Sum of Good Subsequences
#1312 Most AskedHard
Sum of Good Subsequences #1312 Most Asked
Hard
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Topics:
Dynamic ProgrammingArrays
You are given an integer array nums. A good subsequence is defined as a subsequence of nums where the absolute difference between any two consecutive elements in the subsequence is exactly 1.
Return the sum of all possible good subsequences of nums.
Since the answer may be very large, return it modulo 109 + 7.
Note that a subsequence of size 1 is considered good by definition.
Example 1:
Input: nums = [1,2,1]
Output: 14
Explanation:
- Good subsequences are:
[1],[2],[1],[1,2],[2,1],[1,2,1]. - The sum of elements in these subsequences is 14.
Example 2:
Input: nums = [3,4,5]
Output: 40
Explanation:
- Good subsequences are:
[3],[4],[5],[3,4],[4,5],[3,4,5]. - The sum of elements in these subsequences is 40.
Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 105
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 for the numbers in the input array? Can they be negative, zero, or very large?
- What defines a 'good' subsequence? What is the specific criteria or condition that makes a subsequence 'good'?
- How large can the input array be? Is there a limit to the number of elements in the array?
- If there are multiple 'good' subsequences, should I return the sum of all of them, or is there a specific requirement for which one to return (e.g., the subsequence with the largest sum)?
- Is the order of elements within a subsequence important? For example, are [1, 2] and [2, 1] considered the same subsequence?
Brute Force Solution
Approach
The brute force method for this problem involves exploring every possible subsequence within the given sequence. We need to calculate the 'goodness' of each of these subsequences. Then, we sum the 'goodness' values of every subsequence to get the final result.
Here's how the algorithm would work step-by-step:
- First, consider every possible combination of elements from the original sequence. Each combination represents a potential subsequence.
- For each of these subsequences, calculate its 'goodness' score according to the rules of what makes a subsequence 'good'.
- Keep a running total. For each subsequence, add its 'goodness' score to this total.
- After you've checked every possible subsequence and added its 'goodness' to the total, the final total is your answer.
Code Implementation
def sum_of_good_subsequences_brute_force(sequence):
total_goodness = 0
number_of_elements = len(sequence)
# Iterate through all possible subsequences
for i in range(1 << number_of_elements):
subsequence = []
for j in range(number_of_elements):
# Check if the j-th element is included in the current subsequence
if (i >> j) & 1:
subsequence.append(sequence[j])
# Calculate the goodness of the current subsequence
subsequence_goodness = calculate_goodness(subsequence)
# Add the goodness to the total
total_goodness += subsequence_goodness
return total_goodness
def calculate_goodness(subsequence):
# Placeholder function, replace with actual goodness calculation
if not subsequence:
return 0
sum_of_elements = sum(subsequence)
if sum_of_elements % 2 == 0:
return 1
else:
return 0
#Example Usage
#my_sequence = [1, 2, 3]
#result = sum_of_good_subsequences_brute_force(my_sequence)
#print(result)
#my_sequence = [1, 3, 5]
#result = sum_of_good_subsequences_brute_force(my_sequence)
#print(result)
#my_sequence = [2, 4, 6]
#result = sum_of_good_subsequences_brute_force(my_sequence)
#print(result)Big(O) Analysis
Time Complexity
O(2^n) – The algorithm iterates through all possible subsequences of the input sequence. For a sequence of size n, there are 2^n possible subsequences, because each element can either be included or excluded from a subsequence. For each subsequence, some 'goodness' calculation is performed, which contributes a constant amount of work. Thus, the dominant factor is the generation and processing of each subsequence, leading to a time complexity of O(2^n).
Space Complexity
O(1) – The brute force method, as described, iterates through all possible subsequences and calculates the 'goodness' of each one, adding it to a running total. This implies we only need a few variables to store the running total, the 'goodness' of the current subsequence, and possibly some counters for iterating through subsequences. The number of these variables does not depend on the size of the input sequence (N). Therefore, the space complexity is constant.
Optimal Solution
Approach
The goal is to efficiently find all the 'good' subsequences of a given sequence of numbers and sum them up. Instead of checking every possible subsequence, we can use a mathematical trick based on binary representations to quickly compute the final sum.
Here's how the algorithm would work step-by-step:
- Consider each number in the sequence individually.
- For each number, figure out how many times it will appear in the sum of all 'good' subsequences. A good subsequence will be any non-empty subsequence.
- The important insight is that each number will appear in half of all possible subsequences. Since a set of N numbers has 2 to the power of N subsequences, each single number will appear in 2 to the power of N-1 subsequences. Since we are only considering non-empty subsequences, the count would be 2 to the power of N-1.
- Multiply the number with the total amount of times it occurs, which is 2 to the power of number of elements in the sequence, minus 1.
- Add the result to your overall total sum.
- Repeat the process for every number in the sequence and you'll get the final sum.
Code Implementation
def sum_of_good_subsequences(sequence):
total_sum = 0
sequence_length = len(sequence)
# Calculate 2^(N-1), the number of times each element appears
if sequence_length > 0:
occurrences = pow(2, sequence_length - 1)
else:
return 0
# Iterate through each number in the sequence
for number in sequence:
# Add the contribution of the number to the total sum
total_sum += number * occurrences
return total_sumBig(O) Analysis
Time Complexity
O(n) – The algorithm iterates through each of the n numbers in the input sequence. For each number, it calculates 2 to the power of (n-1) (or equivalently multiplies the number by 2^(n-1)), and adds the result to the total sum. Raising 2 to the power of (n-1) can be done in O(log n) time using exponentiation by squaring if we implement it ourselves or with O(1) using native power functions. However, the core logic involves iterating once over n elements, resulting in a complexity dominated by the linear pass. Thus, the overall time complexity is O(n), assuming the power calculation is reasonably efficient.
Space Complexity
O(1) – The algorithm iterates through the input sequence and performs calculations on each number independently. It uses a constant amount of extra space for variables like the overall total sum and the intermediate result of the power calculation, which do not depend on the input size N (the number of elements in the sequence). No additional data structures like lists or hash maps are created to store intermediate results or track visited elements. Therefore, the auxiliary space complexity is constant.
Edge Cases
| Case | How to Handle |
|---|---|
| Empty input array | Return 0, as there are no subsequences to sum. |
| Array with a single element | If the single element meets the 'good' criteria, return that element; otherwise return 0. |
| Array with all zeros | Depending on what defines a good subsequence, handle zero values explicitly, which might involve checking subsequence length or specific value requirements |
| Array with very large numbers (potential integer overflow during summation) | Use a data type that supports larger numbers (e.g., long) or perform the calculations using modular arithmetic if applicable to avoid overflow. |
| Array contains negative numbers | The definition of 'good subsequence' might change; consider how negative numbers interact with the criteria and adjust logic accordingly. |
| Maximum sized input array (performance considerations) | Optimize the solution's time complexity to avoid exceeding time limits, potentially using dynamic programming or efficient data structures. |
| No 'good' subsequences exist in the array | Return 0, as the sum of no elements is 0. |
| Input contains extreme boundary values close to integer limits | Be careful of potential overflows or underflows when calculating sums or intermediate values involving these numbers. |
Empty input array
How to Handle:
Return 0, as there are no subsequences to sum.
Array with a single element
How to Handle:
If the single element meets the 'good' criteria, return that element; otherwise return 0.
Array with all zeros
How to Handle:
Depending on what defines a good subsequence, handle zero values explicitly, which might involve checking subsequence length or specific value requirements
Array with very large numbers (potential integer overflow during summation)
How to Handle:
Use a data type that supports larger numbers (e.g., long) or perform the calculations using modular arithmetic if applicable to avoid overflow.
Array contains negative numbers
How to Handle:
The definition of 'good subsequence' might change; consider how negative numbers interact with the criteria and adjust logic accordingly.
Maximum sized input array (performance considerations)
How to Handle:
Optimize the solution's time complexity to avoid exceeding time limits, potentially using dynamic programming or efficient data structures.
No 'good' subsequences exist in the array
How to Handle:
Return 0, as the sum of no elements is 0.
Input contains extreme boundary values close to integer limits
How to Handle:
Be careful of potential overflows or underflows when calculating sums or intermediate values involving these numbers.
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