How do you solve Split Array With Same Average on LeetCode?

Split Array With Same Average (LeetCode #805) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * n/2) time complexity and O(n) space complexity. Key insight: The average condition can be transformed into a sum condition, allowing for more efficient checks.

What is the brute force solution for Split Array With Same Average?

The brute force approach for Split Array With Same Average is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try every possible way to split the array into two non-empty subsets and check if their averages are equal. This approach is simple but inefficient as it involves checking all combinations. The optimal Optimal Solution approach improves this to O(n * n/2).

What are the interview tips for Split Array With Same Average?

Always clarify the problem requirements and constraints before diving into coding. Think about how mathematical properties can simplify the problem. Practice explaining your thought process clearly, as communication is key during interviews.

What are common mistakes when solving Split Array With Same Average?

Not considering the mathematical relationship between averages and sums. Failing to handle edge cases, such as when the array has only two elements.

#805

Split Array With Same Average

Hard

ArrayHash TableMathDynamic ProgrammingBit ManipulationBitmaskHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n * n/2)Space O(n)

Instead of checking all combinations, we can use a mathematical property. If the average of two subsets is equal, their sums must be proportional to their sizes. We can use dynamic programming to efficiently find valid subsets.

⚙️

Algorithm

5 steps

1Step 1: Calculate the total sum and length of the array.
2Step 2: Iterate through possible sizes of subset A from 1 to n/2.
3Step 3: For each size, check if there exists a subset of that size with a sum that maintains the average condition.
4Step 4: Use a dynamic programming approach to track possible sums for each subset size.
5Step 5: If a valid subset is found, return true; otherwise, return false.

solution.py13 lines

1def splitArraySameAverage(nums):
2    total_sum = sum(nums)
3    n = len(nums)
4    for size in range(1, n // 2 + 1):
5        if total_sum * size % n == 0:
6            target = total_sum * size // n
7            dp = set([0])
8            for num in nums:
9                for s in list(dp):
10                    dp.add(s + num)
11            if target in dp:
12                return True
13    return False

ℹ

Complexity note: The time complexity is O(n * n/2) due to the dynamic programming approach to track possible sums, which is more efficient than the brute force method.

1The average condition can be transformed into a sum condition, allowing for more efficient checks.
2Dynamic programming can be used to track achievable sums for subsets, reducing the need for exhaustive search.

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