#2835

Hard

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Minimum Operations to Form Subsequence With Target Sum

The problem requires finding the minimum number of operations to form a subsequence summing to a target using powers of two.

array greedy bit manipulation

Problem Statement

You are given a 0-indexed array nums consisting of non-negative powers of 2, and an integer target. In one operation, you must choose an element from nums and duplicate it. This can happen multiple times to ensure that nums contains a subsequence whose sum equals the target.

Your task is to return the minimum number of operations required to form a subsequence that sums to the target. If it's not possible to achieve the target sum, return -1.

Examples

Example 1

Input: nums = [1,2,8], target = 7

Output: 1

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

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

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

Solution Approach

Greedy Selection of Powers of Two

The problem revolves around selecting the largest possible powers of two from the array that help achieve the target sum. The greedy approach ensures that the largest numbers are selected first to reach the target faster. By doubling smaller elements, the process can be sped up.

Validation of Invariants

In this approach, we maintain an invariant where at every step, the sum of selected elements matches the target. If this invariant is broken, the process stops and the solution is deemed impossible.

Efficiency Considerations

The challenge lies in optimizing the number of operations. By leveraging bit manipulation and power-of-two properties, the process of finding valid subsequences can be expedited. Each operation manipulates the array's values in a controlled manner, ensuring minimal operations for an efficient solution.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexities of this problem depend on the final approach used. A greedy solution with bit manipulation can optimize the time complexity, while space complexity remains manageable with a careful selection of operations. The operations focus on adjusting values to form the target sum, with each step contributing to minimizing the operations required.

What Interviewers Usually Probe

The candidate demonstrates a solid understanding of greedy approaches and array manipulation.
The candidate identifies the role of invariant validation to ensure correct results.
The candidate balances time and space complexity considerations while optimizing the solution.

Common Pitfalls or Variants

Common pitfalls

Ignoring the constraints related to powers of two, which could lead to incorrect assumptions about the sequence's structure.
Failing to recognize when the target sum is impossible to achieve, such as when the sum of nums is smaller than the target.
Overcomplicating the solution by attempting to apply non-greedy methods when a greedy approach would suffice.

Follow-up variants

What if nums contains negative powers of two?
Can the solution be optimized for extremely large arrays, say nums.length > 1000?
What happens if the target is much larger than the sum of the numbers in nums?

FAQ

What is the primary strategy for solving the Minimum Operations to Form Subsequence With Target Sum?

The primary strategy is to use a greedy approach combined with validation of invariants to minimize the number of operations while ensuring the subsequence sum matches the target.

What if the target sum is impossible to achieve?

If the target sum exceeds the total sum of all elements in the array, or if there is no combination of elements that can sum up to the target, the answer should be -1.

Can you use dynamic programming to solve this problem?

Although dynamic programming can be applied to similar problems, this problem can be more efficiently solved using greedy methods, leveraging powers of two for optimization.

How does bit manipulation play a role in solving this problem?

Bit manipulation allows for efficient management of powers of two and helps in optimizing the process of doubling smaller numbers to form the target sum.

How does GhostInterview help in solving this type of problem?

GhostInterview provides detailed step-by-step guidance, helping you understand the greedy approach, optimize operations, and avoid common pitfalls in achieving the target sum.

terminal

Solution

Solution 1: Greedy + Bit Manipulation

Observing the operation in the problem, we find that each operation actually splits a number greater than $1$ into two equal numbers, which means that the sum of the elements in the array will not change after the operation. Therefore, if the sum of the elements in the array $s$ is less than $target$, it is impossible to obtain a subsequence with a sum of $target$ through the operation described in the problem, and we can directly return $-1$. Otherwise, we can definitely make the sum of some subsequences in the array equal to $target$ through the split operation.

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class Solution:
    def minOperations(self, nums: List[int], target: int) -> int:
        s = sum(nums)
        if s < target:
            return -1
        cnt = [0] * 32
        for x in nums:
            for i in range(32):
                if x >> i & 1:
                    cnt[i] += 1
        i = j = 0
        ans = 0
        while 1:
            while i < 32 and (target >> i & 1) == 0:
                i += 1
            if i == 32:
                break
            while j < i:
                cnt[j + 1] += cnt[j] // 2
                cnt[j] %= 2
                j += 1
            while cnt[j] == 0:
                cnt[j] = 1
                j += 1
            ans += j - i
            cnt[j] -= 1
            j = i
            i += 1
        return ans

Continue Practicing

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