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Problem solution · Python

Sort Array by Moving Items to Empty Space

Sort Array by Moving Items to Empty Space: 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
34 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 Sort Array by Moving Items to Empty Space, 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

  • 34 lines of Python from the credited upstream file 2459.py.
  • The implementation visibly relies on sequence storage.
  • 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 codeSort Array by Moving Items to Empty Space · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def sortArray(self, nums: list[int]) -> int:    n = len(nums)    numToIndex = [0] * n     for i, num in enumerate(nums):      numToIndex[num] = i     def minOps(numToIndex: list[int], zeroInBeginning: bool) -> int:      ops = 0      num = 1      # If zeroInBeginning, the correct index of each num is num.      # If not zeroInBeginning, the correct index of each num is num - 1.      offset = 0 if zeroInBeginning else 1      while num < n:        # 0 is in the correct index, so swap 0 with the first `numInWrongIndex`.        if (zeroInBeginning and numToIndex[0] == 0 or                not zeroInBeginning and numToIndex[0] == n - 1):          while numToIndex[num] == num - offset:  # num is in correct position            num += 1            if num == n:              return ops          numInWrongIndex = num        # 0 is in the wrong index. e.g. numToIndex[0] == 2, that means 2 is not        # in nums[2] because nums[2] == 0.        else:          numInWrongIndex = numToIndex[0] + offset        numToIndex[0], numToIndex[numInWrongIndex] = (            numToIndex[numInWrongIndex], numToIndex[0])        ops += 1     return min(minOps(numToIndex.copy(), True),               minOps(numToIndex.copy(), False))

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