LeetCode #3405 — HARD

Count the Number of Arrays with K Matching Adjacent Elements

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

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.

Example 1:

Input: n = 3, m = 2, k = 1

Output: 4

Explanation:

There are 4 good arrays. They are [1, 1, 2], [1, 2, 2], [2, 1, 1] and [2, 2, 1].
Hence, the answer is 4.

Example 2:

Input: n = 4, m = 2, k = 2

Output: 6

Explanation:

The good arrays are [1, 1, 1, 2], [1, 1, 2, 2], [1, 2, 2, 2], [2, 1, 1, 1], [2, 2, 1, 1] and [2, 2, 2, 1].
Hence, the answer is 6.

Example 3:

Input: n = 5, m = 2, k = 0

Output: 2

Explanation:

The good arrays are [1, 2, 1, 2, 1] and [2, 1, 2, 1, 2]. Hence, the answer is 2.

Constraints:

1 <= n <= 10⁵
1 <= m <= 10⁵
0 <= k <= n - 1

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

3
2
1

Example 2

4
2
2

Example 3

5
2
0

Core Insight

What unlocks the optimal approach

The first position <code>arr[0]</code> has <code>m</code> choices.
For each of the remaining <code>n - 1</code> indices, <code>0 < i < n</code>, select <code>k</code> positions from left to right and set <code>arr[i] = arr[i - 1]</code>.
For all other indices, <code>set arr[i] != arr[i - 1]</code> with (<code>m - 1</code>) choices for each of the <code>n - 1 - k</code> positions.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3405: Count the Number of Arrays with K Matching Adjacent Elements
class Solution {
    private static final int N = (int) 1e5 + 10;
    private static final int MOD = (int) 1e9 + 7;
    private static final long[] f = new long[N];
    private static final long[] g = new long[N];

    static {
        f[0] = 1;
        g[0] = 1;
        for (int i = 1; i < N; ++i) {
            f[i] = f[i - 1] * i % MOD;
            g[i] = qpow(f[i], MOD - 2);
        }
    }

    public static long qpow(long a, int k) {
        long res = 1;
        while (k != 0) {
            if ((k & 1) == 1) {
                res = res * a % MOD;
            }
            k >>= 1;
            a = a * a % MOD;
        }
        return res;
    }

    public static long comb(int m, int n) {
        return (int) f[m] * g[n] % MOD * g[m - n] % MOD;
    }

    public int countGoodArrays(int n, int m, int k) {
        return (int) (comb(n - 1, k) * m % MOD * qpow(m - 1, n - k - 1) % MOD);
    }
}

// Accepted solution for LeetCode #3405: Count the Number of Arrays with K Matching Adjacent Elements
const MX = 1e5 + 10
const MOD = 1e9 + 7

var f [MX]int64
var g [MX]int64

func qpow(a int64, k int) int64 {
	res := int64(1)
	for k != 0 {
		if k&1 == 1 {
			res = res * a % MOD
		}
		a = a * a % MOD
		k >>= 1
	}
	return res
}

func init() {
	f[0], g[0] = 1, 1
	for i := 1; i < MX; i++ {
		f[i] = f[i-1] * int64(i) % MOD
		g[i] = qpow(f[i], MOD-2)
	}
}

func comb(m, n int) int64 {
	return f[m] * g[n] % MOD * g[m-n] % MOD
}

func countGoodArrays(n int, m int, k int) int {
	ans := comb(n-1, k) * int64(m) % MOD * qpow(int64(m-1), n-k-1) % MOD
	return int(ans)
}

# Accepted solution for LeetCode #3405: Count the Number of Arrays with K Matching Adjacent Elements
mx = 10**5 + 10
mod = 10**9 + 7
f = [1] + [0] * mx
g = [1] + [0] * mx

for i in range(1, mx):
    f[i] = f[i - 1] * i % mod
    g[i] = pow(f[i], mod - 2, mod)


def comb(m: int, n: int) -> int:
    return f[m] * g[n] * g[m - n] % mod


class Solution:
    def countGoodArrays(self, n: int, m: int, k: int) -> int:
        return comb(n - 1, k) * m * pow(m - 1, n - k - 1, mod) % mod

// Accepted solution for LeetCode #3405: Count the Number of Arrays with K Matching Adjacent Elements
impl Solution {
    pub fn count_good_arrays(n: i32, m: i32, k: i32) -> i32 {
        const N: usize = 1e5 as usize + 10;
        const MOD: i64 = 1_000_000_007;
        use std::sync::OnceLock;

        static F: OnceLock<Vec<i64>> = OnceLock::new();
        static G: OnceLock<Vec<i64>> = OnceLock::new();

        fn qpow(mut a: i64, mut k: i64, m: i64) -> i64 {
            let mut res = 1;
            while k != 0 {
                if k & 1 == 1 {
                    res = res * a % m;
                }
                a = a * a % m;
                k >>= 1;
            }
            res
        }

        fn init() -> (&'static Vec<i64>, &'static Vec<i64>) {
            F.get_or_init(|| {
                let mut f = vec![1i64; N];
                for i in 1..N {
                    f[i] = f[i - 1] * i as i64 % MOD;
                }
                f
            });

            G.get_or_init(|| {
                let f = F.get().unwrap();
                let mut g = vec![1i64; N];
                for i in 1..N {
                    g[i] = qpow(f[i], MOD - 2, MOD);
                }
                g
            });

            (F.get().unwrap(), G.get().unwrap())
        }

        fn comb(f: &[i64], g: &[i64], m: usize, n: usize) -> i64 {
            f[m] * g[n] % MOD * g[m - n] % MOD
        }

        let (f, g) = init();
        let n = n as usize;
        let m = m as i64;
        let k = k as usize;

        let c = comb(f, g, n - 1, k);
        let pow = qpow(m - 1, (n - 1 - k) as i64, MOD);
        (c * m % MOD * pow % MOD) as i32
    }
}

// Accepted solution for LeetCode #3405: Count the Number of Arrays with K Matching Adjacent Elements
const MX = 1e5 + 10;
const MOD = BigInt(1e9 + 7);

const f: bigint[] = Array(MX).fill(1n);
const g: bigint[] = Array(MX).fill(1n);

function qpow(a: bigint, k: number): bigint {
    let res = 1n;
    while (k !== 0) {
        if ((k & 1) === 1) {
            res = (res * a) % MOD;
        }
        a = (a * a) % MOD;
        k >>= 1;
    }
    return res;
}

(function init() {
    for (let i = 1; i < MX; ++i) {
        f[i] = (f[i - 1] * BigInt(i)) % MOD;
        g[i] = qpow(f[i], Number(MOD - 2n));
    }
})();

function comb(m: number, n: number): bigint {
    return (((f[m] * g[n]) % MOD) * g[m - n]) % MOD;
}

export function countGoodArrays(n: number, m: number, k: number): number {
    const ans = (((comb(n - 1, k) * BigInt(m)) % MOD) * qpow(BigInt(m - 1), n - k - 1)) % MOD;
    return Number(ans);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log (n - k)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Count the Number of Arrays with K Matching Adjacent Elements

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Overflow in intermediate arithmetic