What is the time complexity of Count the Number of Arrays with K Matching Adjacent Elements?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Count the Number of Arrays with K Matching Adjacent Elements?

This problem uses Math, Combinatorics. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Count the Number of Arrays with K Matching Adjacent Elements?

Insight 1: The problem can be solved using combinatorics. We need to choose which `k` pairs of adjacent elements will be equal. Insight 2: The remaining `n - 1 - k` adjacent pairs must be different. Each element must be selected from the range [1,m]

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Count the Number of Arrays with K Matching Adjacent Elements - LeetCode 3405 Solution

Q: How to solve Count the Number of Arrays with K Matching Adjacent Elements?

The solution uses combinatorics to calculate the number of good arrays. First, we calculate the number of ways to choose `k` adjacent pairs to be equal using combinations (nCr). Then, for each of these combinations, the first element has `m` choices, and the other elements that do not form equal adjacent pairs must have `m-1` choices, which we can calculate by taking power (m-1)^(n-1-k). The product of these terms modulo `10^9 + 7` gives the final answer.

Q: Which companies ask LeetCode 3405. Count the Number of Arrays with K Matching Adjacent Elements in interviews?

This problem is frequently asked at PhonePe.

Count the Number of Arrays with K Matching Adjacent Elements - Complete Solution Guide

Count the Number of Arrays with K Matching Adjacent Elements is LeetCode problem 3405, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 10 9 + 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,

Detailed Explanation

The problem requires us to count the number of "good arrays" of size `n` where each element is in the range `[1, m]` and exactly `k` adjacent elements are equal. We need to return this count modulo `10^9 + 7` because the number of such arrays can be very large.

Solution Approach

The solution uses combinatorics to calculate the number of good arrays. First, we calculate the number of ways to choose `k` adjacent pairs to be equal using combinations (nCr). Then, for each of these combinations, the first element has `m` choices, and the other elements that do not form equal adjacent pairs must have `m-1` choices, which we can calculate by taking power (m-1)^(n-1-k). The product of these terms modulo `10^9 + 7` gives the final answer.

Step-by-Step Algorithm

Step 1: Precompute factorials modulo `10^9 + 7` up to `n - 1`. This is to efficiently calculate combinations later.
Step 2: Implement a `power` function to calculate `base^exp` modulo `10^9 + 7` using binary exponentiation. This is used for modular exponentiation.
Step 3: Implement a `modInverse` function to calculate the modular inverse of a number modulo `10^9 + 7` using Fermat's Little Theorem (`a^(p-2) mod p`).
Step 4: Implement an `nCr` function to calculate combinations (`n choose k`) modulo `10^9 + 7`. It uses the precomputed factorials and modular inverses to compute the result.
Step 5: Calculate the number of ways to choose `k` matching adjacent elements by calling `nCr(n - 1, k)`.
Step 6: Calculate the number of ways to fill the remaining `n - 1 - k` non-matching adjacent pairs. The first element has 'm' choices and the subsequent n-1-k non-matching elements have m-1 choices, so, calculate `(m-1)^(n-1-k)` modulo `10^9 + 7`.
Step 7: Multiply the result of steps 5 and 6 with `m` and take the modulo `10^9 + 7` to get the final answer.

Key Insights

Insight 1: The problem can be solved using combinatorics. We need to choose which `k` pairs of adjacent elements will be equal.
Insight 2: The remaining `n - 1 - k` adjacent pairs must be different. Each element must be selected from the range [1,m]
Insight 3: We can precompute factorials for efficiency and use the modular inverse to calculate combinations (nCr) effectively modulo a large prime.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Math, Combinatorics.

Companies

Asked at: PhonePe.

Count the Number of Arrays with K Matching Adjacent Elements | LeetCode 3405

Hard

Math Combinatorics

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int countGoodArrays(int n, int m, int k) {
        final int MOD = 1_000_000_007;

        int max_fact = n - 1;
        long[] fact = new long[max_fact + 1];
        fact[0] = 1;
        for (int i = 1; i <= max_fact; i++) {
            fact[i] = (fact[i - 1] * i) % MOD;
        }

        
        
        
        
        long combinations = nCr(n - 1, k, fact, MOD);

        long power_m_minus_1 = power(m - 1, n - 1 - k, MOD);

        long result = (combinations * m) % MOD;
        result = (result * power_m_minus_1) % MOD;

        return (int) result;
    }

    private long power(long base, long exp, int MOD) {
        long res = 1;
        base %= MOD;
        while (exp > 0) {
            if (exp % 2 == 1)
                res = (res * base) % MOD;
            base = (base * base) % MOD;
            exp /= 2;
        }
        return res;
    }

    private long modInverse(long num, int MOD) {
        return power(num, MOD - 2, MOD);
    }

    private long nCr(int N, int K, long[] fact, int MOD) {
        if (K < 0 || K > N) {
            return 0;
        }

        long numerator = fact[N];
        long denominator = (fact[K] * fact[N - K]) % MOD;
        return (numerator * modInverse(denominator, MOD)) % MOD;
    }
}

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Problem Info

Description

Problem 3405: Count the Number of Arrays with K Matching Adjacent Elements - Detailed explanation and solution approaches.

Topics

Math Combinatorics

Companies

PhonePe

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3405

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Precompute factorials modulo `10^9 + 7` up to `n - 1`. This is to efficiently calculate combinations later.
2
Step 2: Implement a `power` function to calculate `base^exp` modulo `10^9 + 7` using binary exponentiation. This is used for modular exponentiation.
3
Step 3: Implement a `modInverse` function to calculate the modular inverse of a number modulo `10^9 + 7` using Fermat's Little Theorem (`a^(p-2) mod p`).
4
Step 4: Implement an `nCr` function to calculate combinations (`n choose k`) modulo `10^9 + 7`. It uses the precomputed factorials and modular inverses to compute the result.
5
Step 5: Calculate the number of ways to choose `k` matching adjacent elements by calling `nCr(n - 1, k)`.
6
Step 6: Calculate the number of ways to fill the remaining `n - 1 - k` non-matching adjacent pairs. The first element has 'm' choices and the subsequent n-1-k non-matching elements have m-1 choices, so, calculate `(m-1)^(n-1-k)` modulo `10^9 + 7`.
7
Step 7: Multiply the result of steps 5 and 6 with `m` and take the modulo `10^9 + 7` to get the final answer.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant operation is the precomputation of factorials which is O(n). The power and modular inverse functions use a logarithmic number of operations (binary exponentiation) which is O(log n), which is lesser than O(n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming

Dynamic programming could be used. Define dp[i][j] as the number of good arrays of length i, having j adjacent equal elements, and ending with a certain number. However, this method often leads to a higher complexity than the combinatoric approach if implemented naively.

Time: O(n*k*m)Space: O(n*k)

Trade-offs: Dynamic programming is more intuitive but likely less efficient for larger values of n and m. The combinatorics approach is more optimized.

✨

Key Insights

1Insight 1: The problem can be solved using combinatorics. We need to choose which `k` pairs of adjacent elements will be equal.
2Insight 2: The remaining `n - 1 - k` adjacent pairs must be different. Each element must be selected from the range [1,m]
3Insight 3: We can precompute factorials for efficiency and use the modular inverse to calculate combinations (nCr) effectively modulo a large prime.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: `k = 0`. In this case, no adjacent elements can be equal. The solution correctly calculates the result using the formula described above.
Edge case 2: `k = n - 1`. All adjacent elements must be equal. The solution handles this case as well because `(m-1)^(n-1-k)` will be `(m-1)^0`, and nCr(n-1, n-1) will be 1. Then the result would be 1 * m * 1 = m.
Edge case 3: `k > n - 1` or `k < 0`. The number of matching adjacent elements cannot be more than `n-1` or less than zero. The combination function returns 0 in these situations.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not taking the modulo at each step. Intermediate results can become very large and cause overflow if the modulo operation isn't applied frequently.

✗Common mistake 2: Not handling the modular inverse correctly. When dividing in modular arithmetic, you need to multiply by the modular inverse, calculated using Fermat's Little Theorem or the Extended Euclidean Algorithm.

✗Common mistake 3: Not understanding the constraints of combinations (nCr). Ensure that k is within the valid range (0 <= k <= n-1) and handle cases where it is not appropriately.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: This type of counting problem arises in probability calculations related to sequences or patterns, like counting DNA sequences with specific repeating patterns.

🏦Real-world scenario 2: In cryptography, similar combinatoric principles are used when analyzing the probability of certain key patterns or generating keys with particular properties.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 13, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 30

❓

Frequently Asked Questions

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Count the Number of Arrays with K Matching Adjacent Elements"?

For this hard-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether combinatorics techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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