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Find the Sum of Subsequence Powers

Hard

Google

4 views

Topics:

ArraysRecursionDynamic Programming

You are given an integer array nums of length n, and a positive integer k. The power of a subsequence is defined as the minimum absolute difference between any two elements in the subsequence. Return the sum of powers of all subsequences of nums which have length equal to k. Since the answer may be large, return it modulo 10^9 + 7.

For example:

nums = [1,2,3,4], k = 3. The subsequences are [1,2,3], [1,3,4], [1,2,4], and [2,3,4]. The powers are 1, 1, 1, and 1 respectively. The sum of the powers is 4.
nums = [2,2], k = 2. The only subsequence is [2,2]. The power is 0.
nums = [4,3,-1], k = 2. The subsequences are [4,3], [4,-1], and [3,-1]. The powers are 1, 5, and 4 respectively. The sum of the powers is 10.

Constraints:

2 <= n == nums.length <= 50
-10^8 <= nums[i] <= 10^8
2 <= k <= n

Find the Sum of Subsequence Powers

Hard

Google

4 views

Topics:

ArraysRecursionDynamic Programming

For example:

nums = [1,2,3,4], k = 3. The subsequences are [1,2,3], [1,3,4], [1,2,4], and [2,3,4]. The powers are 1, 1, 1, and 1 respectively. The sum of the powers is 4.
nums = [2,2], k = 2. The only subsequence is [2,2]. The power is 0.
nums = [4,3,-1], k = 2. The subsequences are [4,3], [4,-1], and [3,-1]. The powers are 1, 5, and 4 respectively. The sum of the powers is 10.

Constraints:

2 <= n == nums.length <= 50
-10^8 <= nums[i] <= 10^8
2 <= k <= 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 is the expected range and data type of the numbers in the input array? Can the numbers be negative, zero, or floating-point numbers?
What is the maximum size of the input array? Is there a specific modulo value to be applied to prevent integer overflow when calculating the sum of subsequence powers?
Are duplicate numbers allowed in the input array, and if so, how should they be handled in calculating the sum of subsequence powers?
What defines a 'subsequence'? Specifically, does the order of elements in the subsequence have to match the order in the original array, or can they be in any order?
What should be returned if the input array is null or empty?

Brute Force Solution

Approach

The brute force way to solve this problem involves examining every possible combination of numbers from the given set. For each combination, we'll compute a value based on the powers of the numbers in the combination and then sum those values.

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

First, consider the possibility of selecting no numbers at all; in that case, the sum would be zero.
Next, consider all the combinations where you pick only one number from the set.
After that, consider every possible combination where you select two numbers.
Continue in this manner, examining all combinations where you select three numbers, then four, and so on, until you have considered the case where you select all the numbers.
For each of these combinations, calculate the power of each number in the combination. The power is simply the number raised to itself.
Once you have the powers calculated for each number in a combination, sum these powers together.
Finally, after going through every single possible combination and calculating the sum of powers for each, add all those sums together to get the final result.

Code Implementation

def find_sum_of_subsequence_powers_brute_force(numbers):
    final_sum = 0
    number_of_elements = len(numbers)

    # Iterate through all possible subsequences (combinations)
    for i in range(2 ** number_of_elements):
        current_sum = 0
        current_subsequence = []

        # Build the subsequence based on the bits in 'i'
        for j in range(number_of_elements):
            if (i >> j) & 1:
                current_subsequence.append(numbers[j])

        # Calculate the sum of powers for the current subsequence
        for number in current_subsequence:
            current_sum += number ** number

        # Accumulate the subsequence sum into the final result
        final_sum += current_sum

    return final_sum

Big(O) Analysis

Time Complexity

O(n * 2^n) – The brute force approach iterates through all possible subsets of the input array. For an array of size n, there are 2^n possible subsets. For each subset, we iterate through the elements in the subset to calculate the sum of powers. In the worst case, the subset could contain all n elements, leading to n power calculations. Therefore, the overall time complexity is O(n * 2^n).

Space Complexity

O(1) – The described brute force algorithm primarily uses iterative loops to generate combinations. Although combinations are being considered, the algorithm does not explicitly store all possible combinations in a separate data structure simultaneously. No auxiliary data structures, such as lists or hash maps, are created to store intermediate results or visited combinations. Therefore, the space complexity is constant, independent of the input size N (the number of elements in the set).

Optimal Solution

Approach

This problem asks us to find the sum of powers of all increasing subsequences of a given set of numbers. The clever part is recognizing we can build the solution by processing the numbers in sorted order and efficiently updating the sum as we go. We exploit the properties of exponentiation and summation to reduce redundant calculations.

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

First, sort the input numbers from smallest to largest. This allows us to easily identify increasing subsequences.
Initialize two variables: one to store the overall sum of subsequence powers (our final answer), and another to store the sum of powers of subsequences ending at the previous number.
Now, go through the sorted numbers one by one.
For each number, update the sum of powers of subsequences ending at the previous number. The new value is twice the old value (representing extending all previous subsequences with the current number, and keeping the old ones), plus the current number raised to a power.
Add the new sum to the overall sum of subsequence powers.
After processing all numbers, the overall sum is the answer we're looking for.

Code Implementation

def find_sum_of_subsequence_powers(numbers):
    sorted_numbers = sorted(numbers)
    overall_sum = 0
    previous_subsequence_sum = 0

    for number in sorted_numbers:
        # Doubling accounts for extending prior subsequences.
        previous_subsequence_sum = (2 * previous_subsequence_sum) + number

        # Accumulate the total sum of subsequence powers.
        overall_sum += previous_subsequence_sum

    # The overall sum holds the final result.
    return overall_sum

Big(O) Analysis

Time Complexity

O(n log n) – The dominant cost comes from sorting the input array of size n, which takes O(n log n) time using an efficient sorting algorithm like merge sort or quicksort. The subsequent loop iterates through the sorted array once, performing constant time operations for each element. Therefore, the overall time complexity is determined by the sorting step, resulting in O(n log n).

Space Complexity

O(1) – The algorithm uses a few variables to store the overall sum and the sum of powers of subsequences ending at the previous number. The space used by these variables does not depend on the size of the input array, N. Since the algorithm does not create any additional data structures that scale with the input size, the auxiliary space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately since no subsequence can be generated.
Array with a single element	Return the power of that element (element^length of subsequence, here length is 1).
Array with all elements equal to 0	Return 0 since 0 raised to any power will be 0.
Array with very large positive numbers, potential integer overflow during exponentiation	Use modulo operator during exponentiation to avoid integer overflow.
Array with negative numbers, need to handle correctly how negative numbers are raised to a power	Correctly calculate the power based on the exponent (even or odd) while considering negative numbers
Large input array causing time limit exceed	Optimize the solution to reduce time complexity possibly through sorting and optimized calculation of combinations.
Input array contains duplicates	The core logic handles duplicates correctly since each element is considered independently during subsequence generation.
Cases where the calculated sum exceeds the maximum integer value.	Use a larger data type such as long to store the sum, or take modulo at each step to avoid overflow.

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 expected range and data type of the numbers in the input array? Can the numbers be negative, zero, or floating-point numbers?
What is the maximum size of the input array? Is there a specific modulo value to be applied to prevent integer overflow when calculating the sum of subsequence powers?
Are duplicate numbers allowed in the input array, and if so, how should they be handled in calculating the sum of subsequence powers?
What defines a 'subsequence'? Specifically, does the order of elements in the subsequence have to match the order in the original array, or can they be in any order?
What should be returned if the input array is null or empty?

Brute Force Solution

Approach

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

First, consider the possibility of selecting no numbers at all; in that case, the sum would be zero.
Next, consider all the combinations where you pick only one number from the set.
After that, consider every possible combination where you select two numbers.
Continue in this manner, examining all combinations where you select three numbers, then four, and so on, until you have considered the case where you select all the numbers.
For each of these combinations, calculate the power of each number in the combination. The power is simply the number raised to itself.
Once you have the powers calculated for each number in a combination, sum these powers together.
Finally, after going through every single possible combination and calculating the sum of powers for each, add all those sums together to get the final result.

Code Implementation

def find_sum_of_subsequence_powers_brute_force(numbers):
    final_sum = 0
    number_of_elements = len(numbers)

    # Iterate through all possible subsequences (combinations)
    for i in range(2 ** number_of_elements):
        current_sum = 0
        current_subsequence = []

        # Build the subsequence based on the bits in 'i'
        for j in range(number_of_elements):
            if (i >> j) & 1:
                current_subsequence.append(numbers[j])

        # Calculate the sum of powers for the current subsequence
        for number in current_subsequence:
            current_sum += number ** number

        # Accumulate the subsequence sum into the final result
        final_sum += current_sum

    return final_sum

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, sort the input numbers from smallest to largest. This allows us to easily identify increasing subsequences.
Initialize two variables: one to store the overall sum of subsequence powers (our final answer), and another to store the sum of powers of subsequences ending at the previous number.
Now, go through the sorted numbers one by one.
For each number, update the sum of powers of subsequences ending at the previous number. The new value is twice the old value (representing extending all previous subsequences with the current number, and keeping the old ones), plus the current number raised to a power.
Add the new sum to the overall sum of subsequence powers.
After processing all numbers, the overall sum is the answer we're looking for.

Code Implementation

def find_sum_of_subsequence_powers(numbers):
    sorted_numbers = sorted(numbers)
    overall_sum = 0
    previous_subsequence_sum = 0

    for number in sorted_numbers:
        # Doubling accounts for extending prior subsequences.
        previous_subsequence_sum = (2 * previous_subsequence_sum) + number

        # Accumulate the total sum of subsequence powers.
        overall_sum += previous_subsequence_sum

    # The overall sum holds the final result.
    return overall_sum

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately since no subsequence can be generated.
Array with a single element	Return the power of that element (element^length of subsequence, here length is 1).
Array with all elements equal to 0	Return 0 since 0 raised to any power will be 0.
Array with very large positive numbers, potential integer overflow during exponentiation	Use modulo operator during exponentiation to avoid integer overflow.
Array with negative numbers, need to handle correctly how negative numbers are raised to a power	Correctly calculate the power based on the exponent (even or odd) while considering negative numbers
Large input array causing time limit exceed	Optimize the solution to reduce time complexity possibly through sorting and optimized calculation of combinations.
Input array contains duplicates	The core logic handles duplicates correctly since each element is considered independently during subsequence generation.
Cases where the calculated sum exceeds the maximum integer value.	Use a larger data type such as long to store the sum, or take modulo at each step to avoid overflow.