3105: Longest-Strictly-Increasing-or-Strictly-Decreasing-Subarray
Easy

table of contents

Simply just iterate through the array nums, keeping track of both ans and counter variables, which tracks the longest previous strictly monotonic subarray and the length of the current strictly monotonic subarray. When iterating through nums, keep track of the current direction (increasing, decreasing or equal), check the relationship of nums[i-1] with nums[i] and respond accordingly (e.g. if nums[i] was strictly increasing and then suddenly becomes strictly decreasing, then make ans = max(ans, counter), reset the counter variable, and change the direction variable)

code

class Solution {
public:
    int longestMonotonicSubarray(vector<int>& nums) {
        int ans = 1;
        int counter = 1;
        int direction = 0;
        for (int i = 1; i < nums.size(); ++i) {
            if (nums[i-1] != nums[i]) {
                if (direction == ((nums[i-1] < nums[i]) ? 1 : -1)) {
                    ++counter;
                } else {
                    direction = (nums[i-1] < nums[i]) ? 1 : -1;
                    ans = max(ans, counter);
                    counter = 2;
                }
            } else {
                direction = 0;
                ans = max(ans, counter);
                counter = 0;
            }
        }
        ans = max(ans, counter);
        return ans;
    }
};

complexity