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 :



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                        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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