Add solution #2537

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,320 LeetCode solutions in JavaScript
+# 1,321 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1243,6 +1243,7 @@
 2523|[Closest Prime Numbers in Range](./solutions/2523-closest-prime-numbers-in-range.js)|Medium|
 2529|[Maximum Count of Positive Integer and Negative Integer](./solutions/2529-maximum-count-of-positive-integer-and-negative-integer.js)|Easy|
 2535|[Difference Between Element Sum and Digit Sum of an Array](./solutions/2535-difference-between-element-sum-and-digit-sum-of-an-array.js)|Easy|
+2537|[Count the Number of Good Subarrays](./solutions/2537-count-the-number-of-good-subarrays.js)|Medium|
 2542|[Maximum Subsequence Score](./solutions/2542-maximum-subsequence-score.js)|Medium|
 2551|[Put Marbles in Bags](./solutions/2551-put-marbles-in-bags.js)|Hard|
 2559|[Count Vowel Strings in Ranges](./solutions/2559-count-vowel-strings-in-ranges.js)|Medium|

solutions/2537-count-the-number-of-good-subarrays.js

@@ -0,0 +1,42 @@
+/**
+ * 2537. Count the Number of Good Subarrays
+ * https://leetcode.com/problems/count-the-number-of-good-subarrays/
+ * Difficulty: Medium
+ *
+ * Given an integer array nums and an integer k, return the number of good subarrays of nums.
+ *
+ * A subarray arr is good if there are at least k pairs of indices (i, j) such that i < j
+ * and arr[i] == arr[j].
+ *
+ * A subarray is a contiguous non-empty sequence of elements within an array.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+var countGood = function(nums, k) {
+  const frequency = new Map();
+  let result = 0;
+  let currentPairs = 0;
+  let left = 0;
+
+  for (let right = 0; right < nums.length; right++) {
+    const num = nums[right];
+    const prevCount = frequency.get(num) || 0;
+    currentPairs += prevCount;
+    frequency.set(num, prevCount + 1);
+
+    while (currentPairs >= k) {
+      result += nums.length - right;
+      const leftNum = nums[left];
+      const leftCount = frequency.get(leftNum);
+      currentPairs -= leftCount - 1;
+      frequency.set(leftNum, leftCount - 1);
+      left++;
+    }
+  }
+
+  return result;
+};