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 hide this comment? It will become hidden in your post, but will still be visible via the comment's permalink. Hide child comments as well For further actions, you may consider blocking this person and/or reporting abuse 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; } }; Enter fullscreen mode Exit fullscreen mode 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 Enter fullscreen mode Exit fullscreen mode 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); }; Enter fullscreen mode Exit fullscreen mode 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.