Add solution #1525

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,346 LeetCode solutions in JavaScript
+# 1,347 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1165,6 +1165,7 @@
 1521|[Find a Value of a Mysterious Function Closest to Target](./solutions/1521-find-a-value-of-a-mysterious-function-closest-to-target.js)|Hard|
 1523|[Count Odd Numbers in an Interval Range](./solutions/1523-count-odd-numbers-in-an-interval-range.js)|Easy|
 1524|[Number of Sub-arrays With Odd Sum](./solutions/1524-number-of-sub-arrays-with-odd-sum.js)|Medium|
+1525|[Number of Good Ways to Split a String](./solutions/1525-number-of-good-ways-to-split-a-string.js)|Medium|
 1528|[Shuffle String](./solutions/1528-shuffle-string.js)|Easy|
 1534|[Count Good Triplets](./solutions/1534-count-good-triplets.js)|Easy|
 1535|[Find the Winner of an Array Game](./solutions/1535-find-the-winner-of-an-array-game.js)|Medium|

solutions/1525-number-of-good-ways-to-split-a-string.js

@@ -0,0 +1,44 @@
+/**
+ * 1525. Number of Good Ways to Split a String
+ * https://leetcode.com/problems/number-of-good-ways-to-split-a-string/
+ * Difficulty: Medium
+ *
+ * You are given a string s.
+ *
+ * A split is called good if you can split s into two non-empty strings sleft and sright where
+ * their concatenation is equal to s (i.e., sleft + sright = s) and the number of distinct letters
+ * in sleft and sright is the same.
+ *
+ * Return the number of good splits you can make in s.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var numSplits = function(s) {
+  const leftDistinct = new Map();
+  const rightDistinct = new Map();
+  let result = 0;
+
+  for (const char of s) {
+    rightDistinct.set(char, (rightDistinct.get(char) || 0) + 1);
+  }
+
+  for (let i = 0; i < s.length - 1; i++) {
+    const char = s[i];
+    leftDistinct.set(char, (leftDistinct.get(char) || 0) + 1);
+
+    if (rightDistinct.get(char) === 1) {
+      rightDistinct.delete(char);
+    } else {
+      rightDistinct.set(char, rightDistinct.get(char) - 1);
+    }
+
+    if (leftDistinct.size === rightDistinct.size) {
+      result++;
+    }
+  }
+
+  return result;
+};