Add solution #3405

Author: JoshCrozier

README.md

@@ -2278,6 +2278,7 @@
 3397|[Maximum Number of Distinct Elements After Operations](./solutions/3397-maximum-number-of-distinct-elements-after-operations.js)|Medium|
 3402|[Minimum Operations to Make Columns Strictly Increasing](./solutions/3402-minimum-operations-to-make-columns-strictly-increasing.js)|Easy|
 3403|[Find the Lexicographically Largest String From the Box I](./solutions/3403-find-the-lexicographically-largest-string-from-the-box-i.js)|Medium|
+3405|[Count the Number of Arrays with K Matching Adjacent Elements](./solutions/3405-count-the-number-of-arrays-with-k-matching-adjacent-elements.js)|Hard|
 3423|[Maximum Difference Between Adjacent Elements in a Circular Array](./solutions/3423-maximum-difference-between-adjacent-elements-in-a-circular-array.js)|Easy|
 3442|[Maximum Difference Between Even and Odd Frequency I](./solutions/3442-maximum-difference-between-even-and-odd-frequency-i.js)|Easy|
 3445|[Maximum Difference Between Even and Odd Frequency II](./solutions/3445-maximum-difference-between-even-and-odd-frequency-ii.js)|Hard|

solutions/3405-count-the-number-of-arrays-with-k-matching-adjacent-elements.js

@@ -0,0 +1,69 @@
+/**
+ * 3405. Count the Number of Arrays with K Matching Adjacent Elements
+ * https://leetcode.com/problems/count-the-number-of-arrays-with-k-matching-adjacent-elements/
+ * Difficulty: Hard
+ *
+ * You are given three integers n, m, k. A good array arr of size n is defined as follows:
+ * - Each element in arr is in the inclusive range [1, m].
+ * - Exactly k indices i (where 1 <= i < n) satisfy the condition arr[i - 1] == arr[i].
+ *
+ * Return the number of good arrays that can be formed.
+ *
+ * Since the answer may be very large, return it modulo 109 + 7.
+ */
+
+/**
+ * @param {number} n
+ * @param {number} m
+ * @param {number} k
+ * @return {number}
+ */
+var countGoodArrays = function(n, m, k) {
+  const MOD = 1e9 + 7;
+
+  if (m === 1) {
+    return k === n - 1 ? 1 : 0;
+  }
+
+  const choose = binomialCoeff(n - 1, k);
+  const power = modPow(m - 1, n - 1 - k, MOD);
+  const result = Number((BigInt(choose) * BigInt(m) * BigInt(power)) % BigInt(MOD));
+
+  return result;
+
+  function modPow(base, exp, mod) {
+    let result = 1n;
+    base = ((BigInt(base) % BigInt(mod)) + BigInt(mod)) % BigInt(mod);
+    exp = BigInt(exp);
+    mod = BigInt(mod);
+
+    while (exp > 0n) {
+      if (exp & 1n) result = (result * base) % mod;
+      base = (base * base) % mod;
+      exp >>= 1n;
+    }
+    return Number(result);
+  }
+
+  function modInverse(a, mod) {
+    return modPow(a, mod - 2, mod);
+  }
+
+  function binomialCoeff(n, k) {
+    if (k > n || k < 0) return 0;
+    if (k === 0 || k === n) return 1;
+
+    if (k > n - k) k = n - k;
+
+    let numerator = 1n;
+    let denominator = 1n;
+
+    for (let i = 0; i < k; i++) {
+      numerator = (numerator * BigInt(n - i)) % BigInt(MOD);
+      denominator = (denominator * BigInt(i + 1)) % BigInt(MOD);
+    }
+
+    const invDenom = modInverse(Number(denominator), MOD);
+    return Number((numerator * BigInt(invDenom)) % BigInt(MOD));
+  }
+};