1671

1671. Minimum Number of Removals to Make Mountain Array (Hard)

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.

Companies: Apple, Microsoft

Related Topics:
Array, Binary Search, Dynamic Programming, Greedy

Similar Questions:

• Longest Increasing Subsequence (Medium)
• Longest Mountain in Array (Medium)
• Peak Index in a Mountain Array (Medium)
• Valid Mountain Array (Easy)
• Find in Mountain Array (Hard)

Solution 1. Three passes

Intuition

Assume we pick A[i] as the peak of the mountain, then we are looking for the longest increasing subsequence to the left of A[i], and the longest decreasing subsequence to the right of A[i].

We can reuse the O(NlogN) time binary search solution to 300. Longest Increasing Subsequence (Medium).

For the binary search solution to problem 300, please checkout my explanation.

Algorithm

Let left[i] be the length of the longest increasing subsequence in A[0..(i-1)] that can has A[i] appended to it, and right[i] be the length of the longest decreasing subsequence in A[(i+1)..(N-1)] that can has A[i] prepended to it.

We can scan from left to right to set the left[i] values, and scan from right to left to set the right[i] values.

For 1 <= i <= N - 2, the longest mountain size is left[i] + right[i] + 1.

So the answer is the minimum N - (left[i] + right[i] + 1).

Note that we need to skip cases where either left[i] or right[i] is zero because it's invalid.

Test cases that should be added:

• [1,2,1,2,3,4] and [4,3,2,1,2,1]. Correct answer is 3.
• [1,2,1,1,2,3,4,5] and [5,4,3,2,1,1,2,1]. Correct answer is 5.

// OJ: https://leetcode.com/problems/minimum-number-of-removals-to-make-mountain-array/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    int minimumMountainRemovals(vector<int>& A) {
        int N = A.size(), maxLen = 0;
        vector<int> left(N), right(N), tmp;
        for (int i = 0; i < N; ++i) {
            int j = lower_bound(begin(tmp), end(tmp), A[i]) - begin(tmp);
            if (j == tmp.size()) tmp.emplace_back();
            tmp[j] = A[i];
            left[i] = j;
        }
        tmp.clear();
        for (int i = N - 1; i >= 0; --i) {
            int j = lower_bound(begin(tmp), end(tmp), A[i]) - begin(tmp);
            if (j == tmp.size()) tmp.emplace_back();
            tmp[j] = A[i];
            right[i] = j;
        }
        for (int i = 1; i < N - 1; ++i) maxLen = max(maxLen, left[i] && right[i] ? left[i] + right[i] + 1 : 0);
        return N - maxLen;
    }
};

We don't have to store right array. Instead, we can update the answer when we compute the right values.

// OJ: https://leetcode.com/problems/minimum-number-of-removals-to-make-mountain-array/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    int minimumMountainRemovals(vector<int>& A) {
        int N = A.size(), ans = N;
        vector<int> left(N), tmp;
        for (int i = 0; i < N; ++i) {
            int j = lower_bound(begin(tmp), end(tmp), A[i]) - begin(tmp);
            if (j == tmp.size()) tmp.emplace_back();
            tmp[j] = A[i];
            left[i] = j;
        }
        tmp.clear();
        for (int i = N - 1; i >= 0; --i) {
            int j = lower_bound(begin(tmp), end(tmp), A[i]) - begin(tmp);
            if (j == tmp.size()) tmp.emplace_back();
            tmp[j] = A[i];
            if (left[i] && j) ans = min(ans, N - (left[i] + j + 1));
        }
        return ans;
    }
};