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Problem solution · Python

Find in Mountain Array

Find in Mountain 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
55 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 in Mountain 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

  • 55 lines of Python from the credited upstream file 1095.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 in Mountain Array · PythonPython
Use this to learn the idea, then write your own version.
# """# This is MountainArray's API interface.# You should not implement it, or speculate about its implementation# """# Class MountainArray:#   def get(self, index: int) -> int:#   def length(self) -> int: class Solution:  def findInMountainArray(      self,      target: int,      mountain_arr: 'MountainArray',  ) -> int:    n = mountain_arr.length()    peakIndex = self.peakIndexInMountainArray(mountain_arr, 0, n - 1)     leftIndex = self.searchLeft(mountain_arr, target, 0, peakIndex)    if mountain_arr.get(leftIndex) == target:      return leftIndex     rightIndex = self.searchRight(mountain_arr, target, peakIndex + 1, n - 1)    if mountain_arr.get(rightIndex) == target:      return rightIndex     return -1   # 852. Peak Index in a Mountain Array  def peakIndexInMountainArray(self, A: 'MountainArray', l: int, r: int) -> int:    while l < r:      m = (l + r) // 2      if A.get(m) < A.get(m + 1):        l = m + 1      else:        r = m    return l   def searchLeft(self, A: 'MountainArray', target: int, l: int, r: int) -> int:    while l < r:      m = (l + r) // 2      if A.get(m) < target:        l = m + 1      else:        r = m    return l   def searchRight(self, A: 'MountainArray', target: int, l: int, r: int) -> int:    while l < r:      m = (l + r) // 2      if A.get(m) > target:        l = m + 1      else:        r = m    return l

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