Add solution #1562

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,370 LeetCode solutions in JavaScript
+# 1,371 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1193,6 +1193,7 @@
 1559|[Detect Cycles in 2D Grid](./solutions/1559-detect-cycles-in-2d-grid.js)|Medium|
 1560|[Most Visited Sector in a Circular Track](./solutions/1560-most-visited-sector-in-a-circular-track.js)|Easy|
 1561|[Maximum Number of Coins You Can Get](./solutions/1561-maximum-number-of-coins-you-can-get.js)|Medium|
+1562|[Find Latest Group of Size M](./solutions/1562-find-latest-group-of-size-m.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|
 1576|[Replace All ?'s to Avoid Consecutive Repeating Characters](./solutions/1576-replace-all-s-to-avoid-consecutive-repeating-characters.js)|Medium|
 1598|[Crawler Log Folder](./solutions/1598-crawler-log-folder.js)|Easy|

solutions/1562-find-latest-group-of-size-m.js

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+/**
+ * 1562. Find Latest Group of Size M
+ * https://leetcode.com/problems/find-latest-group-of-size-m/
+ * Difficulty: Medium
+ *
+ * Given an array arr that represents a permutation of numbers from 1 to n.
+ *
+ * You have a binary string of size n that initially has all its bits set to zero. At each step
+ * i (assuming both the binary string and arr are 1-indexed) from 1 to n, the bit at position
+ * arr[i] is set to 1.
+ *
+ * You are also given an integer m. Find the latest step at which there exists a group of ones
+ * of length m. A group of ones is a contiguous substring of 1's such that it cannot be extended
+ * in either direction.
+ *
+ * Return the latest step at which there exists a group of ones of length exactly m. If no such
+ * group exists, return -1.
+ */
+
+/**
+ * @param {number[]} arr
+ * @param {number} m
+ * @return {number}
+ */
+var findLatestStep = function(arr, m) {
+  const lengths = new Array(arr.length + 2).fill(0);
+  const count = new Map();
+  let result = -1;
+
+  for (let step = 0; step < arr.length; step++) {
+    const pos = arr[step];
+    const left = lengths[pos - 1];
+    const right = lengths[pos + 1];
+    const newLength = left + right + 1;
+
+    lengths[pos - left] = newLength;
+    lengths[pos + right] = newLength;
+    lengths[pos] = newLength;
+
+    count.set(left, (count.get(left) || 0) - 1);
+    count.set(right, (count.get(right) || 0) - 1);
+    count.set(newLength, (count.get(newLength) || 0) + 1);
+
+    if (count.get(m) > 0) result = step + 1;
+  }
+
+  return result;
+};