Jobs

Interviews

Discussions

Largest Sum of Averages

Medium

Asked by:

20 views

Topics:

ArraysDynamic Programming

You are given an integer array nums and an integer k. You can 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.

Note that the partition must use every integer in nums, and that the score is not necessarily an integer.

Return the maximum score you can achieve of all the possible partitions. Answers within 10^-6 of the actual answer will be accepted.

Example 1:

Input: nums = [9,1,2,3,9], k = 3
Output: 20.00000
Explanation: 
The best choice is to partition nums into [9], [1, 2, 3], [9]. The answer is 9 + (1 + 2 + 3) / 3 + 9 = 20.
We could have also partitioned nums into [9, 1], [2], [3, 9], for example.
That partition would lead to a score of 5 + 2 + 6 = 13, which is worse.

Example 2:

Input: nums = [1,2,3,4,5,6,7], k = 4
Output: 20.50000

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 10⁴
1 <= k <= nums.length

Try coding it in LeetCode

Largest Sum of Averages

Medium

Asked by:

20 views

Topics:

ArraysDynamic Programming

Note that the partition must use every integer in nums, and that the score is not necessarily an integer.

Return the maximum score you can achieve of all the possible partitions. Answers within 10^-6 of the actual answer will be accepted.

Example 1:

Input: nums = [9,1,2,3,9], k = 3
Output: 20.00000
Explanation: 
The best choice is to partition nums into [9], [1, 2, 3], [9]. The answer is 9 + (1 + 2 + 3) / 3 + 9 = 20.
We could have also partitioned nums into [9, 1], [2], [3, 9], for example.
That partition would lead to a score of 5 + 2 + 6 = 13, which is worse.

Example 2:

Input: nums = [1,2,3,4,5,6,7], k = 4
Output: 20.50000

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 10⁴
1 <= k <= nums.length

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the input array `nums` and the value of `k`?
Can the elements in the input array `nums` be negative, zero, or only positive?
If `nums` is empty or `k` is 0 or greater than the length of `nums`, what should the function return?
If there are multiple ways to partition `nums` into `k` non-empty groups, is any optimal solution acceptable or is there a specific method of choosing amongst them?
Should the answer be returned as an integer or a floating-point number? If floating-point, what level of precision is expected?

Brute Force Solution

Approach

The brute force approach to this problem involves trying every possible way to divide the numbers into groups. We want to find the grouping that gives us the highest score when we calculate the average of each group and add those averages together.

Here's how the algorithm would work step-by-step:

First, try making only one group that contains all the numbers. Calculate the average of this group.
Next, try dividing the numbers into two groups in every possible way. Calculate the average of each group and add them together. Keep track of the highest total average you find.
Then, try dividing the numbers into three groups in every possible way. Again, calculate the average of each group and add them together. Compare this total to the highest total you've found so far and update if needed.
Continue this process, increasing the number of groups up to the maximum allowed. Each time, try all possible combinations of numbers in each group.
Finally, after exploring all possible groupings, the highest total average you kept track of is the answer.

Code Implementation

def largest_sum_of_averages_brute_force(numbers, group_size):
    number_count = len(numbers)
    maximum_average_sum = 0.0

    # Iterate through all possible numbers of groups
    for number_of_groups in range(1, group_size + 1):
        maximum_average_sum_for_groups = calculate_largest_average_sum(
            numbers, 0, number_of_groups, {}
        )

        maximum_average_sum = max(maximum_average_sum, maximum_average_sum_for_groups)

    return maximum_average_sum

def calculate_largest_average_sum(
    numbers, current_index, remaining_groups, memo
):
    if (current_index, remaining_groups) in memo:
        return memo[(current_index, remaining_groups)]

    number_count = len(numbers)

    # Base case: If no groups are left, no more average to add
    if remaining_groups == 0:
        if current_index == number_count:
            return 0.0
        else:
            return float('-inf')

    # Base case: Only one group remaining, include all numbers
    if current_index == number_count:
        return float('-inf')

    if remaining_groups == 1:
        current_sum = sum(numbers[current_index:])
        average = current_sum / (number_count - current_index)
        memo[(current_index, remaining_groups)] = average
        return average

    maximum_average_sum = float('-inf')

    # Iterate through all possible sizes of the current group
    for group_end_index in range(current_index + 1, number_count + 1):
        current_sum = sum(numbers[current_index:group_end_index])
        average = current_sum / (group_end_index - current_index)

        # Recursively calculate the largest average sum for remaining groups
        remaining_average_sum = calculate_largest_average_sum(
            numbers, group_end_index, remaining_groups - 1, memo
        )

        maximum_average_sum = max(
            maximum_average_sum, average + remaining_average_sum
        )

    memo[(current_index, remaining_groups)] = maximum_average_sum
    return maximum_average_sum

Big(O) Analysis

Time Complexity

O(n^k) – The brute force approach iterates through all possible groupings of n numbers into k groups. For each k from 1 to k, the algorithm explores all combinations, leading to a computationally expensive process. The number of ways to partition a set of n elements into k non-empty subsets is given by Stirling numbers of the second kind, which grows faster than polynomial but slower than exponential. Since we are exploring *all* such groupings for each possible k from 1 up to k, the runtime is dominated by the largest number of combinations explored. While difficult to express precisely, it involves summing these Stirling numbers which is upper-bounded by something akin to n^k where 'k' is the maximum number of groups, leading to O(n^k).

Space Complexity

O(1) – The plain English explanation focuses on iterative calculations and comparisons of averages. It only describes tracking the highest total average found so far, which can be stored in a single variable. No auxiliary arrays, lists, or other data structures that scale with the input size N (the number of numbers) are mentioned or implied. Thus, the space required is constant regardless of the input size, leading to O(1) space complexity.

Optimal Solution

Approach

The best way to solve this is by figuring out the largest average we can achieve by strategically splitting the numbers into groups. We use a method that calculates the optimal average sum from the end towards the beginning, remembering our previous calculations to avoid repetition.

Here's how the algorithm would work step-by-step:

Start by understanding that the best average sum for the entire set when divided into one group is just the average of all numbers.
Now, consider splitting the set into two groups. To find the best split, try every possible split point and calculate the average sum for each split. Remember the highest average sum you find.
Continue this process for splitting into three, four, and finally up to K groups. Each time, you're using the best average sums you've already calculated for the earlier parts of the list.
The key is to calculate the best average sum for a sub-list from some point until the end. Then, when you want to calculate for the next group before it, you can reuse this calculation to avoid recalculating.
The final answer is the best average sum you found when splitting into K groups.

Code Implementation

def largest_sum_of_averages(numbers, k_groups):
    number_count = len(numbers)
    # This array will store the prefix sums.
    prefix_sums = [0] * (number_count + 1)
    for i in range(1, number_count + 1):
        prefix_sums[i] = prefix_sums[i - 1] + numbers[i - 1]

    # dp[i] is the largest average sum of numbers[i:] into 1 group
    dp = [0] * number_count
    for i in range(number_count):
        dp[i] = (prefix_sums[number_count] - prefix_sums[i]) / (number_count - i)

    for group_count in range(2, k_groups + 1):
        # Iterate backwards, building up the optimal solutions
        for i in range(number_count - 1):
            # Try every possible split point.
            for j in range(i + 1, number_count):
                dp[i] = max(dp[i], (prefix_sums[j] - prefix_sums[i]) / (j - i) + dp[j])

    return dp[0]

Big(O) Analysis

Time Complexity

O(n*K) – The algorithm iterates to find the largest average by splitting the input array of size n into K groups. For each possible value of K (from 1 to the given K), the algorithm iterates through the array to find the optimal split point. The outer loop runs K times, and the inner loop to find the split runs up to n times. This results in a time complexity of O(n*K).

Space Complexity

O(N*K) – The algorithm uses dynamic programming to store the maximum average sums for subproblems. Specifically, it stores the best average sum for splitting a sub-array from a given index to the end into a certain number of groups. This requires a 2D array of size N*K, where N is the number of elements in the input array and K is the maximum number of groups we can split into. Therefore, the space complexity is O(N*K).

Edge Cases

Case	How to Handle
Empty input array	Return 0, as there are no numbers to partition and average.
K is 0	Return negative infinity or an error as it's an invalid partition.
K is greater than the array length	Return the average of the entire array (equivalent to K=1).
Array contains negative numbers	The algorithm should handle negative numbers correctly as it calculates sums and averages.
Array contains a mix of very large positive and very large negative numbers	Ensure sums are calculated using a data type that can accommodate large numbers to prevent integer overflow (e.g., long or double).
Array contains only zeros	The average calculation should correctly handle zero values and avoid division by zero errors.
Array is very large and K is close to the array size	The dynamic programming solution should be optimized for memory and time complexity, considering that the DP table size depends on the array size and K.
Floating-point precision issues	Be mindful of potential precision loss when calculating averages, especially with very large or very small numbers; consider using higher-precision floating-point types if necessary and handle rounding appropriately when comparing results.

Empty input array

How to Handle:

Return 0, as there are no numbers to partition and average.

K is 0

How to Handle:

Return negative infinity or an error as it's an invalid partition.

K is greater than the array length

How to Handle:

Return the average of the entire array (equivalent to K=1).

Array contains negative numbers

How to Handle:

The algorithm should handle negative numbers correctly as it calculates sums and averages.

Array contains a mix of very large positive and very large negative numbers

How to Handle:

Ensure sums are calculated using a data type that can accommodate large numbers to prevent integer overflow (e.g., long or double).

Array contains only zeros

How to Handle:

The average calculation should correctly handle zero values and avoid division by zero errors.

Array is very large and K is close to the array size

How to Handle:

The dynamic programming solution should be optimized for memory and time complexity, considering that the DP table size depends on the array size and K.

Floating-point precision issues

How to Handle:

Be mindful of potential precision loss when calculating averages, especially with very large or very small numbers; consider using higher-precision floating-point types if necessary and handle rounding appropriately when comparing results.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the input array `nums` and the value of `k`?
Can the elements in the input array `nums` be negative, zero, or only positive?
If `nums` is empty or `k` is 0 or greater than the length of `nums`, what should the function return?
If there are multiple ways to partition `nums` into `k` non-empty groups, is any optimal solution acceptable or is there a specific method of choosing amongst them?
Should the answer be returned as an integer or a floating-point number? If floating-point, what level of precision is expected?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

First, try making only one group that contains all the numbers. Calculate the average of this group.
Next, try dividing the numbers into two groups in every possible way. Calculate the average of each group and add them together. Keep track of the highest total average you find.
Then, try dividing the numbers into three groups in every possible way. Again, calculate the average of each group and add them together. Compare this total to the highest total you've found so far and update if needed.
Continue this process, increasing the number of groups up to the maximum allowed. Each time, try all possible combinations of numbers in each group.
Finally, after exploring all possible groupings, the highest total average you kept track of is the answer.

Code Implementation

def largest_sum_of_averages_brute_force(numbers, group_size):
    number_count = len(numbers)
    maximum_average_sum = 0.0

    # Iterate through all possible numbers of groups
    for number_of_groups in range(1, group_size + 1):
        maximum_average_sum_for_groups = calculate_largest_average_sum(
            numbers, 0, number_of_groups, {}
        )

        maximum_average_sum = max(maximum_average_sum, maximum_average_sum_for_groups)

    return maximum_average_sum

def calculate_largest_average_sum(
    numbers, current_index, remaining_groups, memo
):
    if (current_index, remaining_groups) in memo:
        return memo[(current_index, remaining_groups)]

    number_count = len(numbers)

    # Base case: If no groups are left, no more average to add
    if remaining_groups == 0:
        if current_index == number_count:
            return 0.0
        else:
            return float('-inf')

    # Base case: Only one group remaining, include all numbers
    if current_index == number_count:
        return float('-inf')

    if remaining_groups == 1:
        current_sum = sum(numbers[current_index:])
        average = current_sum / (number_count - current_index)
        memo[(current_index, remaining_groups)] = average
        return average

    maximum_average_sum = float('-inf')

    # Iterate through all possible sizes of the current group
    for group_end_index in range(current_index + 1, number_count + 1):
        current_sum = sum(numbers[current_index:group_end_index])
        average = current_sum / (group_end_index - current_index)

        # Recursively calculate the largest average sum for remaining groups
        remaining_average_sum = calculate_largest_average_sum(
            numbers, group_end_index, remaining_groups - 1, memo
        )

        maximum_average_sum = max(
            maximum_average_sum, average + remaining_average_sum
        )

    memo[(current_index, remaining_groups)] = maximum_average_sum
    return maximum_average_sum

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Start by understanding that the best average sum for the entire set when divided into one group is just the average of all numbers.
Now, consider splitting the set into two groups. To find the best split, try every possible split point and calculate the average sum for each split. Remember the highest average sum you find.
Continue this process for splitting into three, four, and finally up to K groups. Each time, you're using the best average sums you've already calculated for the earlier parts of the list.
The key is to calculate the best average sum for a sub-list from some point until the end. Then, when you want to calculate for the next group before it, you can reuse this calculation to avoid recalculating.
The final answer is the best average sum you found when splitting into K groups.

Code Implementation

def largest_sum_of_averages(numbers, k_groups):
    number_count = len(numbers)
    # This array will store the prefix sums.
    prefix_sums = [0] * (number_count + 1)
    for i in range(1, number_count + 1):
        prefix_sums[i] = prefix_sums[i - 1] + numbers[i - 1]

    # dp[i] is the largest average sum of numbers[i:] into 1 group
    dp = [0] * number_count
    for i in range(number_count):
        dp[i] = (prefix_sums[number_count] - prefix_sums[i]) / (number_count - i)

    for group_count in range(2, k_groups + 1):
        # Iterate backwards, building up the optimal solutions
        for i in range(number_count - 1):
            # Try every possible split point.
            for j in range(i + 1, number_count):
                dp[i] = max(dp[i], (prefix_sums[j] - prefix_sums[i]) / (j - i) + dp[j])

    return dp[0]

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0, as there are no numbers to partition and average.
K is 0	Return negative infinity or an error as it's an invalid partition.
K is greater than the array length	Return the average of the entire array (equivalent to K=1).
Array contains negative numbers	The algorithm should handle negative numbers correctly as it calculates sums and averages.
Array contains a mix of very large positive and very large negative numbers	Ensure sums are calculated using a data type that can accommodate large numbers to prevent integer overflow (e.g., long or double).
Array contains only zeros	The average calculation should correctly handle zero values and avoid division by zero errors.
Array is very large and K is close to the array size	The dynamic programming solution should be optimized for memory and time complexity, considering that the DP table size depends on the array size and K.
Floating-point precision issues	Be mindful of potential precision loss when calculating averages, especially with very large or very small numbers; consider using higher-precision floating-point types if necessary and handle rounding appropriately when comparing results.

Empty input array

How to Handle:

Return 0, as there are no numbers to partition and average.

K is 0

How to Handle:

Return negative infinity or an error as it's an invalid partition.

K is greater than the array length

How to Handle:

Return the average of the entire array (equivalent to K=1).

Array contains negative numbers

How to Handle:

The algorithm should handle negative numbers correctly as it calculates sums and averages.

Array contains a mix of very large positive and very large negative numbers

How to Handle:

Ensure sums are calculated using a data type that can accommodate large numbers to prevent integer overflow (e.g., long or double).

Array contains only zeros

How to Handle:

The average calculation should correctly handle zero values and avoid division by zero errors.

Array is very large and K is close to the array size

How to Handle:

The dynamic programming solution should be optimized for memory and time complexity, considering that the DP table size depends on the array size and K.

Floating-point precision issues

How to Handle: