Add solution #3337

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,765 LeetCode solutions in JavaScript
+# 1,766 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1756,6 +1756,7 @@
 3272|[Find the Count of Good Integers](./solutions/3272-find-the-count-of-good-integers.js)|Hard|
 3306|[Count of Substrings Containing Every Vowel and K Consonants II](./solutions/3306-count-of-substrings-containing-every-vowel-and-k-consonants-ii.js)|Medium|
 3335|[Total Characters in String After Transformations I](./solutions/3335-total-characters-in-string-after-transformations-i.js)|Medium|
+3337|[Total Characters in String After Transformations II](./solutions/3337-total-characters-in-string-after-transformations-ii.js)|Hard|
 3341|[Find Minimum Time to Reach Last Room I](./solutions/3341-find-minimum-time-to-reach-last-room-i.js)|Medium|
 3342|[Find Minimum Time to Reach Last Room II](./solutions/3342-find-minimum-time-to-reach-last-room-ii.js)|Medium|
 3343|[Count Number of Balanced Permutations](./solutions/3343-count-number-of-balanced-permutations.js)|Hard|

solutions/3337-total-characters-in-string-after-transformations-ii.js

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+/**
+ * 3337. Total Characters in String After Transformations II
+ * https://leetcode.com/problems/total-characters-in-string-after-transformations-ii/
+ * Difficulty: Hard
+ *
+ * You are given a string s consisting of lowercase English letters, an integer t representing
+ * the number of transformations to perform, and an array nums of size 26. In one transformation,
+ * every character in s is replaced according to the following rules:
+ * - Replace s[i] with the next nums[s[i] - 'a'] consecutive characters in the alphabet. For
+ *   example, if s[i] = 'a' and nums[0] = 3, the character 'a' transforms into the next 3
+ *   consecutive characters ahead of it, which results in "bcd".
+ * - The transformation wraps around the alphabet if it exceeds 'z'. For example, if s[i] = 'y'
+ *   and nums[24] = 3, the character 'y' transforms into the next 3 consecutive characters ahead
+ *   of it, which results in "zab".
+ *
+ * Return the length of the resulting string after exactly t transformations.
+ *
+ * Since the answer may be very large, return it modulo 109 + 7.
+ */
+
+/**
+ * @param {string} s
+ * @param {number} t
+ * @param {number[]} nums
+ * @return {number}
+ */
+var lengthAfterTransformations = function(s, t, nums) {
+  const MOD = 1000000007n;
+  const freq = new Array(26).fill(0n);
+
+  for (const char of s) {
+    freq[char.charCodeAt(0) - 97]++;
+  }
+
+  const matrix = new Array(26).fill().map(() => new Array(26).fill(0n));
+  for (let i = 0; i < 26; i++) {
+    for (let j = 0; j < nums[i]; j++) {
+      matrix[(i + j + 1) % 26][i] = 1n;
+    }
+  }
+
+  const finalMatrix = matrixPower(matrix, t);
+  let total = 0n;
+
+  for (let i = 0; i < 26; i++) {
+    let sum = 0n;
+    for (let j = 0; j < 26; j++) {
+      sum = (sum + finalMatrix[i][j] * freq[j]) % MOD;
+    }
+    total = (total + sum) % MOD;
+  }
+
+  return Number(total);
+
+  function matrixMultiply(a, b) {
+    const result = new Array(26).fill().map(() => new Array(26).fill(0n));
+    for (let i = 0; i < 26; i++) {
+      for (let j = 0; j < 26; j++) {
+        for (let k = 0; k < 26; k++) {
+          result[i][j] = (result[i][j] + a[i][k] * b[k][j]) % MOD;
+        }
+      }
+    }
+    return result;
+  }
+
+  function matrixPower(mat, power) {
+    let result = new Array(26).fill().map(() => new Array(26).fill(0n));
+    for (let i = 0; i < 26; i++) result[i][i] = 1n;
+
+    while (power > 0) {
+      if (power & 1) result = matrixMultiply(result, mat);
+      mat = matrixMultiply(mat, mat);
+      power >>= 1;
+    }
+    return result;
+  }
+};