Add solution #1375

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,261 LeetCode solutions in JavaScript
+# 1,262 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1047,6 +1047,7 @@
 1372|[Longest ZigZag Path in a Binary Tree](./solutions/1372-longest-zigzag-path-in-a-binary-tree.js)|Medium|
 1373|[Maximum Sum BST in Binary Tree](./solutions/1373-maximum-sum-bst-in-binary-tree.js)|Hard|
 1374|[Generate a String With Characters That Have Odd Counts](./solutions/1374-generate-a-string-with-characters-that-have-odd-counts.js)|Easy|
+1375|[Number of Times Binary String Is Prefix-Aligned](./solutions/1375-number-of-times-binary-string-is-prefix-aligned.js)|Medium|
 1380|[Lucky Numbers in a Matrix](./solutions/1380-lucky-numbers-in-a-matrix.js)|Easy|
 1389|[Create Target Array in the Given Order](./solutions/1389-create-target-array-in-the-given-order.js)|Easy|
 1400|[Construct K Palindrome Strings](./solutions/1400-construct-k-palindrome-strings.js)|Medium|

solutions/1375-number-of-times-binary-string-is-prefix-aligned.js

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+/**
+ * 1375. Number of Times Binary String Is Prefix-Aligned
+ * https://leetcode.com/problems/number-of-times-binary-string-is-prefix-aligned/
+ * Difficulty: Medium
+ *
+ * You have a 1-indexed binary string of length n where all the bits are 0 initially. We will
+ * flip all the bits of this binary string (i.e., change them from 0 to 1) one by one. You are
+ * given a 1-indexed integer array flips where flips[i] indicates that the bit at index i will
+ * be flipped in the ith step.
+ *
+ * A binary string is prefix-aligned if, after the ith step, all the bits in the inclusive range
+ * [1, i] are ones and all the other bits are zeros.
+ *
+ * Return the number of times the binary string is prefix-aligned during the flipping process.
+ */
+
+/**
+ * @param {number[]} light
+ * @return {number}
+ */
+var numTimesAllBlue = function(flips) {
+  let maxFlipped = 0;
+  let result = 0;
+
+  flips.forEach((position, step) => {
+    maxFlipped = Math.max(maxFlipped, position);
+    if (maxFlipped === step + 1) result++;
+  });
+
+  return result;
+};