Add solution #1573

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,376 LeetCode solutions in JavaScript
+# 1,377 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1200,6 +1200,7 @@
 1568|[Minimum Number of Days to Disconnect Island](./solutions/1568-minimum-number-of-days-to-disconnect-island.js)|Hard|
 1569|[Number of Ways to Reorder Array to Get Same BST](./solutions/1569-number-of-ways-to-reorder-array-to-get-same-bst.js)|Hard|
 1572|[Matrix Diagonal Sum](./solutions/1572-matrix-diagonal-sum.js)|Easy|
+1573|[Number of Ways to Split a String](./solutions/1573-number-of-ways-to-split-a-string.js)|Medium|
 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|
 1657|[Determine if Two Strings Are Close](./solutions/1657-determine-if-two-strings-are-close.js)|Medium|

solutions/1573-number-of-ways-to-split-a-string.js

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+/**
+ * 1573. Number of Ways to Split a String
+ * https://leetcode.com/problems/number-of-ways-to-split-a-string/
+ * Difficulty: Medium
+ *
+ * Given a binary string s, you can split s into 3 non-empty strings s1, s2, and s3
+ * where s1 + s2 + s3 = s.
+ *
+ * Return the number of ways s can be split such that the number of ones is the same in s1,
+ * s2, and s3. Since the answer may be too large, return it modulo 109 + 7.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var numWays = function(s) {
+  const MOD = 1e9 + 7;
+  let ones = 0;
+
+  for (const char of s) {
+    if (char === '1') ones++;
+  }
+
+  if (ones % 3 !== 0) return 0;
+  if (ones === 0) {
+    return Number((BigInt(s.length - 1) * BigInt(s.length - 2) / 2n) % BigInt(MOD));
+  }
+
+  const targetOnes = ones / 3;
+  let firstCount = 0;
+  let secondCount = 0;
+  let currentOnes = 0;
+
+  for (let i = 0; i < s.length; i++) {
+    if (s[i] === '1') currentOnes++;
+    if (currentOnes === targetOnes) firstCount++;
+    else if (currentOnes === 2 * targetOnes) secondCount++;
+  }
+
+  return Number((BigInt(firstCount) * BigInt(secondCount)) % BigInt(MOD));
+};