3637: Trionic Array I

Problem Statement

Given an integer array nums of length n, determine if it is a trionic array.

An array is trionic if there exist indices p and q such that:

• 0 < p < q < n - 1 (all indices are valid and segments have at least 2 elements)
• The segment nums[0...p] is strictly increasing
• The segment nums[p...q] is strictly decreasing
• The segment nums[q...n-1] is strictly increasing

Return true if the array is trionic, otherwise return false.

Key Insight

The solution relies on identifying the characteristic "mountain" pattern in the array:

1. The array starts by strictly increasing to a peak (at index p)
2. Then strictly decreases to a valley (at index q)
3. Finally strictly increases again to the end

Brute Force Approach (Check All Possible p and q)

This approach checks all possible combinations of p and q indices:

Algorithm Steps

1. Iterate through all possible p indices (from 1 to n-3)
2. For each p, iterate through all possible q indices (from p+1 to n-2)
3. For each (p, q) pair:
   • Check if [0, p] is strictly increasing
   • Check if [p, q] is strictly decreasing
   • Check if [q, n-1] is strictly increasing
4. If any combination satisfies all three conditions, return true
5. If no combination works, return false

Complexity Analysis

This solution becomes inefficient for larger arrays (n ≈ 100) as it performs up to 100*100*100 = 1,000,000 operations in worst case.

class Solution {
public:
    bool isTrionic(vector<int>& nums) {
        int n = nums.size();
        if (n < 4) return false;
        
        for (int p = 1; p <= n-3; p++) {
            for (int q = p+1; q <= n-2; q++) {
                // Check first segment [0, p]
                bool valid = true;
                for (int i = 1; i <= p; i++) {
                    if (nums[i] <= nums[i-1]) {
                        valid = false;
                        break;
                    }
                }
                if (!valid) continue;
                
                // Check second segment [p, q]
                for (int i = p+1; i <= q; i++) {
                    if (nums[i] >= nums[i-1]) {
                        valid = false;
                        break;
                    }
                }
                if (!valid) continue;
                
                // Check third segment [q, n-1]
                for (int i = q+1; i < n; i++) {
                    if (nums[i] <= nums[i-1]) {
                        valid = false;
                        break;
                    }
                }
                if (valid) return true;
            }
        }
        return false;
    }
};

class Solution {
    public boolean isTrionic(int[] nums) {
        int n = nums.length;
        if (n < 4) return false;
        
        for (int p = 1; p <= n-3; p++) {
            for (int q = p+1; q <= n-2; q++) {
                // Check first segment [0, p]
                boolean valid = true;
                for (int i = 1; i <= p; i++) {
                    if (nums[i] <= nums[i-1]) {
                        valid = false;
                        break;
                    }
                }
                if (!valid) continue;
                
                // Check second segment [p, q]
                for (int i = p+1; i <= q; i++) {
                    if (nums[i] >= nums[i-1]) {
                        valid = false;
                        break;
                    }
                }
                if (!valid) continue;
                
                // Check third segment [q, n-1]
                for (int i = q+1; i < n; i++) {
                    if (nums[i] <= nums[i-1]) {
                        valid = false;
                        break;
                    }
                }
                if (valid) return true;
            }
        }
        return false;
    }
}

class Solution:
    def isTrionic(self, nums: List[int]) -> bool:
        n = len(nums)
        if n < 4:
            return False
            
        for p in range(1, n-2):
            for q in range(p+1, n-1):
                valid = True
                
                # Check first segment [0, p]
                for i in range(1, p+1):
                    if nums[i] <= nums[i-1]:
                        valid = False
                        break
                if not valid:
                    continue
                    
                # Check second segment [p, q]
                for i in range(p+1, q+1):
                    if nums[i] >= nums[i-1]:
                        valid = False
                        break
                if not valid:
                    continue
                    
                # Check third segment [q, n-1]
                for i in range(q+1, n):
                    if nums[i] <= nums[i-1]:
                        valid = False
                        break
                if valid:
                    return True
                    
        return False

Optimized Approach: Single Pass Through Array

This approach traverses the array once to find valid p and q indices:

Algorithm Steps

1. Start from index 0 and traverse while strictly increasing (find peak p)
2. From p, traverse while strictly decreasing (find valley q)
3. From q, traverse while strictly increasing to the end
4. Verify that:
   • p and q are valid (0 < p < q < n-1)
   • The entire array is traversed

Why This Works

The trionic array has a specific pattern: increase → decrease → increase. By traversing the array once, we can validate this pattern exists.

Complexity Analysis

This solution efficiently handles the maximum constraint (n=100) with a single pass through the array.

class Solution {
public:
    bool isTrionic(vector<int>& nums) {
        int n = nums.size();
        int i = 0;

        // Strictly increasing to peak p
        while (i+1 < n && nums[i] < nums[i+1]) {
            i++;
        }
        
        // Check if we have valid p (must have increased at least once)
        if (i == 0 || i >= n-2) {
            return false;
        }
        int p = i;

        // Strictly decreasing to valley q
        while (i+1 < n && nums[i] > nums[i+1]) {
            i++;
        }
        
        // Check if we have valid q (must have decreased at least once)
        if (i == p || i >= n-1) {
            return false;
        }
        int q = i;

        // Strictly increasing to the end
        while (i+1 < n && nums[i] < nums[i+1]) {
            i++;
        }
        
        // Must reach the last element
        return i == n-1;
    }
};

class Solution {
    public boolean isTrionic(int[] nums) {
        int n = nums.length;
        int i = 0;

        // Strictly increasing to peak p
        while (i+1 < n && nums[i] < nums[i+1]) {
            i++;
        }
        
        // Check valid p (must have increased at least once)
        if (i == 0 || i >= n-2) {
            return false;
        }
        int p = i;

        // Strictly decreasing to valley q
        while (i+1 < n && nums[i] > nums[i+1]) {
            i++;
        }
        
        // Check valid q (must have decreased at least once)
        if (i == p || i >= n-1) {
            return false;
        }
        int q = i;

        // Strictly increasing to the end
        while (i+1 < n && nums[i] < nums[i+1]) {
            i++;
        }
        
        // Must reach the last element
        return i == n-1;
    }
}

class Solution:
    def isTrionic(self, nums: List[int]) -> bool:
        n = len(nums)
        i = 0
        
        # Strictly increasing to peak p
        while i+1 < n and nums[i] < nums[i+1]:
            i += 1
            
        # Check valid p (must have increased at least once)
        if i == 0 or i >= n-2:
            return False
        p = i
        
        # Strictly decreasing to valley q
        while i+1 < n and nums[i] > nums[i+1]:
            i += 1
            
        # Check valid q (must have decreased at least once)
        if i == p or i >= n-1:
            return False
        q = i
        
        # Strictly increasing to the end
        while i+1 < n and nums[i] < nums[i+1]:
            i += 1
            
        # Must reach the last element
        return i == n-1

Edge Cases and Testing

1. Minimum Length Array

Input: [1,2,3] → Output: false
Explanation: Array has only 3 elements, cannot form three segments

2. Constant Values

Input: [5,5,5,5] → Output: false
Explanation: No strictly increasing or decreasing segments

3. Multiple Peaks

Input: [1,3,2,4,3] → Output: false
Explanation: Has two peaks but doesn't follow the required pattern

4. Valid Complex Pattern

Input: [1,2,5,4,3,2,1,2,3,4] → Output: true
Explanation: 
  p=2 (value 5), q=6 (value 1)
  [1,2,5] → increasing
  [5,4,3,2,1] → decreasing
  [1,2,3,4] → increasing

5. Only Increasing

Input: [1,2,3,4,5] → Output: false
Explanation: Never decreases so cannot form the middle segment

Detailed Explanation of Optimal Approach

The optimal solution works by traversing the array once and verifying the required pattern as it goes. Here's a step-by-step breakdown:

1. Initialization: Start at index 0
2. First Segment (Increasing): Move forward as long as each element is greater than the previous
3. Peak Validation: After the first segment, we must have moved at least one step (i > 0) and have at least two elements remaining (i < n-2)
4. Second Segment (Decreasing): Continue moving forward as long as each element is less than the previous
5. Valley Validation: After the second segment, we must have moved at least one step (i > p) and have at least one element remaining (i < n-1)
6. Third Segment (Increasing): Continue moving forward as long as each element is greater than the previous
7. Final Validation: We must reach the last element of the array

This approach efficiently verifies the trionic pattern in O(n) time with O(1) space by leveraging the sequential nature of the required pattern. The key insight is that the trionic array has a single peak followed by a single valley, after which it increases again.

Related Topics Explained

Array Traversal

Array traversal is fundamental to solving this problem:

• We examine elements sequentially from start to end
• At each step, we compare the current element with the next
• Different traversal patterns (single pass, nested loops) have significant performance implications

Pattern Recognition

This problem is essentially about recognizing a specific pattern in data:

• The trionic pattern has three distinct phases: up, down, up
• Real-world applications include analyzing stock prices, signal processing, and topographic data
• Pattern recognition problems often have efficient solutions when the pattern is well-defined

Algorithm Optimization

Moving from O(n³) to O(n) demonstrates important optimization principles:

• Recognize that brute force checking is inefficient
• Look for patterns that can be verified sequentially
• Leverage constraints to eliminate unnecessary checks