# 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.

Constraints

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

Example 1

Input

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

Output

1

Explanation

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())
lis.push_back(i);
else
*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());
lis.clear();
for (auto i : nums) {
auto it = lower_bound(lis.begin(), lis.end(), i);
if (it == lis.end())
lis.push_back(i);
else
*it = i;

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

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.append(num)
else:
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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