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Problem solution · Python

Minimum Operations to Form Subsequence With Target Sum

Minimum Operations to Form Subsequence With Target Sum: a Python solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
Length
29 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Hash-based lookup

For Minimum Operations to Form Subsequence With Target Sum, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 29 lines of Python from the credited upstream file 2835.py.
  • The implementation visibly relies on sequence storage, hash lookup.
  • No explicit loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Operations to Form Subsequence With Target Sum · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def minOperations(self, nums: list[int], target: int) -> int:    NO_MISSING_BIT = 31    maxBit = 31    ans = 0    minMissingBit = NO_MISSING_BIT    # count[i] := the number of occurrences of 2^i    count = collections.Counter(int(math.log2(num)) for num in nums)     for bit in range(maxBit):      # Check if `bit` is in the target.      if target >> bit & 1:        # If there are available bits, use one bit.        if count[bit] > 0:          count[bit] -= 1        else:          minMissingBit = min(minMissingBit, bit)      # If we previously missed a bit and there are available bits.      if minMissingBit != NO_MISSING_BIT and count[bit] > 0:        count[bit] -= 1        # Count the operations to break `bit` into `minMissingBit`.        ans += bit - minMissingBit        minMissingBit = NO_MISSING_BIT  # Set it to an the invalid value.      # Combining smaller numbers costs nothing.      count[bit + 1] += count[bit] // 2     # Check if all target bits have been covered, otherwise return -1.    return ans if minMissingBit == maxBit else -1

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