Maximum Sum Obtained of Any Permutation

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ArraysGreedy Algorithms

We have an array of integers, nums, and an array of requests where requests[i] = [start_i, end_i]. The i^th request asks for the sum of nums[start_i] + nums[start_i + 1] + ... + nums[end_i - 1] + nums[end_i]. Both start_i and end_i are 0-indexed.

Return the maximum total sum of all requests among all permutations of nums.

Since the answer may be too large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [1,2,3,4,5], requests = [[1,3],[0,1]]
Output: 19
Explanation: One permutation of nums is [2,1,3,4,5] with the following result: 
requests[0] -> nums[1] + nums[2] + nums[3] = 1 + 3 + 4 = 8
requests[1] -> nums[0] + nums[1] = 2 + 1 = 3
Total sum: 8 + 3 = 11.
A permutation with a higher total sum is [3,5,4,2,1] with the following result:
requests[0] -> nums[1] + nums[2] + nums[3] = 5 + 4 + 2 = 11
requests[1] -> nums[0] + nums[1] = 3 + 5  = 8
Total sum: 11 + 8 = 19, which is the best that you can do.

Example 2:

Input: nums = [1,2,3,4,5,6], requests = [[0,1]]
Output: 11
Explanation: A permutation with the max total sum is [6,5,4,3,2,1] with request sums [11].

Example 3:

Input: nums = [1,2,3,4,5,10], requests = [[0,2],[1,3],[1,1]]
Output: 47
Explanation: A permutation with the max total sum is [4,10,5,3,2,1] with request sums [19,18,10].

Constraints:

n == nums.length
1 <= n <= 10⁵
0 <= nums[i] <= 10⁵
1 <= requests.length <= 10⁵
requests[i].length == 2
0 <= start_i <= end_i < n

Maximum Sum Obtained of Any Permutation

Medium

PayPal

1 view

Topics:

ArraysGreedy Algorithms

We have an array of integers, nums, and an array of requests where requests[i] = [start_i, end_i]. The i^th request asks for the sum of nums[start_i] + nums[start_i + 1] + ... + nums[end_i - 1] + nums[end_i]. Both start_i and end_i are 0-indexed.

Return the maximum total sum of all requests among all permutations of nums.

Since the answer may be too large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [1,2,3,4,5], requests = [[1,3],[0,1]]
Output: 19
Explanation: One permutation of nums is [2,1,3,4,5] with the following result: 
requests[0] -> nums[1] + nums[2] + nums[3] = 1 + 3 + 4 = 8
requests[1] -> nums[0] + nums[1] = 2 + 1 = 3
Total sum: 8 + 3 = 11.
A permutation with a higher total sum is [3,5,4,2,1] with the following result:
requests[0] -> nums[1] + nums[2] + nums[3] = 5 + 4 + 2 = 11
requests[1] -> nums[0] + nums[1] = 3 + 5  = 8
Total sum: 11 + 8 = 19, which is the best that you can do.

Example 2:

Input: nums = [1,2,3,4,5,6], requests = [[0,1]]
Output: 11
Explanation: A permutation with the max total sum is [6,5,4,3,2,1] with request sums [11].

Example 3:

Input: nums = [1,2,3,4,5,10], requests = [[0,2],[1,3],[1,1]]
Output: 47
Explanation: A permutation with the max total sum is [4,10,5,3,2,1] with request sums [19,18,10].

Constraints:

n == nums.length
1 <= n <= 10⁵
0 <= nums[i] <= 10⁵
1 <= requests.length <= 10⁵
requests[i].length == 2
0 <= start_i <= end_i < n

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 `nums` and `requests` arrays, and what is the range of values within the `nums` array and the `start` and `end` indices in the `requests` array?
Can the elements in the `nums` array be negative, zero, or only positive?
Are the start and end indices in the `requests` array inclusive or exclusive? And what happens if a start index is greater than the end index in a request?
If the sum of all requests exceeds the maximum integer value that can be represented, how should I handle the overflow (e.g., return modulo a specific number)?
If `nums` is empty, or `requests` is empty, or both are empty, what should the return value be?

Brute Force Solution

Approach

The brute force approach to this problem involves exploring absolutely every possible arrangement of numbers to find the best one. We will generate all possible orderings of the numbers and then, for each ordering, calculate a score based on the specific requests made for different number ranges. We then keep track of the highest score we've seen and return that at the end.

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

First, come up with every single possible order in which you can arrange the given numbers.
For each of these arrangements, go through the requests one by one.
For each request, figure out what range of numbers the request refers to.
Sum up the numbers in that range for the current arrangement.
Add this sum to a running total which is the score for this particular arrangement.
Once you've gone through all the requests for that arrangement, compare its score to the highest score you have seen so far.
If the current arrangement's score is higher, remember that score and which arrangement it came from.
After you have scored all possible arrangements, return the highest score that you found.

Code Implementation

import itertools

def maximum_sum_obtained_of_any_permutation_brute_force(numbers, requests):
    maximum_sum = 0

    # Iterate through all possible permutations
    for permutation in itertools.permutations(numbers):
        current_sum = 0

        # Calculate the sum for the current permutation based on requests
        for start_index, end_index in requests:
            # We need to sum a sub-array based on the request indices
            current_sum += sum(permutation[start_index:end_index + 1])

        # Update the maximum sum if the current sum is greater
        maximum_sum = max(maximum_sum, current_sum)

    return maximum_sum

Big(O) Analysis

Time Complexity

O(n! * m * n) – The described brute force approach first generates all permutations of the input array of size n. Generating all permutations takes O(n!) time. For each permutation, the algorithm iterates through m requests. For each request, it sums up the numbers within a specified range of the permutation. Summing the range takes O(n) time in the worst case. Therefore, the total time complexity is O(n! * m * n), where n! is the cost of generating all possible permutations, m is the number of requests, and n is the length of the range for which the sum is calculated.

Space Complexity

O(N!) – The algorithm generates all possible permutations of the input numbers, which requires storing each permutation. Since there are N! possible permutations of N numbers, the auxiliary space used to store these permutations grows factorially with the input size. Therefore, the space complexity is O(N!).

Optimal Solution

Approach

The best arrangement of numbers to get the highest possible sum involves figuring out which numbers get used the most often. Instead of checking every possible order, we focus on matching the biggest numbers with the parts of the list that get used repeatedly based on the 'requests'.

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

Figure out how many times each position in the number list is 'requested' by looking at all the 'request' ranges. Keep track of these counts for each position.
Sort the number list from smallest to largest.
Sort the counts of the positions from smallest to largest as well.
Match the largest number with the position that's used the most, the second largest number with the position used second most, and so on. Essentially, assign the biggest values to the slots called upon most frequently.
Arrange the original number list according to how you've matched numbers to counts.
Finally, add up the numbers within each 'request' range using this new arrangement. The total sum will be the maximum possible.

Code Implementation

def maximum_sum(numbers, requests):
    list_length = len(numbers)
    counts = [0] * list_length

    # Count frequency of each index using requests.
    for start, end in requests:
        for i in range(start, end + 1):
            counts[i] += 1

    # This sorts counts in prep for maximizing sum
    counts.sort()
    numbers.sort()

    max_sum = 0

    # Multiply the largest number with the largest count.
    for i in range(list_length):
        max_sum += numbers[i] * counts[i]

    return max_sum

Big(O) Analysis

Time Complexity

O(n log n) – The first step iterates through the requests array to calculate the frequency of each position, which takes O(m) time where m is the number of requests. The second step sorts the numbers array of size n, taking O(n log n) time. Similarly, sorting the frequency counts array of size n also takes O(n log n) time. The final summation iterates through the requests again (O(m)), but the dominant operations are the two sorts, leading to a time complexity of O(n log n) + O(n log n) + O(m) = O(n log n + m). Since the number of requests m can be at most n^2 (every pair of indices is a request), but is usually much smaller and often a fraction of n, we consider the case when m <= n, resulting in O(n log n).

Space Complexity

O(N) – The algorithm creates an auxiliary array to store the frequency counts of each position in the number list, which grows linearly with the size of the number list, denoted as N. Sorting these frequency counts also requires O(N) space in the worst-case scenario, depending on the sorting algorithm. The sorted number list and position counts are then matched, but this matching occurs in place, and doesn't add extra space. Therefore, the dominant space usage is proportional to N, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
nums is null or empty	Return 0 since there are no elements to sum.
requests is null or empty	Return 0 since there are no requests to process, regardless of the nums array.
nums contains only negative numbers	The greedy approach of assigning largest nums to most frequent request ranges will still work correctly, potentially yielding a negative sum.
nums contains zeros	Zeros will be assigned to less frequent ranges, which minimizes their impact on the final sum.
Large input arrays for nums and requests	Use an efficient sorting algorithm (e.g., mergesort or quicksort) with O(n log n) time complexity to sort nums and the prefix sum array.
requests with overlapping ranges	The prefix sum difference array correctly accounts for overlaps in requests, incrementing the count for all included indices.
requests with start index equal to end index	The prefix sum difference array can handle requests with equal start and end indices by incrementing and decrementing at the same index.
Integer overflow when calculating the sum	Use long data type to store intermediate sums to prevent overflow and take modulo at the end.