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Partition Array for Maximum Sum

Medium

Google

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Topics:

ArraysDynamic Programming

Given an integer array arr, partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.

Return the largest sum of the given array after partitioning. Test cases are generated so that the answer fits in a 32-bit integer.

For example:

Consider the array arr = [1,15,7,9,2,5,10] and k = 3. The optimal partitioning would be [15,15,15,9,10,10,10]. The sum is 15 + 15 + 15 + 9 + 10 + 10 + 10 = 84. Therefore, the function should return 84.

Another example, if arr = [1,4,1,5,7,3,6,1,9,9,3] and k = 4, what is the maximum sum after partitioning?

Could you write a function to solve this problem efficiently?

Partition Array for Maximum Sum

Medium

Google

2 views

Topics:

ArraysDynamic Programming

Return the largest sum of the given array after partitioning. Test cases are generated so that the answer fits in a 32-bit integer.

For example:

Another example, if arr = [1,4,1,5,7,3,6,1,9,9,3] and k = 4, what is the maximum sum after partitioning?

Could you write a function to solve this problem efficiently?

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 possible values within the array `arr`? Could they be negative, zero, or very large?
What is the maximum size of the array `arr`? This will help me think about space complexity.
If there are multiple ways to partition the array that result in the same maximum sum, is any valid partitioning acceptable, or is there a specific rule for choosing one?
What should the function return if the input array is empty or null?
Are there any constraints on the size of each chunk in the partition? Must each chunk have at least one element?

Brute Force Solution

Approach

The brute force approach to maximizing the sum after partitioning an array involves exploring absolutely every possible way to divide the array into smaller sub-arrays. For each division, we calculate the sum based on the rules (replacing each sub-array with its largest element multiplied by its size), and finally, we pick the division that gives the largest sum.

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

Consider the array. Start by assuming we do not partition it at all. Calculate the sum in that case.
Now, imagine splitting the array after the first element. You have one sub-array with just the first element, and another sub-array containing the rest. Calculate the sum resulting from this division.
Next, split the original array after the second element. Calculate the sum resulting from this division.
Keep doing this, splitting the array after each element, one at a time, all the way to splitting it after the second to last element.
Now, go back and consider splitting the array into three parts. Try all possible locations where you could place the two splitting points.
Then try all the ways to split the array into four parts, five parts, and so on, until each element is in its own sub-array.
For every possible splitting arrangement we have considered, calculate the resulting sum according to the given rules.
Compare all the sums from all the different ways you split the array, and pick the biggest sum. That's the answer.

Code Implementation

def partition_array_for_maximum_sum_brute_force(array, max_subarray_length):
    array_length = len(array)

    # Initialize max sum to track the optimal partition
    maximum_sum = 0

    # Iterate through all possible partition arrangements
    for i in range(1 << (array_length - 1)):
        current_sum = 0
        subarray_start = 0
        
        # Iterate through array to calculate sum for the current partition
        for j in range(array_length):
            # Check if we should end current subarray
            if (i >> j) & 1 or j == array_length - 1:
                subarray_end = j
                
                # Extract the current subarray
                subarray = array[subarray_start:subarray_end+1]

                # Calculate max value of subarray
                maximum_value_in_subarray = max(subarray)
                
                # Update current sum
                current_sum += maximum_value_in_subarray * len(subarray)
                
                # Update subarray start
                subarray_start = j + 1

        # Update the maximum sum if current partition gives better result
        maximum_sum = max(maximum_sum, current_sum)

    return maximum_sum

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach considers all possible partitions of the array. For an array of size n, there are 2^(n-1) possible ways to partition it, since each of the n-1 spaces between elements can either be a partition or not. For each of these partitions, we need to calculate the sum, which takes O(n) time in the worst case (when all elements are in their own sub-array). Therefore, the total time complexity is O(n * 2^(n-1)), which simplifies to O(2^n) since exponential terms dominate polynomial ones. Thus, the algorithm exhibits exponential time complexity.

Space Complexity

O(1) – The brute force approach explores all possible partitioning arrangements of the input array. It does not explicitly mention using any auxiliary data structures like arrays, hash maps, or sets to store intermediate results or visited states. The described operations primarily involve calculations and comparisons of sums based on different partitionings. Therefore, the space complexity is dominated by a few constant variables for tracking maximum sums and partition indices, leading to O(1) auxiliary space.

Optimal Solution

Approach

The goal is to split the array into groups to maximize a sum. Instead of trying every possible split, we use a method where we make the best decision for each smaller chunk of the array, ultimately leading to the best overall answer. This avoids a lot of unnecessary calculations.

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

Consider each possible ending point for the first group of numbers.
For each ending point, find the largest number within that group.
Multiply that largest number by the number of elements in that group, as this is the contribution of this group to the total sum.
Now, we need to consider the rest of the array. Think of using results calculated earlier to calculate current result, by choosing the best ending points of previous groups to add to current group's calculated contribution.
Remember the best possible sum we can achieve up to that point.
Repeat for all the numbers in the array until the very last one.

Code Implementation

def partition_array_for_maximum_sum(array, max_group_size):
    array_length = len(array)
    dp = [0] * (array_length + 1)

    for i in range(1, array_length + 1):
        maximum_value = 0
        for k in range(1, min(max_group_size, i) + 1):
            # Find the maximum value in the current group.
            maximum_value = max(maximum_value, array[i - k])

            # Calculate the score of current group and add
            # previously computed result to get best sum.
            dp[i] = max(dp[i], dp[i - k] + maximum_value * k)

    # The last element holds the answer.
    return dp[array_length]

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the array of size n, considering each index as a potential end of a partition. For each of these n indices, it iterates backwards from that index to the beginning of the array to find the maximum value within that partition and calculate the sum. Thus, for each element, we perform a loop that can go up to n times. Therefore, the total operations approximate to n * n, which simplifies to O(n²).

Space Complexity

O(N) – The algorithm uses dynamic programming to store the best possible sum achievable up to each index in the array. This is done by keeping track of intermediate results using an auxiliary array, where each element corresponds to the maximum sum up to that index. Therefore, an array of size N (where N is the length of the input array) is used to store these intermediate maximum sums. Thus the space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty array	Return 0 immediately as there's nothing to partition and sum.
k = 1 (partition every element)	The result would be the sum of all elements in the array.
k is greater than the array size	Treat k as the array size since we cannot have groups larger than the array.
Array with all identical values	The maximum sum will be the length of the array times the repeated value.
Large array size (scalability)	Dynamic programming approach should scale linearly O(n) with array size, avoid recursion.
Array with very large numbers (potential overflow)	Use a data type that can handle large values (e.g., long long in C++, long in Java) to prevent integer overflow during calculations of max values.
k is a large number close to array size	Dynamic programming would still work and iterate only up to k elements during each step
Array containing only negative numbers	The algorithm will still find the optimal partition to maximize the sum, which might be a negative value.

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 possible values within the array `arr`? Could they be negative, zero, or very large?
What is the maximum size of the array `arr`? This will help me think about space complexity.
If there are multiple ways to partition the array that result in the same maximum sum, is any valid partitioning acceptable, or is there a specific rule for choosing one?
What should the function return if the input array is empty or null?
Are there any constraints on the size of each chunk in the partition? Must each chunk have at least one element?

Brute Force Solution

Approach

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

Consider the array. Start by assuming we do not partition it at all. Calculate the sum in that case.
Now, imagine splitting the array after the first element. You have one sub-array with just the first element, and another sub-array containing the rest. Calculate the sum resulting from this division.
Next, split the original array after the second element. Calculate the sum resulting from this division.
Keep doing this, splitting the array after each element, one at a time, all the way to splitting it after the second to last element.
Now, go back and consider splitting the array into three parts. Try all possible locations where you could place the two splitting points.
Then try all the ways to split the array into four parts, five parts, and so on, until each element is in its own sub-array.
For every possible splitting arrangement we have considered, calculate the resulting sum according to the given rules.
Compare all the sums from all the different ways you split the array, and pick the biggest sum. That's the answer.

Code Implementation

def partition_array_for_maximum_sum_brute_force(array, max_subarray_length):
    array_length = len(array)

    # Initialize max sum to track the optimal partition
    maximum_sum = 0

    # Iterate through all possible partition arrangements
    for i in range(1 << (array_length - 1)):
        current_sum = 0
        subarray_start = 0
        
        # Iterate through array to calculate sum for the current partition
        for j in range(array_length):
            # Check if we should end current subarray
            if (i >> j) & 1 or j == array_length - 1:
                subarray_end = j
                
                # Extract the current subarray
                subarray = array[subarray_start:subarray_end+1]

                # Calculate max value of subarray
                maximum_value_in_subarray = max(subarray)
                
                # Update current sum
                current_sum += maximum_value_in_subarray * len(subarray)
                
                # Update subarray start
                subarray_start = j + 1

        # Update the maximum sum if current partition gives better result
        maximum_sum = max(maximum_sum, current_sum)

    return maximum_sum

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Consider each possible ending point for the first group of numbers.
For each ending point, find the largest number within that group.
Multiply that largest number by the number of elements in that group, as this is the contribution of this group to the total sum.
Now, we need to consider the rest of the array. Think of using results calculated earlier to calculate current result, by choosing the best ending points of previous groups to add to current group's calculated contribution.
Remember the best possible sum we can achieve up to that point.
Repeat for all the numbers in the array until the very last one.

Code Implementation

def partition_array_for_maximum_sum(array, max_group_size):
    array_length = len(array)
    dp = [0] * (array_length + 1)

    for i in range(1, array_length + 1):
        maximum_value = 0
        for k in range(1, min(max_group_size, i) + 1):
            # Find the maximum value in the current group.
            maximum_value = max(maximum_value, array[i - k])

            # Calculate the score of current group and add
            # previously computed result to get best sum.
            dp[i] = max(dp[i], dp[i - k] + maximum_value * k)

    # The last element holds the answer.
    return dp[array_length]

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty array	Return 0 immediately as there's nothing to partition and sum.
k = 1 (partition every element)	The result would be the sum of all elements in the array.
k is greater than the array size	Treat k as the array size since we cannot have groups larger than the array.
Array with all identical values	The maximum sum will be the length of the array times the repeated value.
Large array size (scalability)	Dynamic programming approach should scale linearly O(n) with array size, avoid recursion.
Array with very large numbers (potential overflow)	Use a data type that can handle large values (e.g., long long in C++, long in Java) to prevent integer overflow during calculations of max values.
k is a large number close to array size	Dynamic programming would still work and iterate only up to k elements during each step
Array containing only negative numbers	The algorithm will still find the optimal partition to maximize the sum, which might be a negative value.