Find the Score of All Prefixes of an Array

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Arrays

We define the conversion array conver of an array arr as follows:

conver[i] = arr[i] + max(arr[0..i]) where max(arr[0..i]) is the maximum value of arr[j] over 0 <= j <= i.

We also define the score of an array arr as the sum of the values of the conversion array of arr.

Given a 0-indexed integer array nums of length n, return an array ans of length n where ans[i] is the score of the prefix nums[0..i].

Example 1:

Input: nums = [2,3,7,5,10]
Output: [4,10,24,36,56]
Explanation: 
For the prefix [2], the conversion array is [4] hence the score is 4
For the prefix [2, 3], the conversion array is [4, 6] hence the score is 10
For the prefix [2, 3, 7], the conversion array is [4, 6, 14] hence the score is 24
For the prefix [2, 3, 7, 5], the conversion array is [4, 6, 14, 12] hence the score is 36
For the prefix [2, 3, 7, 5, 10], the conversion array is [4, 6, 14, 12, 20] hence the score is 56

Example 2:

Input: nums = [1,1,2,4,8,16]
Output: [2,4,8,16,32,64]
Explanation: 
For the prefix [1], the conversion array is [2] hence the score is 2
For the prefix [1, 1], the conversion array is [2, 2] hence the score is 4
For the prefix [1, 1, 2], the conversion array is [2, 2, 4] hence the score is 8
For the prefix [1, 1, 2, 4], the conversion array is [2, 2, 4, 8] hence the score is 16
For the prefix [1, 1, 2, 4, 8], the conversion array is [2, 2, 4, 8, 16] hence the score is 32
For the prefix [1, 1, 2, 4, 8, 16], the conversion array is [2, 2, 4, 8, 16, 32] hence the score is 64

Constraints:

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

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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 for elements in the input array `nums`?
Can the input array `nums` be empty or null?
Can you provide an example of the expected output for the input [1, 2, 3]?
Is the input array `nums` read-only, or am I allowed to modify it in place?
Could you clarify what should happen if the sum of the scores of all prefixes exceeds the maximum value that an integer data type can hold?

Brute Force Solution

Approach

The brute force method to find the score of all prefixes involves calculating each prefix's score independently. We go through all possible prefix combinations, calculate their scores by finding the largest value and summing it with the prefix value, and then store these scores. We then return the list of prefix scores.

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

Start at the very beginning of the list of numbers.
Consider the first number by itself. Find the biggest number within this very small list (which is just the number itself). Add that biggest number to the original number to get the score for this first prefix.
Now, consider the first two numbers together. Find the biggest number within this slightly larger list. Add that biggest number to the second number to get the score for the second prefix.
Next, consider the first three numbers together. Find the biggest number within this expanded list. Add that biggest number to the third number to get the score for the third prefix.
Keep doing this, each time including one more number from the beginning of the list. Each time, find the biggest number in the growing list, and add it to the most recently added number to get its corresponding prefix score.
Continue until you have considered all the numbers in the list, each time calculating the score for the corresponding prefix.
Finally, present the list of scores you've calculated for each prefix from start to finish.

Code Implementation

def find_prefix_scores(numbers):
    prefix_scores = []
    
    for i in range(len(numbers)): # Iterate through all possible prefix lengths
        current_prefix = numbers[:i+1]
        
        maximum_value = max(current_prefix)
        # Finding the maximum value in the current prefix is critical
        
        prefix_sum = sum(current_prefix)
        score = prefix_sum + maximum_value
        # The score is defined as the sum of prefix plus maximum prefix value
        
        prefix_scores.append(score)

    return prefix_scores

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all prefixes of the input array of size n. For each prefix, it finds the maximum element within that prefix. Finding the maximum within a prefix of size i takes O(i) time. Since we do this for prefixes of size 1, 2, ..., n, the total time complexity is 1 + 2 + ... + n, which is the sum of an arithmetic series. This sum is equal to n*(n+1)/2, which simplifies to O(n²).

Space Complexity

O(N) – The algorithm stores the prefix scores in a list, where the length of the list corresponds to the number of elements in the input array. Therefore, an auxiliary list of size N is created to store the prefix scores, where N is the number of elements in the input array. No other significant auxiliary data structures are used. This results in an auxiliary space complexity of O(N).

Optimal Solution

Approach

To efficiently calculate the score for each part of the array, we'll keep track of the running total and the largest number we've seen so far. We'll build up the answer piece by piece, using information from the previous steps.

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

Start by looking at the very first number in the array.
The score for this first number is simply the number itself plus the largest number we have seen up until now, which is also just the first number.
Move to the second number in the array.
Update the largest number seen so far; it's either the previous largest number or the current number, whichever is bigger.
The score for the first two numbers is the sum of the second number and the largest number seen so far.
Continue moving through the array, one number at a time.
For each number, update the largest number seen so far if necessary.
Calculate the score for the part of the array up to the current number by adding the current number and the largest number seen so far.
The final result is a new array containing all the calculated scores for each part of the original array.

Code Implementation

def find_prefix_scores(numbers):
    prefix_scores = []
    running_maximum = 0
    running_sum = 0

    for number in numbers:
        # Update running maximum with current number.
        running_maximum = max(running_maximum, number)

        # The current score is the updated max plus the running sum.
        current_score = running_maximum + running_sum
        prefix_scores.append(current_score)

        # Update running sum to prepare for the next iteration.
        running_sum += running_maximum

    return prefix_scores

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array nums once, where n is the length of nums. Inside the loop, the algorithm performs constant time operations: updating the maximum value seen so far and calculating the score for the current prefix. Since the dominant operation is a single pass through the array, the time complexity is O(n).

Space Complexity

O(N) – The algorithm creates a new array to store the calculated scores for each prefix of the input array. The size of this new array is directly proportional to the size of the input array (N). Therefore, the auxiliary space used by the algorithm grows linearly with the input size. No other significant data structures contribute to the space complexity.

Edge Cases

Case	How to Handle
Empty input array	Return an empty array since there are no prefixes to score.
Array with a single element	Calculate the score of the single-element prefix (which is just the element itself) and return an array containing that score.
Array with all positive numbers	The maximum-so-far will increase monotonically, leading to each element in the score representing cumulative sums of increasing numbers.
Array with all negative numbers	The maximum-so-far will likely change frequently, and the score will still be the cumulative sum of these maximums.
Array containing a mix of large positive and large negative numbers	Ensure integer overflow is handled by using appropriate data types (e.g., long) for storing the running sum and the maximum.
Array with a large number of elements (scalability check)	The solution should have a time complexity of O(n) to efficiently process large arrays; avoid nested loops or operations that scale quadratically.
Array containing only zeros	The score array will contain a series of zeros representing the sum of maximums, where maximum remains as zero always.
Array with extreme values (Integer.MAX_VALUE, Integer.MIN_VALUE)	Handle these extreme values carefully to prevent overflow or underflow when calculating maximums and sums.

Empty input array

How to Handle:

Return an empty array since there are no prefixes to score.

Array with a single element

How to Handle:

Calculate the score of the single-element prefix (which is just the element itself) and return an array containing that score.

Array with all positive numbers

How to Handle:

The maximum-so-far will increase monotonically, leading to each element in the score representing cumulative sums of increasing numbers.

Array with all negative numbers

How to Handle:

The maximum-so-far will likely change frequently, and the score will still be the cumulative sum of these maximums.

Array containing a mix of large positive and large negative numbers

How to Handle:

Ensure integer overflow is handled by using appropriate data types (e.g., long) for storing the running sum and the maximum.

Array with a large number of elements (scalability check)

How to Handle:

The solution should have a time complexity of O(n) to efficiently process large arrays; avoid nested loops or operations that scale quadratically.

Array containing only zeros

How to Handle:

The score array will contain a series of zeros representing the sum of maximums, where maximum remains as zero always.

Array with extreme values (Integer.MAX_VALUE, Integer.MIN_VALUE)

How to Handle:

Handle these extreme values carefully to prevent overflow or underflow when calculating maximums and sums.