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

The optimal solution for Count the Number of Arrays with K Matching Adjacent Elements runs in O(n) time and uses O(1) space. The optimal solution runs in O(n) due to the power calculation, making it efficient for large inputs.

What is the brute force solution for Count the Number of Arrays with K Matching Adjacent Elements?

The brute force approach for Count the Number of Arrays with K Matching Adjacent Elements is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all possible arrays and count those that meet the criteria. This method is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count the Number of Arrays with K Matching Adjacent Elements?

Count the Number of Arrays with K Matching Adjacent Elements uses the following concepts: Math, Combinatorics. The recommended patterns to study are: Combinatorics, Dynamic Programming.

What are the interview tips for Count the Number of Arrays with K Matching Adjacent Elements?

Explain your thought process clearly. Use examples to illustrate your approach. Practice combinatorial problems to build intuition.

What are common mistakes when solving Count the Number of Arrays with K Matching Adjacent Elements?

Not considering edge cases like k > n-1. Misunderstanding how to fill remaining positions.

#3405

Count the Number of Arrays with K Matching Adjacent Elements

Hard

MathCombinatoricsCombinatoricsDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Use combinatorial counting to determine valid configurations without generating all arrays. This leverages the structure of the problem.

⚙️

Algorithm

3 steps

1Step 1: Choose the first element (m options).
2Step 2: Choose k positions to have matching adjacent elements.
3Step 3: Fill the remaining positions with different elements (m-1 options).

solution.py5 lines

1def count_good_arrays(n, m, k):
2    MOD = 10**9 + 7
3    if k > n - 1:
4        return 0
5    return (m * pow(m - 1, n - k - 1, MOD)) % MOD

ℹ

Complexity note: The optimal solution runs in O(n) due to the power calculation, making it efficient for large inputs.

1Choosing the first element defines the rest of the array structure.
2The number of ways to choose positions for matches is combinatorial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.