Add solution #1759

JoshCrozier · JoshCrozier · commit ae02d4ac1841 · 2025-04-28T21:36:52.000-05:00
diff --git a/README.md b/README.md
@@ -1,4 +1,4 @@
-# 1,524 LeetCode solutions in JavaScript
+# 1,525 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1354,6 +1354,7 @@
 1754|[Largest Merge Of Two Strings](./solutions/1754-largest-merge-of-two-strings.js)|Medium|
 1755|[Closest Subsequence Sum](./solutions/1755-closest-subsequence-sum.js)|Hard|
 1758|[Minimum Changes To Make Alternating Binary String](./solutions/1758-minimum-changes-to-make-alternating-binary-string.js)|Easy|
+1759|[Count Number of Homogenous Substrings](./solutions/1759-count-number-of-homogenous-substrings.js)|Medium|
 1764|[Form Array by Concatenating Subarrays of Another Array](./solutions/1764-form-array-by-concatenating-subarrays-of-another-array.js)|Medium|
 1765|[Map of Highest Peak](./solutions/1765-map-of-highest-peak.js)|Medium|
 1768|[Merge Strings Alternately](./solutions/1768-merge-strings-alternately.js)|Easy|
diff --git a/solutions/1759-count-number-of-homogenous-substrings.js b/solutions/1759-count-number-of-homogenous-substrings.js
@@ -0,0 +1,35 @@
+/**
+ * 1759. Count Number of Homogenous Substrings
+ * https://leetcode.com/problems/count-number-of-homogenous-substrings/
+ * Difficulty: Medium
+ *
+ * Given a string s, return the number of homogenous substrings of s. Since the answer may be
+ * too large, return it modulo 109 + 7.
+ *
+ * A string is homogenous if all the characters of the string are the same.
+ *
+ * A substring is a contiguous sequence of characters within a string.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var countHomogenous = function(s) {
+  const mod = 1e9 + 7;
+  let result = 0;
+  let streak = 1;
+
+  for (let i = 1; i < s.length; i++) {
+    if (s[i] === s[i - 1]) {
+      streak++;
+    } else {
+      result = (result + (streak * (streak + 1) / 2)) % mod;
+      streak = 1;
+    }
+  }
+
+  result = (result + (streak * (streak + 1) / 2)) % mod;
+
+  return result;
+};
