main.cpp

// Source: https://leetcode.com/problems/trionic-array-i
// Title: Trionic Array I
// Difficulty: Easy
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// You are given an integer array `nums` of length `n`.
//
// An array is **trionic** if there exist indices `0 < p < q < n-1` such that:
//
// - `nums[0...p]` is **strictly** increasing,
// - `nums[p...q]` is **strictly** decreasing,
// - `nums[q...n-1]` is **strictly** increasing.
//
// Return `true` if `nums` is trionic, otherwise return `false`.
//
// **Example 1:**
//
// ```
// Input: nums = [1,3,5,4,2,6]
// Output: true
// Explanation:
// Pick `p = 2`, `q = 4`:
// - `nums[0...2] = [1, 3, 5]` is strictly increasing (`1 < 3 < 5`).
// - `nums[2...4] = [5, 4, 2]` is strictly decreasing (`5 > 4 > 2`).
// - `nums[4...5] = [2, 6]` is strictly increasing (`2 < 6`).
// ```
//
// **Example 2:**
// Input: nums = [2,1,3]
// Output: false
// Explanation:
// There is no way to pick `p` and `q` to form the required three segments.
// ```
//
// **Constraints:**
//
// - `3 <= n <= 100`
// - `-1000 <= nums[i] <= 1000`
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

#include <vector>

using namespace std;

class Solution {
 public:
  bool isTrionic(vector<int>& nums) {
    int n = nums.size();

    // Find p
    int p;
    for (p = 0; p < n - 1 && nums[p] < nums[p + 1]; ++p);

    // Find q
    int q;
    for (q = p; q < n - 1 && nums[q] > nums[q + 1]; ++q);

    // Find r (end)
    int r;
    for (r = q; r < n - 1 && nums[r] < nums[r + 1]; ++r);
    return (0 != p) && (p != q) && (q != r) && (r == n - 1);
  }
};