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

Number of ZigZag Arrays I (LeetCode #3699) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * (r - l)^2) time complexity and O(n * (r - l)) space complexity. Key insight: Adjacent elements must differ.

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

The optimal solution for Number of ZigZag Arrays I runs in O(n * (r - l)^2) time and uses O(n * (r - l)) space. The complexity arises from filling the DP table, iterating through all possible values in the range for each length.

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

The brute force approach for Number of ZigZag Arrays I is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all possible arrays of length n within the range [l, r] and check for ZigZag properties. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n * (r - l)^2).

What algorithm or data structure is used to solve Number of ZigZag Arrays I?

Number of ZigZag Arrays I uses the following concepts: Dynamic Programming, Prefix Sum. The recommended patterns to study are: Dynamic Programming, Prefix Sum.

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

Explain your thought process clearly. Discuss edge cases. Practice dynamic programming problems.

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

Forgetting to check all adjacent elements. Not considering the range limits.

#3699

Number of ZigZag Arrays I

Hard

Dynamic ProgrammingPrefix SumDynamic ProgrammingPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * (r - l)^2)
Space	O(1)	O(n * (r - l))

💡

Intuition

Time O(n * (r - l)^2)Space O(n * (r - l))

Use dynamic programming to count valid sequences based on the last element and direction of the last move, leveraging prefix sums for efficiency.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP table where dp[i][dir][x] counts sequences of length i ending at x with direction dir.
2Step 2: Fill the DP table using the rules for ZigZag arrays based on the last direction and valid transitions.
3Step 3: Sum the counts from the last row of the DP table to get the total valid ZigZag arrays.

solution.py13 lines

1def zigzagArrays(n, l, r):
2    mod = 10**9 + 7
3    dp = [[[0] * (r + 1) for _ in range(2)] for _ in range(n + 1)]
4    for x in range(l, r + 1):
5        dp[1][0][x] = dp[1][1][x] = 1
6    for i in range(1, n):
7        for x in range(l, r + 1):
8            for y in range(l, r + 1):
9                if x < y:
10                    dp[i + 1][0][y] = (dp[i + 1][0][y] + dp[i][1][x]) % mod
11                elif x > y:
12                    dp[i + 1][1][y] = (dp[i + 1][1][y] + dp[i][0][x]) % mod
13    return sum(dp[n][0][l:r + 1]) + sum(dp[n][1][l:r + 1]) % mod

ℹ

Complexity note: The complexity arises from filling the DP table, iterating through all possible values in the range for each length.

1Adjacent elements must differ.
2Three consecutive elements cannot form increasing/decreasing sequences.

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