Maximum Balanced Subsequence Sum

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ArraysDynamic Programming

You are given a 0-indexed integer array nums.

A subsequence of nums having length k and consisting of indices i₀ < i₁ < ... < i_k-1 is balanced if the following holds:

nums[i_j] - nums[i_j-1] >= i_j - i_j-1, for every j in the range [1, k - 1].

A subsequence of nums having length 1 is considered balanced.

Return an integer denoting the maximum possible sum of elements in a balanced subsequence of nums.

A subsequence of an array is a new non-empty array that is formed from the original array by deleting some (possibly none) of the elements without disturbing the relative positions of the remaining elements.

Example 1:

Input: nums = [3,3,5,6]
Output: 14
Explanation: In this example, the subsequence [3,5,6] consisting of indices 0, 2, and 3 can be selected.
nums[2] - nums[0] >= 2 - 0.
nums[3] - nums[2] >= 3 - 2.
Hence, it is a balanced subsequence, and its sum is the maximum among the balanced subsequences of nums.
The subsequence consisting of indices 1, 2, and 3 is also valid.
It can be shown that it is not possible to get a balanced subsequence with a sum greater than 14.

Example 2:

Input: nums = [5,-1,-3,8]
Output: 13
Explanation: In this example, the subsequence [5,8] consisting of indices 0 and 3 can be selected.
nums[3] - nums[0] >= 3 - 0.
Hence, it is a balanced subsequence, and its sum is the maximum among the balanced subsequences of nums.
It can be shown that it is not possible to get a balanced subsequence with a sum greater than 13.

Example 3:

Input: nums = [-2,-1]
Output: -1
Explanation: In this example, the subsequence [-1] can be selected.
It is a balanced subsequence, and its sum is the maximum among the balanced subsequences of nums.

Constraints:

1 <= nums.length <= 10⁵
-10⁹ <= nums[i] <= 10⁹

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 is the range of values within the input array? Can I expect negative numbers, zeros, or only positive numbers?
What is the maximum size of the input array?
If no balanced subsequence exists, what value should I return?
Are duplicate numbers allowed in the input array, and if so, how should they be handled when finding the maximum sum?
Could you clarify what you mean by 'balanced'? Specifically, what relationship between the index and the value of an element qualifies it as balanced (e.g., arr[i] <= i, arr[i] == i, or something else)?

Brute Force Solution

Approach

The brute force approach solves this problem by checking absolutely every possible subsequence. We examine each possible combination of numbers to see if it fits our rules and gives us the best result.

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

First, consider every possible group of numbers you can make from the given set of numbers.
For each group, check if it follows the balance rule: whether the value of each number is greater or equal to its position within the group.
If a group is balanced, calculate its total sum by adding up all the numbers in the group.
Keep track of the biggest sum you've found from any balanced group.
After checking every possible group, the biggest sum you saved is your final answer.

Code Implementation

def maximum_balanced_subsequence_sum_brute_force(numbers):
    max_subsequence_sum = float('-inf')

    number_of_numbers = len(numbers)

    # Iterate through all possible subsequences
    for i in range(1 << number_of_numbers):
        subsequence = []
        for j in range(number_of_numbers):
            # Check if the j-th bit is set in i
            if (i >> j) & 1:
                subsequence.append(numbers[j])

        is_balanced = True
        for index, number in enumerate(subsequence):
            # Check if the subsequence is balanced
            if number < subsequence.index(number) + 1:
                is_balanced = False
                break

        if is_balanced:
            current_subsequence_sum = sum(subsequence)

            # Keep track of the largest balanced subsequence
            max_subsequence_sum = max(max_subsequence_sum, current_subsequence_sum)

    if max_subsequence_sum == float('-inf'):
        return 0

    return max_subsequence_sum

Big(O) Analysis

Time Complexity

O(2^n * n) – The brute force approach examines all possible subsequences of the input array of size n. There are 2^n possible subsequences. For each subsequence, we need to iterate through its elements (in the worst case, the subsequence has length n) to check if it's balanced according to the problem definition. Therefore, the time complexity is O(2^n * n).

Space Complexity

O(1) – The brute force approach described checks all possible subsequences, calculating the sum of each balanced subsequence, and keeping track of the maximum balanced sum seen so far. This involves no auxiliary data structures whose size scales with the input size N (the number of elements in the input list). Only a constant number of variables (like the current subsequence sum and the maximum balanced subsequence sum) are used, so the auxiliary space is constant.

Optimal Solution

Approach

The most efficient approach is to use a data structure that helps us keep track of the best possible sums we can achieve up to a certain point, while also considering the constraints. This way, we avoid recomputing the same information repeatedly and arrive at the final answer quicker. We maintain a record of the best sums we have found so far for all possible 'end points' that satisfy a condition.

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

First, transform the problem slightly to make it easier to work with. Instead of the original numbers, consider the differences between each number and its position.
Now, the problem turns into finding the largest sum of these differences, but only including the differences that are greater than or equal to zero (non-negative differences).
To solve this efficiently, use a structure that lets you quickly find and update the best possible sums seen so far. Think of it like a scoreboard where you're always tracking the highest score possible up to any given point.
Go through the transformed differences one by one. For each difference, look at the scoreboard to see the best possible sum you could have before this difference. Add the current difference to that best sum.
Update the scoreboard with this new possible sum, but only if it's better than any sum already recorded for that point. This makes sure you always keep the best possible path to any position.
After you've processed all the differences, the highest score on your scoreboard represents the maximum possible sum of a subsequence that satisfies the required condition.

Code Implementation

def maximum_balanced_subsequence_sum(numbers):
    modified_numbers = [numbers[i] - i for i in range(len(numbers))]

    best_sums = {}

    for i, modified_number in enumerate(modified_numbers):

        if modified_number < 0:
            continue

        current_best_sum = 0
        # Find the best sum achievable before this number
        for key in best_sums:
            if key <= modified_number:
                current_best_sum = max(current_best_sum, best_sums[key])

        new_sum = current_best_sum + modified_number

        # Update the best sums with the new sum if it's better
        update = True
        for key in list(best_sums.keys()):
            if key >= modified_number and best_sums[key] >= new_sum:
                update = False
                break
            if key <= modified_number and best_sums[key] <= new_sum:
                del best_sums[key]
        if update:
            best_sums[modified_number] = new_sum

    max_sum = 0
    # Find the maximum sum among all achievable sums
    for key in best_sums:
        max_sum = max(max_sum, best_sums[key])

    return max_sum

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm iterates through the n elements of the input array once to compute the differences. The core logic involves using a data structure (implied to be a balanced tree or segment tree based on the scoreboard analogy) to maintain and query the maximum sum seen so far. Both querying and updating this data structure take O(log n) time. Therefore, processing each of the n elements requires O(log n) operations, resulting in a total time complexity of O(n log n).

Space Complexity

O(N) – The solution utilizes a 'scoreboard' data structure to keep track of the best possible sums encountered so far for each position. This scoreboard needs to store a value for each element in the input array, which translates to an auxiliary array of size N, where N is the number of elements in the input. Therefore, the space complexity is directly proportional to the input size. The space used by other variables are constant and not dependent on the input size. Hence, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Empty input array	Return 0, as there's no subsequence to sum.
Array with a single element	Return the element if it's non-negative; otherwise, return 0.
Array with all negative numbers	Return 0, as an empty subsequence is always a valid solution with sum 0.
Array with all zeros	Return 0 if the balance condition requires strictly increasing indices, otherwise return the sum of all zeros.
Large input array with numbers close to the maximum integer value	Use long to prevent integer overflow during summation and address scalability constraints with efficient algorithms.
Large input array with numbers close to the minimum integer value	Use long to prevent integer overflow during summation and address scalability constraints with efficient algorithms.
Array with elements that satisfy nums[i] > i (no valid subsequence)	Return 0 as no single element satisfies the condition and therefore no subsequence does either.
Array with a mix of large positive and large negative values that nearly cancel out	The algorithm must correctly handle negative intermediate sums in the dynamic programming or greedy approach.

Empty input array

How to Handle:

Return 0, as there's no subsequence to sum.

Array with a single element

How to Handle:

Return the element if it's non-negative; otherwise, return 0.

Array with all negative numbers

How to Handle:

Return 0, as an empty subsequence is always a valid solution with sum 0.

Array with all zeros

How to Handle:

Return 0 if the balance condition requires strictly increasing indices, otherwise return the sum of all zeros.

Large input array with numbers close to the maximum integer value

How to Handle:

Use long to prevent integer overflow during summation and address scalability constraints with efficient algorithms.

Large input array with numbers close to the minimum integer value

How to Handle:

Use long to prevent integer overflow during summation and address scalability constraints with efficient algorithms.

Array with elements that satisfy nums[i] > i (no valid subsequence)

How to Handle:

Return 0 as no single element satisfies the condition and therefore no subsequence does either.

Array with a mix of large positive and large negative values that nearly cancel out

How to Handle:

The algorithm must correctly handle negative intermediate sums in the dynamic programming or greedy approach.