Add solution #2134

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,768 LeetCode solutions in JavaScript
+# 1,769 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1635,6 +1635,7 @@
 2131|[Longest Palindrome by Concatenating Two Letter Words](./solutions/2131-longest-palindrome-by-concatenating-two-letter-words.js)|Medium|
 2132|[Stamping the Grid](./solutions/2132-stamping-the-grid.js)|Hard|
 2133|[Check if Every Row and Column Contains All Numbers](./solutions/2133-check-if-every-row-and-column-contains-all-numbers.js)|Easy|
+2134|[Minimum Swaps to Group All 1's Together II](./solutions/2134-minimum-swaps-to-group-all-1s-together-ii.js)|Medium|
 2140|[Solving Questions With Brainpower](./solutions/2140-solving-questions-with-brainpower.js)|Medium|
 2145|[Count the Hidden Sequences](./solutions/2145-count-the-hidden-sequences.js)|Medium|
 2154|[Keep Multiplying Found Values by Two](./solutions/2154-keep-multiplying-found-values-by-two.js)|Easy|

solutions/2134-minimum-swaps-to-group-all-1s-together-ii.js

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+/**
+ * 2134. Minimum Swaps to Group All 1's Together II
+ * https://leetcode.com/problems/minimum-swaps-to-group-all-1s-together-ii/
+ * Difficulty: Medium
+ *
+ * A swap is defined as taking two distinct positions in an array and swapping the values in them.
+ *
+ * A circular array is defined as an array where we consider the first element and the last element
+ * to be adjacent.
+ *
+ * Given a binary circular array nums, return the minimum number of swaps required to group all
+ * 1's present in the array together at any location.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minSwaps = function(nums) {
+  const onesCount = nums.reduce((sum, num) => sum + num, 0);
+  const n = nums.length;
+  let result = Infinity;
+  let currentZeros = 0;
+
+  for (let i = 0; i < onesCount; i++) {
+    if (nums[i] === 0) currentZeros++;
+  }
+
+  result = currentZeros;
+
+  for (let i = 1; i < n; i++) {
+    if (nums[(i - 1) % n] === 0) currentZeros--;
+    if (nums[(i + onesCount - 1) % n] === 0) currentZeros++;
+    result = Math.min(result, currentZeros);
+  }
+
+  return result;
+};