How do you solve Number of ZigZag Arrays II on LeetCode?

Number of ZigZag Arrays II (LeetCode #3700) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: ZigZag arrays require careful state management.

What is the time complexity of Number of ZigZag Arrays II?

The optimal solution for Number of ZigZag Arrays II runs in O(n) time and uses O(n) space. The complexity is linear due to the dynamic programming approach, where we only store the last state.

What is the brute force solution for Number of ZigZag Arrays II?

The brute force approach for Number of ZigZag Arrays II is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all possible arrays and filter out invalid ones. 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 Number of ZigZag Arrays II?

Number of ZigZag Arrays II uses the following concepts: Math, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Matrix Exponentiation.

What are the interview tips for Number of ZigZag Arrays II?

Explain your thought process clearly. Discuss edge cases and how they affect your solution. Practice similar problems to build confidence.

What are common mistakes when solving Number of ZigZag Arrays II?

Not accounting for edge cases in array generation. Overlooking the modulo operation in large sums.

#3700

Number of ZigZag Arrays II

Hard

MathDynamic ProgrammingDynamic ProgrammingMatrix Exponentiation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Use dynamic programming to count valid transitions between states, leveraging matrix exponentiation for efficiency.

⚙️

Algorithm

3 steps

1Step 1: Define a state transition matrix based on ZigZag rules.
2Step 2: Use matrix exponentiation to compute the number of valid arrays of length n efficiently.
3Step 3: Sum the results from the last state to get the total count.

solution.py12 lines

1def countZigZagArrays(n, l, r):
2    mod = 10**9 + 7
3    m = r - l + 1
4    dp = [[0] * (2 * m) for _ in range(n)]
5    for i in range(m):
6        dp[0][i] = 1
7        dp[0][m + i] = 1
8    for i in range(1, n):
9        for j in range(m):
10            dp[i][j] = (sum(dp[i - 1][m + k] for k in range(m) if k != j) % mod)
11            dp[i][m + j] = (sum(dp[i - 1][k] for k in range(m) if k != j) % mod)
12    return sum(dp[n - 1]) % mod

ℹ

Complexity note: The complexity is linear due to the dynamic programming approach, where we only store the last state.

1ZigZag arrays require careful state management.
2Dynamic programming can significantly reduce computation time.

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