# Longest Strictly Increasing Then Decreasing Sublist - Microsoft Top Interview Questions

### Problem Statement :

```You are given a list of integers nums.

Return the length of the longest sublist such that its length is at least 3 and its values are strictly increasing and then decreasing.

Both the increasing part and the decreasing part must be non-empty.

Constraints

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

Example 1

Input

nums = [7, 1, 3, 5, 2, 0]

Output

5

Explanation

The sublist [1, 3, 5, 2, 0] is strictly increasing then decreasing.

Example 2

Input

nums = [1, 2, 3]

Output

0

Example 3

Input

nums = [3, 2, 1]

Output

0

Example 4

Input

nums = [1, 2, 1, 1]

Output

3```

### Solution :

```                        ```Solution in C++ :

int solve(vector<int>& nums) {
int size = nums.size();
int res = INT_MIN;
int i = 0;
int len_inc;
int len_dec;

while (i < size - 2) {
int reference = i;
len_inc = 0;
len_dec = 0;

while (i < size - 1 and nums[i] < nums[i + 1]) {
len_inc++;
i++;
}

while (i < size - 1 and nums[i] > nums[i + 1]) {
len_dec++;
i++;
}

if (len_dec > 0 and len_inc > 0) {
res = max(res, i - reference + 1);
}

while (i < size - 1 and nums[i] == nums[i + 1]) {
i++;
}
}

return res >= 0 ? res : 0;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(int[] A) {
int res = 0;
int left[] = new int[A.length];
int right[] = new int[A.length];
left[0] = 1;
right[A.length - 1] = 1;
for (int i = 1; i < A.length; i++) {
if (A[i] > A[i - 1])
left[i] = 1 + left[i - 1];
else
left[i] = 1;
}

for (int i = right.length - 2; i >= 0; i--) {
if (A[i] > A[i + 1])
right[i] = right[i + 1] + 1;
else
right[i] = 1;
}

for (int i = 1; i < A.length - 1; i++) {
if (A[i] > A[i + 1] && left[i] > 1)
res = Math.max(res, left[i] + right[i + 1]);
}
return res;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, nums):

n, res = len(nums), 0
i, j = 0, 1
increased, decreased = False, False

while j < n:
# increasing sublist
if nums[j - 1] < nums[j]:
increased = True
while j < n and nums[j - 1] < nums[j]:
j += 1
# decreasing sublist after decreasing sublist
if j < n and increased and nums[j - 1] > nums[j]:
decreased = True
while j < n and nums[j - 1] > nums[j]:
j += 1
# j - i gives the len of the sublist that is increased first and then decreased
if increased and decreased:
res = max(res, j - i)
# shifting the pointers to the next part
i = j
j += 1
# initialising increased and decreased to False
increased, decreased = False, False

return res```
```

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