Minimum Removals to Make Mountain List - Microsoft Top Interview Questions

Problem Statement :

You are given a list of integers nums. 

Return the minimum number of elements we can remove from nums such that the list becomes a mountain list.

A mountain list is a list that is first strictly increasing and then strictly decreasing. 

Each of the increasing and decreasing parts should be non-empty. You can assume that a solution exists.


n ≤ 100,000 where n is the length of nums

Example 1


nums = [1, 6, 5, 7, 4]




We can remove 6 to get [1, 5, 7, 4]

Solution :


                        Solution in C++ :

int solve(vector<int>& nums) {
    int n = nums.size();

    vector<int> lis;
    vector<int> dp(n);
    int idx = 0;
    for (auto i : nums) {
        auto it = lower_bound(lis.begin(), lis.end(), i);
        if (it == lis.end())
            *it = i;
        dp[idx++] = lis.size();

    for (auto i : dp) cout << i << " ";
    cout << endl;

    int ans = INT_MAX;
    idx = 0;
    reverse(nums.begin(), nums.end());
    for (auto i : nums) {
        auto it = lower_bound(lis.begin(), lis.end(), i);
        if (it == lis.end())
            *it = i;

        if (dp[n - idx - 1] > 1 && lis.size() > 1) {
            ans = min(ans, n - (int)lis.size() - dp[n - idx - 1] + 1);

    return ans == INT_MAX ? -1 : ans;

                        Solution in Python : 
class Solution:
    def solve(self, nums):
        n = len(nums)

        def lis(nums):
            lengths = []
            mono_list = [math.inf]
            for num in nums:
                index = bisect_left(mono_list, num)
                if index == len(mono_list):
                    mono_list[index] = num
                lengths.append(len(mono_list) if index > 0 else -math.inf)
            return lengths

        increasing = lis(nums)
        decreasing = lis(reversed(nums))[::-1]

        return n - max(increasing[i] + decreasing[i] - 1 for i in range(1, n - 1))

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