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Leetcode #1671: Minimum Number of Removals to Make Mountain Array

In this guide, we solve Leetcode #1671 Minimum Number of Removals to Make Mountain Array in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

You may recall that an array arr is a mountain array if and only if: arr.length >= 3 There exists some index i (0-indexed) with 0 < i < arr.length - 1 such that: arr[0] < arr[1] < ... < arr[i - 1] < arr[i] arr[i] > arr[i + 1] > ...

Quick Facts

  • Difficulty: Hard
  • Premium: No
  • Tags: Greedy, Array, Binary Search, Dynamic Programming

Intuition

The problem structure suggests a monotonic decision, which makes binary search a natural fit.

By halving the search space each step, we reach the answer efficiently.

Approach

Search either directly on a sorted array or on the answer space using a check function.

Each check is fast, and the logarithmic search keeps the overall runtime low.

Steps:

  • Define the search bounds.
  • Check the mid point condition.
  • Narrow the bounds until convergence.

Example

Input: nums = [1,3,1] Output: 0 Explanation: The array itself is a mountain array so we do not need to remove any elements.

Python Solution

class Solution: def minimumMountainRemovals(self, nums: List[int]) -> int: n = len(nums) left = [1] * n right = [1] * n for i in range(1, n): for j in range(i): if nums[i] > nums[j]: left[i] = max(left[i], left[j] + 1) for i in range(n - 2, -1, -1): for j in range(i + 1, n): if nums[i] > nums[j]: right[i] = max(right[i], right[j] + 1) return n - max(a + b - 1 for a, b in zip(left, right) if a > 1 and b > 1)

Complexity

The time complexity is O(n2)O(n^2)O(n2), and the space complexity is O(n)O(n)O(n). The space complexity is O(n)O(n)O(n).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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