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Problem Statement

You are given an integer array A and an integer K. You need to partition the array into at most K non-empty adjacent subarrays. The score of a partition is the sum of the averages of each subarray. Return the largest score you can achieve.

Clarifying Questions

To fully understand the problem, we need to clarify:

  1. What are the constraints on the values of the array A and the integer K?
  2. Are all the elements in array A positive integers, or can they be negative or zero as well?
  3. Can K be greater than the length of the array A?
  4. Do we need to consider any edge cases such as empty array or K equals 1?

Assumptions and Constraints

Based on typical coding interview settings:

  1. Array A can have values ranging from negative to positive integers.
  2. The length of array A is at least 1.
  3. (1 \leq K \leq \text{length of } A).

Strategy

We’ll employ a dynamic programming approach to solve this problem:

  1. Define ( \text{dp}[i][k] ) as the largest score we can achieve for the first i elements with k partitions.
  2. Use a prefix sum array to store the sum of elements up to any given index, which helps in quickly calculating the average of any subarray.
  3. Use nested loops:
    • Outer loop for the number of partitions ( k ) from 1 to K.
    • Inner loop for the length of the array ( i ) from 1 to n.
    • The innermost loop to choose the partition point and calculate the maximum achievable score at each step.

Time Complexity

The dynamic programming approach has a time complexity of ( O(n^2 \times K) ) where n is the length of the array and K is the number of partitions.

This is because:

Implementation

Here is the Python code for this solution:

def largestSumOfAverages(A, K):
    n = len(A)
    prefix_sum = [0] * (n + 1)
    
    # Calculate prefix sums
    for i in range(n):
        prefix_sum[i + 1] = prefix_sum[i] + A[i]
    
    # Initialize dp array
    dp = [[0] * (K + 1) for _ in range(n + 1)]
    
    # Base case: 1 partition
    for i in range(1, n + 1):
        dp[i][1] = prefix_sum[i] / i
        
    # Fill dp array
    for k in range(2, K + 1):  # Number of partitions
        for i in range(1, n + 1):  # Length of array considered
            for j in range(i):  # Partition point
                dp[i][k] = max(dp[i][k], dp[j][k-1] + (prefix_sum[i] - prefix_sum[j]) / (i - j))
    
    return dp[n][K]

# Example usage:
A = [9, 1, 2, 3, 9]
K = 3
print(largestSumOfAverages(A, K))  # Output should be 20

Explanation

  1. Prefix Sum Calculation: We calculate the prefix sums to help in quick subarray sum calculation.
  2. DP Initialization: The base case where there is only one partition (i.e., the entire array).
  3. Main DP Loop: We fill the dp table considering each partition incrementally.
    • For each number of partitions k, we consider the possible subarrays and calculate the best partition point to maximize the average.

Let me know if you have further questions or need additional explanations!

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