Add solution #1638

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,439 LeetCode solutions in JavaScript
+# 1,440 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1262,6 +1262,7 @@
 1632|[Rank Transform of a Matrix](./solutions/1632-rank-transform-of-a-matrix.js)|Hard|
 1636|[Sort Array by Increasing Frequency](./solutions/1636-sort-array-by-increasing-frequency.js)|Easy|
 1637|[Widest Vertical Area Between Two Points Containing No Points](./solutions/1637-widest-vertical-area-between-two-points-containing-no-points.js)|Easy|
+1638|[Count Substrings That Differ by One Character](./solutions/1638-count-substrings-that-differ-by-one-character.js)|Medium|
 1657|[Determine if Two Strings Are Close](./solutions/1657-determine-if-two-strings-are-close.js)|Medium|
 1668|[Maximum Repeating Substring](./solutions/1668-maximum-repeating-substring.js)|Easy|
 1669|[Merge In Between Linked Lists](./solutions/1669-merge-in-between-linked-lists.js)|Medium|

solutions/1638-count-substrings-that-differ-by-one-character.js

@@ -0,0 +1,39 @@
+/**
+ * 1638. Count Substrings That Differ by One Character
+ * https://leetcode.com/problems/count-substrings-that-differ-by-one-character/
+ * Difficulty: Medium
+ *
+ * Given two strings s and t, find the number of ways you can choose a non-empty substring of s
+ * and replace a single character by a different character such that the resulting substring is
+ * a substring of t. In other words, find the number of substrings in s that differ from some
+ * substring in t by exactly one character.
+ *
+ * For example, the underlined substrings in "computer" and "computation" only differ by the
+ * 'e'/'a', so this is a valid way.
+ *
+ * Return the number of substrings that satisfy the condition above.
+ *
+ * A substring is a contiguous sequence of characters within a string.
+ */
+
+/**
+ * @param {string} s
+ * @param {string} t
+ * @return {number}
+ */
+var countSubstrings = function(s, t) {
+  let result = 0;
+
+  for (let i = 0; i < s.length; i++) {
+    for (let j = 0; j < t.length; j++) {
+      let diffCount = 0;
+      for (let k = 0; i + k < s.length && j + k < t.length; k++) {
+        if (s[i + k] !== t[j + k]) diffCount++;
+        if (diffCount === 1) result++;
+        if (diffCount > 1) break;
+      }
+    }
+  }
+
+  return result;
+};