What is the brute force solution for Minimum Operations to Form Subsequence With Target Sum?

The brute force approach for Minimum Operations to Form Subsequence With Target Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach tries to find all combinations of the elements in the array to see if any subsequence sums to the target. This method is straightforward but inefficient, as it explores all possible combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Operations to Form Subsequence With Target Sum?

Minimum Operations to Form Subsequence With Target Sum uses the following concepts: Array, Greedy, Bit Manipulation. The recommended patterns to study are: Greedy, Bit Manipulation.

What are the interview tips for Minimum Operations to Form Subsequence With Target Sum?

Always clarify the problem requirements and constraints before jumping into coding. Think about how you can break down the problem into smaller, manageable parts. Practice explaining your thought process while coding, as communication is key in interviews.

What are common mistakes when solving Minimum Operations to Form Subsequence With Target Sum?

Failing to account for the need to create smaller powers from larger ones. Not considering all bits of the target, leading to missed opportunities to form the required sum.

#2835

Minimum Operations to Form Subsequence With Target Sum

Hard

ArrayGreedyBit ManipulationGreedyBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach focuses on the bits of the target and uses a greedy strategy to break down larger powers of 2 into smaller ones until we can form the target sum. This is efficient because it reduces the problem size significantly with each operation.

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Algorithm

4 steps

1Step 1: Count occurrences of each power of 2 in nums.
2Step 2: For each bit in the target (from least to most significant), check if we can form that bit using available powers of 2.
3Step 3: If a required power is missing, perform operations to create it from larger powers, counting each operation.
4Step 4: If we can form the target sum using available powers, return the total operations; otherwise, return -1.

solution.py22 lines

1# Full working Python code
2from collections import Counter
3
4def min_operations_optimal(nums, target):
5    count = Counter(nums)
6    operations = 0
7    for i in range(32):  # Powers of 2 up to 2^31
8        if (target >> i) & 1:
9            if count[1 << i] > 0:
10                count[1 << i] -= 1
11            else:
12                # Need to create this power
13                while (1 << i) > 1:
14                    if count[1 << (i + 1)] > 0:
15                        operations += 1
16                        count[1 << (i + 1)] -= 1
17                        count[1 << i] += 2
18                        break
19                    i += 1
20                else:
21                    return -1
22    return operations

ℹ

Complexity note: This complexity is efficient because we only traverse the list of numbers and the bits of the target, leading to a linear time complexity relative to the size of the input.

1Understanding how to break down larger powers of 2 is crucial for forming the target sum efficiently.
2Using a greedy approach allows us to minimize operations by always trying to use the largest available power of 2.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.