LeetCode #1671 — HARD

Minimum Number of Removals to Make Mountain Array

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

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] > ... > arr[arr.length - 1]

Given an integer array nums, return the minimum number of elements to remove to make nums a mountain array.

Example 1:

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

Example 2:

Input: nums = [2,1,1,5,6,2,3,1]
Output: 3
Explanation: One solution is to remove the elements at indices 0, 1, and 5, making the array nums = [1,5,6,3,1].

Constraints:

3 <= nums.length <= 1000
1 <= nums[i] <= 10⁹
It is guaranteed that you can make a mountain array out of nums.

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: 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] > ... > arr[arr.length - 1] Given an integer array nums, return the minimum number of elements to remove to make nums a mountain array.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Dynamic Programming · Greedy

Example 1

[1,3,1]

Example 2

[2,1,1,5,6,2,3,1]

Related Problems

Longest Increasing Subsequence (longest-increasing-subsequence)
Longest Mountain in Array (longest-mountain-in-array)
Peak Index in a Mountain Array (peak-index-in-a-mountain-array)
Valid Mountain Array (valid-mountain-array)
Find in Mountain Array (find-in-mountain-array)

Step 02

Core Insight

What unlocks the optimal approach

Think the opposite direction instead of minimum elements to remove the maximum mountain subsequence
Think of LIS it's kind of close

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1671: Minimum Number of Removals to Make Mountain Array
class Solution {
    public int minimumMountainRemovals(int[] nums) {
        int n = nums.length;
        int[] left = new int[n];
        int[] right = new int[n];
        Arrays.fill(left, 1);
        Arrays.fill(right, 1);
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (nums[i] > nums[j]) {
                    left[i] = Math.max(left[i], left[j] + 1);
                }
            }
        }
        for (int i = n - 2; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                if (nums[i] > nums[j]) {
                    right[i] = Math.max(right[i], right[j] + 1);
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (left[i] > 1 && right[i] > 1) {
                ans = Math.max(ans, left[i] + right[i] - 1);
            }
        }
        return n - ans;
    }
}

// Accepted solution for LeetCode #1671: Minimum Number of Removals to Make Mountain Array
func minimumMountainRemovals(nums []int) int {
	n := len(nums)
	left, right := make([]int, n), make([]int, n)
	for i := range left {
		left[i], right[i] = 1, 1
	}
	for i := 1; i < n; i++ {
		for j := 0; j < i; j++ {
			if nums[i] > nums[j] {
				left[i] = max(left[i], left[j]+1)
			}
		}
	}
	for i := n - 2; i >= 0; i-- {
		for j := i + 1; j < n; j++ {
			if nums[i] > nums[j] {
				right[i] = max(right[i], right[j]+1)
			}
		}
	}
	ans := 0
	for i := range left {
		if left[i] > 1 && right[i] > 1 {
			ans = max(ans, left[i]+right[i]-1)
		}
	}
	return n - ans
}

# Accepted solution for LeetCode #1671: Minimum Number of Removals to Make Mountain Array
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)

// Accepted solution for LeetCode #1671: Minimum Number of Removals to Make Mountain Array
impl Solution {
    pub fn minimum_mountain_removals(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut left = vec![1; n];
        let mut right = vec![1; n];
        for i in 1..n {
            for j in 0..i {
                if nums[i] > nums[j] {
                    left[i] = left[i].max(left[j] + 1);
                }
            }
        }
        for i in (0..n - 1).rev() {
            for j in i + 1..n {
                if nums[i] > nums[j] {
                    right[i] = right[i].max(right[j] + 1);
                }
            }
        }

        let mut ans = 0;
        for i in 0..n {
            if left[i] > 1 && right[i] > 1 {
                ans = ans.max(left[i] + right[i] - 1);
            }
        }

        (n as i32) - ans
    }
}

// Accepted solution for LeetCode #1671: Minimum Number of Removals to Make Mountain Array
function minimumMountainRemovals(nums: number[]): number {
    const n = nums.length;
    const left = Array(n).fill(1);
    const right = Array(n).fill(1);
    for (let i = 1; i < n; ++i) {
        for (let j = 0; j < i; ++j) {
            if (nums[i] > nums[j]) {
                left[i] = Math.max(left[i], left[j] + 1);
            }
        }
    }
    for (let i = n - 2; i >= 0; --i) {
        for (let j = i + 1; j < n; ++j) {
            if (nums[i] > nums[j]) {
                right[i] = Math.max(right[i], right[j] + 1);
            }
        }
    }
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        if (left[i] > 1 && right[i] > 1) {
            ans = Math.max(ans, left[i] + right[i] - 1);
        }
    }
    return n - ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.