
+/**
+ * @param {number} n
+ * @param {number} k
+ * @return {character}
+ */
+var findKthBit = function(n, k) {
+  let position = k - 1;
+  let invertCount = 0;
+  let length = (1 << n) - 1;
+
+  while (position !== 0) {
+    const mid = length >> 1;
+
+    if (position === mid) {
+      return invertCount % 2 === 0 ? '1' : '0';
+    }
+
+    if (position > mid) {
+      position = length - position - 1;
+      invertCount++;
+    } else {
+      length = mid;
+    }
+  }
+
+  return invertCount % 2 === 0 ? '0' : '1';
+};