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Problem solution · Python

Find the Minimum Possible Sum of a Beautiful Array

Find the Minimum Possible Sum of a Beautiful Array: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
19 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

Direct simulation

For Find the Minimum Possible Sum of a Beautiful Array, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 19 lines of Python from the credited upstream file 2834.py.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 codeFind the Minimum Possible Sum of a Beautiful Array · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  # Same as 2829. Determine the Minimum Sum of a k-avoiding Array  def minimumPossibleSum(self, n: int, target: int) -> int:    # These are the unique pairs that sum up to k (target):    # (1, k - 1), (2, k - 2), ..., (ceil(k // 2), floor(k // 2)).    # Our optimal strategy is to select 1, 2, ..., floor(k // 2), and then    # choose k, k + 1, ... if necessary, as selecting any number in the range    # [ceil(k // 2), k - 1] will result in a pair summing up to k.    MOD = 1_000_000_007     def trapezoid(a: int, b: int) -> int:      """Returns sum(a..b)."""      return (a + b) * (b - a + 1) // 2     mid = target // 2  # floor(k // 2)    if n <= mid:      return trapezoid(1, n)    return (trapezoid(1, mid) + trapezoid(target, target + (n - mid - 1))) % MOD

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