How do you solve Pizza With 3n Slices on LeetCode?

Pizza With 3n Slices (LeetCode #1388) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n^2) time complexity and O(n) space complexity. Key insight: Understanding the circular nature of the problem is crucial for avoiding adjacent selections.

What is the brute force solution for Pizza With 3n Slices?

The brute force approach for Pizza With 3n Slices is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will try every possible combination of pizza slices we can pick, ensuring that no two picked slices are adjacent. This method guarantees that we explore all options, but it can be very slow for larger inputs. The optimal Optimal Solution approach improves this to O(n^2).

What algorithm or data structure is used to solve Pizza With 3n Slices?

Pizza With 3n Slices uses the following concepts: Array, Dynamic Programming, Greedy, Heap (Priority Queue). The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Pizza With 3n Slices?

Always clarify the problem constraints and edge cases before diving into coding. Explain your thought process and approach clearly to the interviewer as you work through the problem. Practice writing clean and efficient code, as readability can be just as important as functionality.

What are common mistakes when solving Pizza With 3n Slices?

Forgetting to account for the circular arrangement of the slices when checking adjacency. Not properly initializing the DP table, leading to incorrect results.

#1388

Pizza With 3n Slices

Hard

ArrayDynamic ProgrammingGreedyHeap (Priority Queue)Dynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n^2)
Space	O(1)	O(n)

💡

Intuition

Time O(n^2)Space O(n)

The optimal solution uses dynamic programming to efficiently calculate the maximum sum of slices we can pick without selecting adjacent slices. This approach reduces the number of combinations we need to check by leveraging previously computed results.

⚙️

Algorithm

3 steps

1Step 1: Define a DP table where dp[i][j] represents the maximum sum we can get by picking j slices from the first i slices.
2Step 2: Iterate through the slices, updating the DP table based on whether we pick the current slice or not.
3Step 3: Ensure that when picking a slice, we skip the adjacent slices by using the previous results in the DP table.

solution.py7 lines

1def maxSizeSlices(slices):
2    n = len(slices) // 3
3    dp = [[0] * (n + 1) for _ in range(len(slices) + 1)]
4    for i in range(1, len(slices) + 1):
5        for j in range(1, min(i, n) + 1):
6            dp[i][j] = max(dp[i-1][j], dp[i-2][j-1] + slices[i-1])
7    return dp[len(slices)][n]

ℹ

Complexity note: The time complexity is O(n²) due to the nested loops for filling the DP table. The space complexity is O(n) because we store results in a 2D array proportional to the number of slices.

1Understanding the circular nature of the problem is crucial for avoiding adjacent selections.
2Dynamic programming can significantly reduce the number of computations by reusing previously calculated results.

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