Closest Subsequence Sum

Hard

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Topics:

ArraysBinary SearchRecursion

You are given an integer array nums and an integer goal.

You want to choose a subsequence of nums such that the sum of its elements is the closest possible to goal. That is, if the sum of the subsequence's elements is sum, then you want to minimize the absolute difference abs(sum - goal).

Return the minimum possible value of abs(sum - goal).

Note that a subsequence of an array is an array formed by removing some elements (possibly all or none) of the original array.

Example 1:

Input: nums = [5,-7,3,5], goal = 6
Output: 0
Explanation: Choose the whole array as a subsequence, with a sum of 6.
This is equal to the goal, so the absolute difference is 0.

Example 2:

Input: nums = [7,-9,15,-2], goal = -5
Output: 1
Explanation: Choose the subsequence [7,-9,-2], with a sum of -4.
The absolute difference is abs(-4 - (-5)) = abs(1) = 1, which is the minimum.

Example 3:

Input: nums = [1,2,3], goal = -7
Output: 7

Constraints:

1 <= nums.length <= 40
-10⁷ <= nums[i] <= 10⁷
-10⁹ <= goal <= 10⁹

Try coding it in LeetCode

Closest Subsequence Sum

Hard

Asked by:

4 views

Topics:

ArraysBinary SearchRecursion

You are given an integer array nums and an integer goal.

You want to choose a subsequence of nums such that the sum of its elements is the closest possible to goal. That is, if the sum of the subsequence's elements is sum, then you want to minimize the absolute difference abs(sum - goal).

Return the minimum possible value of abs(sum - goal).

Note that a subsequence of an array is an array formed by removing some elements (possibly all or none) of the original array.

Example 1:

Input: nums = [5,-7,3,5], goal = 6
Output: 0
Explanation: Choose the whole array as a subsequence, with a sum of 6.
This is equal to the goal, so the absolute difference is 0.

Example 2:

Input: nums = [7,-9,15,-2], goal = -5
Output: 1
Explanation: Choose the subsequence [7,-9,-2], with a sum of -4.
The absolute difference is abs(-4 - (-5)) = abs(1) = 1, which is the minimum.

Example 3:

Input: nums = [1,2,3], goal = -7
Output: 7

Constraints:

1 <= nums.length <= 40
-10⁷ <= nums[i] <= 10⁷
-10⁹ <= goal <= 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 are the constraints on the input array's size and the range of integer values within it? Can the numbers be negative, zero, or positive?
If there are multiple subsequences with the same minimum absolute difference from the target, should I return any one, or is there a specific criterion for choosing among them?
If no subsequence sum exists that is close to the target (e.g., all subsequence sums are far away), what value should I return as an indicator of this scenario?
Is an empty subsequence considered a valid subsequence, and if so, does its sum of 0 influence the closest value?
Is there a specific definition of 'closest' that I should be aware of, or is it simply minimizing the absolute difference between the subsequence sum and the target?

Brute Force Solution

Approach

The brute force approach tries every single possible combination of numbers from the input to find the subsequence sum that's closest to the target. We consider all possible subsequences, calculate their sums, and compare each sum to the target value.

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

Start by considering each number individually as a possible subsequence.
Then, consider every possible pair of numbers.
Next, consider every possible group of three numbers, then four, and so on, up to using all the numbers.
For each group of numbers, add them together to get a sum.
Compare this sum to the target value and remember how close it is.
After checking all possible groups of numbers, find the sum that was closest to the target value.

Code Implementation

def closest_subsequence_sum_brute_force(numbers, target):
    closest_sum = float('inf')
    
    # Iterate through all possible subsequences
    for i in range(1 << len(numbers)):
        current_subsequence_sum = 0

        # Build the subsequence and calculate its sum
        for j in range(len(numbers)):
            # Check if the j-th element is included in the subsequence
            if (i >> j) & 1:
                current_subsequence_sum += numbers[j]
        
        # Check if this subsequence sum is closer to the target
        if abs(current_subsequence_sum - target) < abs(closest_sum - target):
            closest_sum = current_subsequence_sum

    return closest_sum

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach generates all possible subsequences of the input array. For an array of size n, there are 2^n possible subsequences (each element can either be included or excluded). For each subsequence, we compute its sum, which takes O(n) in the worst case when the subsequence contains all n elements, however, subsequence generation dominates the time complexity. Thus, the overall time complexity is O(2^n * n). Because 2^n grows much faster than n, we can simplify the overall time complexity to O(2^n).

Space Complexity

O(1) – The brute force approach, as described, does not use any significant auxiliary space. It iterates through all possible subsequences and calculates their sums, but it does not store these subsequences or sums in any explicit data structure beyond what is needed for temporary calculations and comparison. Therefore, the auxiliary space used remains constant regardless of the input size N, which is the number of elements in the input array. This leads to a space complexity of O(1).

Optimal Solution

Approach

The key is to split the problem into two smaller, more manageable parts. Then, we cleverly combine results from each part to find the closest possible sum without checking every single option.

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

First, divide the original list of numbers into two roughly equal halves.
Consider the first half. For every possible group of numbers you can pick from this half, calculate the sum.
Do the same thing for the second half: calculate the sum of every possible group of numbers you can pick.
Now, you have two sets of sums, one for each half of the original list.
For each sum in the first set, find the sum in the second set that, when combined, gets you closest to your target number. Because the second set is sorted, finding the closest sum can be done very efficiently without checking every value.
Keep track of the combined sum that's closest to your target. You only need to store the closest one found so far.
After doing this for all sums in the first set, you'll have found the overall closest possible sum to your target.

Code Implementation

def closest_subsequence_sum(numbers, target):
    def generate_all_sums(sub_array):
        sums = {0}
        for number in sub_array:
            new_sums = set()
            for current_sum in sums:
                new_sums.add(current_sum + number)
            sums.update(new_sums)
        return sorted(list(sums))

    length = len(numbers)
    first_half = numbers[:length // 2]
    second_half = numbers[length // 2:]

    # Generate all possible subset sums for the first and second halves
    first_half_sums = generate_all_sums(first_half)
    second_half_sums = generate_all_sums(second_half)

    closest_sum = float('inf')

    # Iterate through sums of first half, finding closest match in second half.
    for first_half_sum in first_half_sums:
        left_index = 0
        right_index = len(second_half_sums) - 1

        while left_index <= right_index:
            middle_index = (left_index + right_index) // 2
            current_sum = first_half_sum + second_half_sums[middle_index]

            if abs(current_sum - target) < abs(closest_sum - target):
                closest_sum = current_sum

            # Binary search: Adjust indices based on comparison to target.
            if current_sum < target:
                left_index = middle_index + 1
            else:
                right_index = middle_index - 1

    # Return the closest sum found after considering all possibilities.
    return closest_sum

Big(O) Analysis

Time Complexity

O(2^(n/2) * log(2^(n/2))) – The algorithm divides the input array of size n into two halves. Generating all possible subsequence sums for each half takes O(2^(n/2)) time, as there are 2^(n/2) possible subsets for a set of size n/2. The crucial step involves iterating through the sums of the first half (2^(n/2) sums) and, for each sum, performing a binary search (log(2^(n/2)) time) on the sorted sums of the second half to find the closest match. Therefore, the overall time complexity is dominated by 2^(n/2) iterations each containing a binary search that takes log(2^(n/2)) time. This gives O(2^(n/2) * log(2^(n/2))).

Space Complexity

O(2^(N/2)) – The algorithm divides the input array into two halves. It then generates all possible subsequence sums for each half. This involves creating lists to store these sums. In the worst case, there are 2^(N/2) possible subsequences for each half, where N is the size of the original input array. Therefore, the algorithm requires auxiliary space to store two lists of size up to 2^(N/2), one for each half, resulting in O(2^(N/2) + 2^(N/2)) which simplifies to O(2^(N/2)).

Edge Cases

Case	How to Handle
Empty input array	Return 0 as no subsequence sum can be calculated.
Array with a single element	Return the value of the single element if its absolute difference with goal is smallest, otherwise 0 if goal is closer to 0.
Array with all positive numbers and a negative target	Return 0, the closest possible sum.
Array with all negative numbers and a positive target	Return 0, the closest possible sum.
Large input array with numbers having a wide range of values (both positive and negative)	The algorithm needs to be efficient enough to avoid Time Limit Exceeded (TLE) errors; consider using dynamic programming or a meet-in-the-middle approach.
Target is extremely large positive/negative number	The closest subsequence sum would likely be the sum of all positive/negative numbers respectively, and the algorithm should handle this without overflow.
Integer overflow when calculating subsequence sums	Use a data type with a larger range (e.g., long) to store the subsequence sums, or check for potential overflows before performing the addition.
Multiple subsequences with the same closest sum	The algorithm should arbitrarily return any one of them as per the problem statement, or specify that it returns one of the closest if the problem statement asked to return all.