Minimum Operations to Form Subsequence With Target Sum

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Greedy AlgorithmsBit Manipulation

You are given a 0-indexed array nums consisting of non-negative powers of 2, and an integer target.

In one operation, you must apply the following changes to the array:

Choose any element of the array nums[i] such that nums[i] > 1.
Remove nums[i] from the array.
Add two occurrences of nums[i] / 2 to the end of nums.

Return the minimum number of operations you need to perform so that nums contains a subsequence whose elements sum to target. If it is impossible to obtain such a subsequence, return -1.

A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

Example 1:

Input: nums = [1,2,8], target = 7
Output: 1
Explanation: In the first operation, we choose element nums[2]. The array becomes equal to nums = [1,2,4,4].
At this stage, nums contains the subsequence [1,2,4] which sums up to 7.
It can be shown that there is no shorter sequence of operations that results in a subsequnce that sums up to 7.

Example 2:

Input: nums = [1,32,1,2], target = 12
Output: 2
Explanation: In the first operation, we choose element nums[1]. The array becomes equal to nums = [1,1,2,16,16].
In the second operation, we choose element nums[3]. The array becomes equal to nums = [1,1,2,16,8,8]
At this stage, nums contains the subsequence [1,1,2,8] which sums up to 12.
It can be shown that there is no shorter sequence of operations that results in a subsequence that sums up to 12.

Example 3:

Input: nums = [1,32,1], target = 35
Output: -1
Explanation: It can be shown that no sequence of operations results in a subsequence that sums up to 35.

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 2³⁰
nums consists only of non-negative powers of two.
1 <= target < 2³¹

Try coding it in LeetCode

Minimum Operations to Form Subsequence With Target Sum

Hard

Asked by:

6 views

Topics:

Greedy AlgorithmsBit Manipulation

You are given a 0-indexed array nums consisting of non-negative powers of 2, and an integer target.

In one operation, you must apply the following changes to the array:

Choose any element of the array nums[i] such that nums[i] > 1.
Remove nums[i] from the array.
Add two occurrences of nums[i] / 2 to the end of nums.

Return the minimum number of operations you need to perform so that nums contains a subsequence whose elements sum to target. If it is impossible to obtain such a subsequence, return -1.

A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

Example 1:

Input: nums = [1,2,8], target = 7
Output: 1
Explanation: In the first operation, we choose element nums[2]. The array becomes equal to nums = [1,2,4,4].
At this stage, nums contains the subsequence [1,2,4] which sums up to 7.
It can be shown that there is no shorter sequence of operations that results in a subsequnce that sums up to 7.

Example 2:

Input: nums = [1,32,1,2], target = 12
Output: 2
Explanation: In the first operation, we choose element nums[1]. The array becomes equal to nums = [1,1,2,16,16].
In the second operation, we choose element nums[3]. The array becomes equal to nums = [1,1,2,16,8,8]
At this stage, nums contains the subsequence [1,1,2,8] which sums up to 12.
It can be shown that there is no shorter sequence of operations that results in a subsequence that sums up to 12.

Example 3:

Input: nums = [1,32,1], target = 35
Output: -1
Explanation: It can be shown that no sequence of operations results in a subsequence that sums up to 35.

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 2³⁰
nums consists only of non-negative powers of two.
1 <= target < 2³¹

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 possible ranges for the values in the input array and the target sum? Can they be negative, zero, or floating-point numbers?
What should I return if it's impossible to form the target sum as a subsequence from the given array? Should I return -1, null, or throw an exception?
Are duplicate numbers allowed in the input array, and if so, how should they be handled when trying to form the subsequence?
Is the order of the elements in the formed subsequence important? Does it need to match the order in the original array?
Is the goal to minimize the number of elements used in the subsequence to achieve the target sum, or to minimize the number of operations needed to transform the original array such that a subsequence sums to the target?

Brute Force Solution

Approach

The brute force approach is like trying every single possible combination of numbers to see if any of them add up to the target. We explore every possible group of numbers from the given list. It's thorough, but can be slow.

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

First, consider using only the very first number in the list. See if that number alone equals the target sum.
Then, consider using the first two numbers. Check if either of those numbers individually equals the target sum, and also check if the sum of the two numbers equals the target sum.
Next, consider using the first three numbers. Check all the possibilities: each number individually, every pair of numbers, and the sum of all three numbers.
Keep doing this, adding one more number from the list each time, and checking all possible combinations of the numbers you're currently considering.
Each time you find a combination of numbers that adds up to the target sum, record the number of operations (how many times you had to split the number) you used to achieve that sum.
After you've checked every possible combination of numbers from the entire list, compare all the operation counts you've recorded.
The smallest operation count among all the successful combinations is your answer.

Code Implementation

def minimum_operations_to_form_subsequence_brute_force(numbers, target):
    minimum_number_of_operations = float('inf')

    for i in range(1 << len(numbers)):
        current_sum = 0
        operations_count = 0
        selected_numbers = []

        for j in range(len(numbers)):
            if (i >> j) & 1:
                selected_numbers.append(numbers[j])

        # Process each of the selected numbers by splitting if necessary
        for number in selected_numbers:
            current_number = number
            while current_number > target:
                current_number //= 2
                operations_count += 1

            if current_number > 0:
                current_sum += current_number

        # Compare with global best if a sum match is found
        if current_sum == target:
            minimum_number_of_operations = min(minimum_number_of_operations, operations_count)

    # If no combination sums to target return -1
    if minimum_number_of_operations == float('inf'):
        return -1
    
    return minimum_number_of_operations

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach considers all possible subsets of the input array of size n. For each element, we have two choices: either include it in the subset or exclude it. This leads to 2*2*...*2 (n times) which equals 2^n possible subsets. For each subset, we need to compute its sum and potentially calculate the minimum number of operations to reach a target if the sum is equal to the target. Therefore, the time complexity is exponential with respect to the input size.

Space Complexity

O(1) – The brute force approach, as described, doesn't explicitly create auxiliary data structures that scale with the input size, N. It iteratively considers sub-arrays and calculates their sums. The only extra memory used are a few variables to keep track of the current sub-array, sum, target, and the minimum operations. Thus, the space used remains constant regardless of the input size.

Optimal Solution

Approach

The problem requires minimizing operations (splitting numbers) to achieve a target sum using a subsequence. We can solve this efficiently by working backward, greedily using the largest possible numbers to reach the target, splitting them if necessary.

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

Start by considering the target sum we need to achieve.
Go through the given numbers from left to right, looking for the largest number that's less than or equal to the remaining part of our target sum.
If we find such a number, we 'use' it towards the target, and subtract it from the target sum.
If the current number is bigger than the remaining target, we need to split it until it's small enough to use. We keep track of each split as an 'operation'.
Once we've made a number small enough, we 'use' it, subtract it from the target, and continue.
Repeat steps 2-5 until the target sum becomes zero. The total number of splits we performed is our answer.

Code Implementation

def min_operations_to_form_subsequence_with_target_sum(numbers, target): 
    operations_count = 0
    numbers.sort(reverse=True)

    for number in numbers:
        # Iterate through the array trying to reach the target.
        if number <= target:
            target -= number
        else:
            # We must split the current number until it can be used.
            while number > target:
                number //= 2
                operations_count += 1

                if number == 0:
                    break

            if number > 0:
                #If number is positive, subtract and continue.
                target -= number

    if target > 0:
        #If the target is unreachable, return -1.
        return -1

    return operations_count

Big(O) Analysis

Time Complexity

O(n log(maxVal)) – We iterate through the input array of size n. In the worst case, for each element, we might need to repeatedly split it until it's small enough to be used, with maxVal being the maximum value in the input array. Each split effectively halves the number, meaning the maximum number of splits for a single element is proportional to log(maxVal). Therefore, the overall time complexity is O(n log(maxVal)).

Space Complexity

O(1) – The algorithm operates primarily in place, modifying the input implicitly during the splitting operations, but does not create new data structures of significant size directly correlated to the input array's size. The primary memory usage is due to a few integer variables used to keep track of the current number and the remaining target, which occupies constant space. Therefore, the auxiliary space complexity is independent of the input size N (where N is the number of elements in the input array) and the target value. Hence, the overall space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 if the input array is null or empty as no operations are possible.
Target sum is zero and the input array contains zero	If target is 0 and array contains 0 return 0 because no operations are required.
Target sum is zero and the input array does not contain zero	Return -1 (or indicate no solution) since it's impossible to form a subsequence with sum 0 without a zero in the array.
Input array contains only 1s and the target sum is much larger than array length	The number of operations will approach array length and the algorithm needs to efficiently divide the 1s until the sum approaches zero.
Input array contains large numbers and target sum is also very large leading to potential integer overflow	Use long data type to prevent integer overflow during intermediate calculations such as summing up the numbers in the array.
Target sum cannot be reached with given numbers	Return -1 or an appropriate error code indicating that no solution exists.
The input array contains only zeros and the target sum is non-zero	Return -1 as it's impossible to reach a non-zero target from only zeros.
Array contains negative numbers and the target sum is positive	Handle negative numbers correctly by potentially including them to reduce the sum if needed.