Split Array With Same Average

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class Solution:
    def splitArraySameAverage(self, nums: List[int]) -> bool:
        total_sum = sum(nums)
        N = len(nums)

        # Possible sums for each length (1 to N//2 + 1)
        sums = [set() for _ in range(N//2 + 1)]
        sums[0].add(0)

        for num in nums:
            for i in range(N//2, 0, -1):
                for prev in sums[i - 1]:
                    sums[i].add(prev + num)

        for i in range(1, N//2 + 1):
            if total_sum * i % N == 0 and total_sum * i // N in sums[i]:
                return True

        return False

Problem Classification

This problem can be categorized into the domain of “Array Manipulation and Mathematical Calculation”. The main theme of the problem is to partition an array into two such that the average of both sub-arrays is the same.

What components:

  1. Input: An integer array nums. The array’s length is at least 1 and at most 30, and each element in the array is in the range [0, 104].

  2. Output: A boolean value (true or false). true indicates that it’s possible to partition the array into two sub-arrays having the same average, and false indicates it’s not possible.

  3. Constraints: The problem requires partitioning the array such that both resultant arrays are non-empty and their averages are equal.

The problem can be further classified as a Partition Problem, which is a subset of combinatorics and optimization problems in computer science. The task is to determine if given a set (or multiset) of integers, whether the numbers can be divided into two subsets such that the sum of numbers in both subsets is the same.

In this problem, we have a twist to the classical partition problem, i.e., we’re required to check if the set can be divided into two subsets such that the averages (not sums) of both subsets are equal.

Constraints

Given the constraints and the problem statement, here are some specific characteristics and conditions that can be exploited:

  1. Array Size: The size of the array is quite small (maximum length is 30). This suggests that even solutions with higher time complexity (like ones involving recursion or backtracking) might be feasible.

  2. Number Range: The values of the array elements are also quite small (ranging from 0 to 10^4). This fact may allow us to make use of some kind of bucketing or counting technique to speed up calculations.

  3. Equal Averages: If the averages of two sub-arrays are to be equal, it means the sums of these sub-arrays should be proportional to their sizes. In other words, sumA/sizeA = sumB/sizeB implies sumA * sizeB = sumB * sizeA. This mathematical relation can be used to guide our approach to the solution.

  4. All Elements Used: Given that all elements from the original array should be distributed into either of the two sub-arrays (which must be non-empty), we cannot discard any values. This suggests that we may need to examine all potential combinations or permutations of the elements.

  5. Return Type: Since the problem only asks for a boolean response (i.e., whether it is possible or not to partition), we do not have to keep track of the actual partitions. This can simplify our solution approach.

These observations give us valuable insights that we can use to optimize our solution. They can help us determine a strategy for solving the problem and may suggest particular algorithmic paradigms that might be effective, such as dynamic programming or backtracking.

Thought Process

Given the problem statement, one of the primary cues is the need to split the array such that the averages of the two resulting arrays are equal. This gives us a mathematical relationship between the sums and lengths of the two arrays. More specifically, the sums of the two arrays divided by their respective lengths must be equal.

This translates to the following equality:

(sum(A) / len(A)) = (sum(B) / len(B))

Which we can rearrange as:

sum(A) * len(B) = sum(B) * len(A)

With this equation, we now have a solid direction to move towards. We need to find a subset of the original array that satisfies this equation.

Given that the length of the array is relatively small (up to 30), a brute force approach could work. We can generate all possible subsets of the array and check if any of them satisfy the equation.

However, a more efficient approach would be to iterate over the array and maintain a set of possible sums at each step. This set represents the sums of all subsets we can form using the elements we have seen so far. We can then check if any of these sums satisfy the equation.

Here is the code:

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class Solution:
    def splitArraySameAverage(self, nums: List[int]) -> bool:
        total_sum = sum(nums)
        N = len(nums)

        # Possible sums for each length (1 to N//2 + 1)
        sums = [set() for _ in range(N//2 + 1)]
        sums[0].add(0)

        for num in nums:
            for i in range(N//2, 0, -1):
                for prev in sums[i - 1]:
                    sums[i].add(prev + num)

        for i in range(1, N//2 + 1):
            if total_sum * i % N == 0 and total_sum * i // N in sums[i]:
                return True

        return False

This code first calculates the total sum and the length of the array. It then creates an array of sets to store the possible sums for each length. It iterates over the array and for each number, it updates the possible sums. Finally, it checks if there exists a length such that the total sum times the length divided by N is in the possible sums for that length. If such a length exists, it returns True. Otherwise, it returns False.

This problem helps highlight the power of dynamic programming and how precomputing certain information (in this case, the possible sums for each length) can significantly speed up the solution.

Observe that avg(B)=avg(C)=avg(A). For each a[i], set a[i]=a[i]-avg(A). Then we can reducing to finding a subset with sum 0. (using this trick, we don’t need to worry about the cardinality of the subset.) I also want to add the following. If n = A.size, avg(A) is the average of A, sum(A) is the sum of all elements of A. Then if we do B[i] = A[i] * n, then avg(B) == sum(A). Now doing the same trick with B: B[i] = B[i] - sum(A), we get: avg(B) = 0 No need to work with floating point numbers.

You’re correct. Shifting the elements by the average or the total sum is a clever approach to simplify the problem.

Doing so, indeed, converts the problem from finding a subset with the same average to finding a subset with sum 0, which is a common problem that can be solved using dynamic programming.

Also, shifting the elements by multiplying with the length and subtracting the total sum eliminates the need to deal with floating-point numbers, and therefore, the potential for precision errors, which is a key consideration when dealing with average values.

Using this insight, we can modify the Python code as follows:

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class Solution:
    def splitArraySameAverage(self, nums: List[int]) -> bool:
        N = len(nums)
        total_sum = sum(nums)
        avg = total_sum / N
        nums = [num*N - total_sum for num in nums]

        # Initialize DP table
        sums = [set() for _ in range(N//2 + 1)]
        sums[0].add(0)

        # Update DP table
        for num in nums:
            for i in range(N//2, 0, -1):
                for prev in sums[i - 1]:
                    sums[i].add(prev + num)
        
        # Check if there exists a subset with sum 0
        for i in range(1, N//2 + 1):
            if 0 in sums[i]:
                return True

        return False

This revised solution uses dynamic programming to maintain possible sums for each subset size and then checks if a zero sum subset exists. If it does, it returns True, else it returns False. This is a more elegant solution that also has the advantage of being more numerically stable due to avoidance of floating point operations.

Solution Approach and Analysis

Given the problem, our task is to determine if we can split the array into two non-empty arrays with equal averages. This essentially means that the sum of the elements in both arrays, divided by the number of elements in each array, should be the same.

Here’s a detailed step-by-step approach to solving the problem:

  1. Calculate Total Sum and Length: Firstly, calculate the total sum and the length of the input array. We’ll need this information to determine if there exists a pair of numbers whose sum times the length is equal to the total sum.

  2. Edge Cases Check: If the total sum is an integer (it’s evenly divisible by the array’s length), this means it’s possible to split the array into two arrays with equal averages. In contrast, if the total sum is a decimal (it’s not evenly divisible by the array’s length), then it’s impossible to split the array into two arrays with equal averages.

  3. Create a Set for Possible Sums: We create a set for each possible sum of the subsets. We initialize the first set with just a zero in it.

  4. Iterate Over the Array: We iterate over the array and for each number, we create a new set and add the current number to each sum in the last set. Then, we add this new set to our list of sets.

  5. Check Each Possible Subset Sum: Finally, we check each possible subset sum if there exists a pair of numbers whose sum times the length is equal to the total sum.

Consider the analogy of putting balls into two boxes (A and B). The total weight of balls (sum of array) and the total number of balls (length of array) is fixed. Our task is to decide how to distribute the balls so that the average weight (sum/number of balls) in both boxes is the same. If we cannot divide the total weight evenly among all balls, then it’s impossible to achieve equal average weight in both boxes.

Let’s see an example:

Input: nums = [1,2,3,4,5,6,7,8]

total_sum = 36, length = 8

The total sum of 36 is evenly divisible by the length 8, which means it's possible to split the array into two non-empty sub-arrays with equal averages. We iterate over the array and check if there exists a pair of numbers whose sum times the length is equal to the total sum.

This approach can handle changes in the problem’s parameters effectively. Adding or removing elements from the array would simply involve updating the total sum and length and re-executing the checks. Changes to individual elements might change the total sum but would not affect the length.

In terms of time complexity, the dominant factor is the generation and checking of all possible subset sums, which is exponential in the number of elements in the array. However, given the problem constraints (specifically, the relatively small maximum length of the array), this approach is feasible.

nums = [1,2,3,4,5,6,7,8] So the possible subsets are 256,

In that [4,5] and [3,6] has also same average.

I confusing in how to make a subset. For Ex) nums = [1,6,1] , It expect false. But my POV is that [1] and [1] comes out same average. So it became true naa???

For the array [1,6,1], indeed, the subsets [1] and [1] have the same average. However, according to the problem statement, you need to divide the original array into two subsets A and B such that each element from the array is in exactly one of these subsets. In this case, the number 6 would be left out, which isn’t allowed by the problem constraints.

Therefore, for the array [1,6,1], you can’t find two subsets A and B where every number from the array is used exactly once, and the averages of A and B are equal, so the correct output would indeed be False.

Language Agnostic Coding Drills

  1. Identification of Coding Concepts:

a. Mathematical Operations: Simple mathematical operations are used to compute the sum and the average of the array. They are also used to transform each number in the array for ease of further computations.

b. List Comprehensions: A Python-specific syntax that creates a new list based on an existing list. In this case, it’s used to generate the transformed array and to initialize the DP table.

c. Dynamic Programming (DP): The core of this solution is a DP strategy that keeps track of all possible sums of subsets of different sizes. The goal is to find if there’s a subset with a sum equal to 0.

d. Nested Loops: The algorithm uses nested loops to iterate over the transformed array and the DP table. Each element in the array is considered with each sum in the DP table.

e. Set Operations: Sets are used in this solution to track the unique possible sums of subsets. They offer efficient membership tests which are exploited to check for the presence of 0 in each subset sum list.

  1. Difficulty Classification:

a. Mathematical Operations (Easy): Basic knowledge required in every programming task.

b. List Comprehensions (Medium): Specific to Python and requires understanding of how it works and when to use it.

c. Dynamic Programming (Hard): Involves understanding the concept of DP, setting up the DP table, and figuring out the transition function.

d. Nested Loops (Medium): Requires a good understanding of iteration in Python and careful handling to avoid infinite loops or off-by-one errors.

e. Set Operations (Medium): Requires knowledge of set data structure and its operations.

  1. Problem-solving approach:

The problem essentially requires us to check if it’s possible to split the array into two non-empty subsets with the same average. The code starts by computing the overall average and then transforming each number in a way that a subset with an average equal to the overall average will have a sum of 0 in the transformed array.

It then sets up a DP table, where each entry is a set that holds the possible sums of subsets of a certain size. It iteratively updates the DP table by considering each number in the transformed array.

Finally, it checks the DP table for a subset with a sum of 0. If it finds such a subset, it means there exists a subset with an average equal to the overall average, hence it returns True. If no such subset is found after checking all subsets, it returns False.

Targeted Drills in Python

  1. Mathematical Operations:
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# Addition
total = 3 + 2
print(total)  # Output: 5

# Division
avg = total / 2
print(avg)  # Output: 2.5

# Multiplication and Subtraction
transformed_num = 2 * 3 - total
print(transformed_num)  # Output: 1
  1. List Comprehensions:
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# Original list
original_list = [1, 2, 3, 4, 5]

# Using list comprehension to create a new list
new_list = [num * 2 for num in original_list]
print(new_list)  # Output: [2, 4, 6, 8, 10]
  1. Dynamic Programming (DP):
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# A simple example of using DP to compute Fibonacci numbers
def fib(n):
    dp = [0, 1] + [0]*(n-1)
    for i in range(2, n+1):
        dp[i] = dp[i-1] + dp[i-2]
    return dp[n]

print(fib(10))  # Output: 55
  1. Nested Loops:
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# A simple example of nested loops to print a 2D array
arr = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
for row in arr:
    for num in row:
        print(num, end=' ')
    print()
  1. Set Operations:
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# Create a set
numbers = set()

# Add elements to the set
numbers.add(1)
numbers.add(2)

# Check if an element is in the set
print(1 in numbers)  # Output: True

To solve our problem, we’ll first compute the sum and average of the array, then create a transformed array using a list comprehension. Then, we’ll initialize our DP table, which will be a list of sets, with the first set containing only 0.

Next, we’ll use nested loops and set operations to update our DP table. For each number in the transformed array, we’ll iterate over the DP table in reverse, and for each set in the DP table, we’ll add the current number to every element in the set, and add the result to the next set.

Finally, we’ll check if there’s a subset with a sum of 0 in our DP table. If there is, we’ll return True, otherwise, we’ll return False.