Tools: 🛩 Beginner-Friendly Guide 'Trionic Array II' - Problem 3640 (C++, Python, JavaScript)

You're given: ## Your goal: ## Intuition ## Walkthrough: Understanding the Examples ## C++ Solution ## Python Solution ## JavaScript Solution ## Key Takeaways ## Final Thoughts Navigating complex array patterns often feels like climbing a mountain range. In this problem, we are looking for a very specific "up-down-up" shape within an array to find the maximum possible sum. This challenge tests your ability to identify structural patterns while optimizing for the best numerical outcome. The core of a trionic subarray is the "valley" formed by indices and . To satisfy the condition, we need a peak at and a trough at . Example 2: `nums = [1, 4, 2, 7]` Example 1: `nums = [0, -2, -1, -3, 0, 2, -1]` Problems like "Trionic Array" are classic "pattern matching" questions often found in interviews for companies like Google or Amazon. They mimic real-world scenarios in signal processing or financial trend analysis, where a system needs to identify specific "bull" or "bear" market patterns before they complete. Mastering the ability to "walk" through an array while keeping track of multiple state changes is a vital skill for any software engineer. Templates let you quickly answer FAQs or store snippets for re-use. Are you sure you want to ? It will become hidden in your post, but will still be visible via the comment's permalink. as well , this person and/or COMMAND_BLOCK: class Solution { public: long long maxSumTrionic(vector<int>& nums) { int n = nums.size(); int i = 0; long long ans = -2e18; // Use a very small value for initialization while (i < n) { int l_bound = i; i += 1; // 1. Find the first increasing segment while (i < n && nums[i - 1] < nums[i]) { i += 1; } if (i == l_bound + 1) continue; int p = i - 1; long long current_sum = (long long)nums[p - 1] + nums[p]; // 2. Find the strictly decreasing segment while (i < n && nums[i - 1] > nums[i]) { current_sum += nums[i]; i += 1; } // Check if a valid trough 'q' exists if (i == p + 1 || i == n || nums[i - 1] == nums[i]) continue; int q = i - 1; // 3. Find the second increasing segment and its max prefix sum current_sum += nums[i]; i += 1; long long max_suffix = 0, temp_suffix = 0; while (i < n && nums[i - 1] < nums[i]) { temp_suffix += nums[i]; i += 1; max_suffix = max(max_suffix, temp_suffix); } current_sum += max_suffix; // 4. Look backward from p to find the max prefix sum for the first segment long long max_prefix = 0, temp_prefix = 0; for (int j = p - 2; j >= l_bound; j--) { temp_prefix += nums[j]; max_prefix = max(max_prefix, temp_prefix); } current_sum += max_prefix; ans = max(ans, current_sum); i = q; // Reset pointer to trough to find overlapping patterns } return ans; } }; COMMAND_BLOCK: class Solution { public: long long maxSumTrionic(vector<int>& nums) { int n = nums.size(); int i = 0; long long ans = -2e18; // Use a very small value for initialization while (i < n) { int l_bound = i; i += 1; // 1. Find the first increasing segment while (i < n && nums[i - 1] < nums[i]) { i += 1; } if (i == l_bound + 1) continue; int p = i - 1; long long current_sum = (long long)nums[p - 1] + nums[p]; // 2. Find the strictly decreasing segment while (i < n && nums[i - 1] > nums[i]) { current_sum += nums[i]; i += 1; } // Check if a valid trough 'q' exists if (i == p + 1 || i == n || nums[i - 1] == nums[i]) continue; int q = i - 1; // 3. Find the second increasing segment and its max prefix sum current_sum += nums[i]; i += 1; long long max_suffix = 0, temp_suffix = 0; while (i < n && nums[i - 1] < nums[i]) { temp_suffix += nums[i]; i += 1; max_suffix = max(max_suffix, temp_suffix); } current_sum += max_suffix; // 4. Look backward from p to find the max prefix sum for the first segment long long max_prefix = 0, temp_prefix = 0; for (int j = p - 2; j >= l_bound; j--) { temp_prefix += nums[j]; max_prefix = max(max_prefix, temp_prefix); } current_sum += max_prefix; ans = max(ans, current_sum); i = q; // Reset pointer to trough to find overlapping patterns } return ans; } }; COMMAND_BLOCK: class Solution { public: long long maxSumTrionic(vector<int>& nums) { int n = nums.size(); int i = 0; long long ans = -2e18; // Use a very small value for initialization while (i < n) { int l_bound = i; i += 1; // 1. Find the first increasing segment while (i < n && nums[i - 1] < nums[i]) { i += 1; } if (i == l_bound + 1) continue; int p = i - 1; long long current_sum = (long long)nums[p - 1] + nums[p]; // 2. Find the strictly decreasing segment while (i < n && nums[i - 1] > nums[i]) { current_sum += nums[i]; i += 1; } // Check if a valid trough 'q' exists if (i == p + 1 || i == n || nums[i - 1] == nums[i]) continue; int q = i - 1; // 3. Find the second increasing segment and its max prefix sum current_sum += nums[i]; i += 1; long long max_suffix = 0, temp_suffix = 0; while (i < n && nums[i - 1] < nums[i]) { temp_suffix += nums[i]; i += 1; max_suffix = max(max_suffix, temp_suffix); } current_sum += max_suffix; // 4. Look backward from p to find the max prefix sum for the first segment long long max_prefix = 0, temp_prefix = 0; for (int j = p - 2; j >= l_bound; j--) { temp_prefix += nums[j]; max_prefix = max(max_prefix, temp_prefix); } current_sum += max_prefix; ans = max(ans, current_sum); i = q; // Reset pointer to trough to find overlapping patterns } return ans; } }; COMMAND_BLOCK: class Solution: def maxSumTrionic(self, nums: list[int]) -> int: n = len(nums) i = 0 ans = float('-inf') while i < n: l_bound = i i += 1 # 1. Find the first increasing segment while i < n and nums[i - 1] < nums[i]: i += 1 if i == l_bound + 1: continue p = i - 1 current_sum = nums[p - 1] + nums[p] # 2. Find the strictly decreasing segment while i < n and nums[i - 1] > nums[i]: current_sum += nums[i] i += 1 if i == p + 1 or i == n or nums[i - 1] == nums[i]: continue q = i - 1 # 3. Find the second increasing segment current_sum += nums[i] i += 1 max_suffix, temp_suffix = 0, 0 while i < n and nums[i - 1] < nums[i]: temp_suffix += nums[i] i += 1 max_suffix = max(max_suffix, temp_suffix) current_sum += max_suffix # 4. Find max prefix backward max_prefix, temp_prefix = 0, 0 for j in range(p - 2, l_bound - 1, -1): temp_prefix += nums[j] max_prefix = max(max_prefix, temp_prefix) current_sum += max_prefix ans = max(ans, current_sum) i = q return ans COMMAND_BLOCK: class Solution: def maxSumTrionic(self, nums: list[int]) -> int: n = len(nums) i = 0 ans = float('-inf') while i < n: l_bound = i i += 1 # 1. Find the first increasing segment while i < n and nums[i - 1] < nums[i]: i += 1 if i == l_bound + 1: continue p = i - 1 current_sum = nums[p - 1] + nums[p] # 2. Find the strictly decreasing segment while i < n and nums[i - 1] > nums[i]: current_sum += nums[i] i += 1 if i == p + 1 or i == n or nums[i - 1] == nums[i]: continue q = i - 1 # 3. Find the second increasing segment current_sum += nums[i] i += 1 max_suffix, temp_suffix = 0, 0 while i < n and nums[i - 1] < nums[i]: temp_suffix += nums[i] i += 1 max_suffix = max(max_suffix, temp_suffix) current_sum += max_suffix # 4. Find max prefix backward max_prefix, temp_prefix = 0, 0 for j in range(p - 2, l_bound - 1, -1): temp_prefix += nums[j] max_prefix = max(max_prefix, temp_prefix) current_sum += max_prefix ans = max(ans, current_sum) i = q return ans COMMAND_BLOCK: class Solution: def maxSumTrionic(self, nums: list[int]) -> int: n = len(nums) i = 0 ans = float('-inf') while i < n: l_bound = i i += 1 # 1. Find the first increasing segment while i < n and nums[i - 1] < nums[i]: i += 1 if i == l_bound + 1: continue p = i - 1 current_sum = nums[p - 1] + nums[p] # 2. Find the strictly decreasing segment while i < n and nums[i - 1] > nums[i]: current_sum += nums[i] i += 1 if i == p + 1 or i == n or nums[i - 1] == nums[i]: continue q = i - 1 # 3. Find the second increasing segment current_sum += nums[i] i += 1 max_suffix, temp_suffix = 0, 0 while i < n and nums[i - 1] < nums[i]: temp_suffix += nums[i] i += 1 max_suffix = max(max_suffix, temp_suffix) current_sum += max_suffix # 4. Find max prefix backward max_prefix, temp_prefix = 0, 0 for j in range(p - 2, l_bound - 1, -1): temp_prefix += nums[j] max_prefix = max(max_prefix, temp_prefix) current_sum += max_prefix ans = max(ans, current_sum) i = q return ans COMMAND_BLOCK: /** * @param {number[]} nums * @return {number} */ var maxSumTrionic = function(nums) { const n = nums.length; let i = 0; let ans = -Infinity; while (i < n) { let lBound = i; i += 1; // 1. Increasing while (i < n && nums[i - 1] < nums[i]) { i += 1; } if (i === lBound + 1) continue; let p = i - 1; let currentSum = BigInt(nums[p - 1]) + BigInt(nums[p]); // 2. Decreasing while (i < n && nums[i - 1] > nums[i]) { currentSum += BigInt(nums[i]); i += 1; } if (i === p + 1 || i === n || nums[i - 1] === nums[i]) continue; let q = i - 1; // 3. Second Increasing currentSum += BigInt(nums[i]); i += 1; let maxSuffix = 0n, tempSuffix = 0n; while (i < n && nums[i - 1] < nums[i]) { tempSuffix += BigInt(nums[i]); i += 1; if (tempSuffix > maxSuffix) maxSuffix = tempSuffix; } currentSum += maxSuffix; // 4. Backward Prefix let maxPrefix = 0n, tempPrefix = 0n; for (let j = p - 2; j >= lBound; j--) { tempPrefix += BigInt(nums[j]); if (tempPrefix > maxPrefix) maxPrefix = tempPrefix; } currentSum += maxPrefix; if (ans === -Infinity || currentSum > ans) { ans = currentSum; } i = q; } return Number(ans); }; COMMAND_BLOCK: /** * @param {number[]} nums * @return {number} */ var maxSumTrionic = function(nums) { const n = nums.length; let i = 0; let ans = -Infinity; while (i < n) { let lBound = i; i += 1; // 1. Increasing while (i < n && nums[i - 1] < nums[i]) { i += 1; } if (i === lBound + 1) continue; let p = i - 1; let currentSum = BigInt(nums[p - 1]) + BigInt(nums[p]); // 2. Decreasing while (i < n && nums[i - 1] > nums[i]) { currentSum += BigInt(nums[i]); i += 1; } if (i === p + 1 || i === n || nums[i - 1] === nums[i]) continue; let q = i - 1; // 3. Second Increasing currentSum += BigInt(nums[i]); i += 1; let maxSuffix = 0n, tempSuffix = 0n; while (i < n && nums[i - 1] < nums[i]) { tempSuffix += BigInt(nums[i]); i += 1; if (tempSuffix > maxSuffix) maxSuffix = tempSuffix; } currentSum += maxSuffix; // 4. Backward Prefix let maxPrefix = 0n, tempPrefix = 0n; for (let j = p - 2; j >= lBound; j--) { tempPrefix += BigInt(nums[j]); if (tempPrefix > maxPrefix) maxPrefix = tempPrefix; } currentSum += maxPrefix; if (ans === -Infinity || currentSum > ans) { ans = currentSum; } i = q; } return Number(ans); }; COMMAND_BLOCK: /** * @param {number[]} nums * @return {number} */ var maxSumTrionic = function(nums) { const n = nums.length; let i = 0; let ans = -Infinity; while (i < n) { let lBound = i; i += 1; // 1. Increasing while (i < n && nums[i - 1] < nums[i]) { i += 1; } if (i === lBound + 1) continue; let p = i - 1; let currentSum = BigInt(nums[p - 1]) + BigInt(nums[p]); // 2. Decreasing while (i < n && nums[i - 1] > nums[i]) { currentSum += BigInt(nums[i]); i += 1; } if (i === p + 1 || i === n || nums[i - 1] === nums[i]) continue; let q = i - 1; // 3. Second Increasing currentSum += BigInt(nums[i]); i += 1; let maxSuffix = 0n, tempSuffix = 0n; while (i < n && nums[i - 1] < nums[i]) { tempSuffix += BigInt(nums[i]); i += 1; if (tempSuffix > maxSuffix) maxSuffix = tempSuffix; } currentSum += maxSuffix; // 4. Backward Prefix let maxPrefix = 0n, tempPrefix = 0n; for (let j = p - 2; j >= lBound; j--) { tempPrefix += BigInt(nums[j]); if (tempPrefix > maxPrefix) maxPrefix = tempPrefix; } currentSum += maxPrefix; if (ans === -Infinity || currentSum > ans) { ans = currentSum; } i = q; } return Number(ans); }; - An array of integers called nums with a length . - A requirement to find a "trionic" subarray, which follows a specific three-part sequence: strictly increasing, then strictly decreasing, then strictly increasing again. - Identify all possible trionic subarrays that meet the index requirements . - Calculate the sum of these subarrays and return the maximum sum found. - Identify the Core: We look for a segment where the numbers go up to a peak (), down to a trough (), and then back up. - The Pivot Points: The sequence between and (the decreasing part) is fixed once we identify a peak and a trough. - Greedy Expansion: To maximize the sum, we don't just take any and . We want the "prefix" before the peak and the "suffix" after the trough to be as large as possible. Since the values can be negative, we use a greedy approach to find the maximum prefix sum ending at and the maximum suffix sum starting after . - Step 1: We find an increasing sequence: . So, could be at index 1 (value 4). - Step 2: We find a decreasing sequence: . So, could be at index 2 (value 2). - Step 3: We find an increasing sequence: . So, could be at index 3 (value 7). - Verification: . The segments are , , and . - The algorithm identifies the peak at index 2 (value -1) and the trough at index 3 (value -3). - It then expands forward to include index 5 (value 2) because it continues the increasing trend. - It expands backward to index 1 (value -2) because that provides the best sum for the initial increasing part. - Two-Pointer/Sliding Window: This problem uses a variation of the sliding window technique to find specific shapes (mountains and valleys) in linear time. - Greedy Sub-problems: By using max(0, temp_sum), we effectively decide whether including extra elements helps us or hurts us. This is a mini-application of Kadane's logic within a structured pattern. - Linear Complexity: Even though there is a nested loop to look backward, the main pointer i only moves forward through the array, ensuring time complexity.