Tome Voyage Tracker 28 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing tome and voyage metrics, construct an optimal algorithm to evaluate and compute the target tracker value under given operational constraints. The algorithm should consider all possible combinations of 3 elements in each window and select the maximum sum.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Voyage Tracker 28"
WHY DOES IT MATTER?
Sliding‑window patterns turn repeated work into constant‑time updates.
OPTIMIZATION CHALLENGE
The key is to reduce per‑window computation from O(k) to O(1) where k is the window size.
REAL-WORLD CONNECTION
Network throughput monitors often compute moving averages over fixed‑size intervals using the same technique.
Keep a running sum variable and update it in‑place; avoid creating new sub‑arrays inside the loop.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The naive solution enumerates every combination of three indices in the array, leading to O(n³) time, which quickly becomes infeasible for large n. By recognizing that the problem only asks for the maximum sum of any three consecutive elements, we can transform it into a sliding‑window problem where each window’s sum is derived from the previous one in O(1) time, yielding an overall linear algorithm. The optimal paradigm leverages prefix sums or a rolling sum to avoid recomputation, exploiting the overlapping nature of consecutive windows. This reduction from cubic to linear time is the hallmark of efficient array‑processing techniques used throughout competitive programming and system design.
Interview Questions on This Problem
Q1How would you compute the maximum sum of any three consecutive elements in a single pass?
Initialize the sum of the first three elements, then slide the window by subtracting the element exiting and adding the new one. Update the maximum after each slide.
Q2What edge cases must you handle when the input length is less than three?
If the array has fewer than three elements, the problem may be undefined or require returning a sentinel value. You should explicitly check the length before processing.
Q3Why is a sliding‑window approach preferable to recomputing each window’s sum from scratch?
Recomputing each sum costs O(3) per window, which multiplies to O(n) but with a higher constant factor. Sliding the window reuses previous work, giving true O(n) with minimal overhead.
Examples
Input
[1, 2, 3, 4, 5]
Output
9
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5], we need to find the maximum sum of all possible combinations of 3 elements. We can achieve this by considering all possible combinations of 3 elements in each window and selecting the maximum sum. For the input [1, 2, 3, 4, 5], the possible combinations are [1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], [2, 3, 4], [2, 3, 5], and [2, 4, 5]. The maximum sum is 9 (3+4+2).
Input
[3, 4, 2, 1, 5]
Output
9
Explanation: Step-by-step: Given the input [3, 4, 2, 1, 5], we need to find the maximum sum of all possible combinations of 3 elements. We can achieve this by considering all possible combinations of 3 elements in each window and selecting the maximum sum. For the input [3, 4, 2, 1, 5], the possible combinations are [3, 4, 2], [3, 4, 1], [3, 4, 5], [3, 2, 1], [3, 2, 5], [3, 1, 5], [4, 2, 1], [4, 2, 5], and [4, 1, 5]. The maximum sum is 9 (3+4+2).
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a sliding window of size three, updating the sum in O(1) per step and tracking the max, achieving O(n) time.
Brute Force Approach
Iterate over all i < j < k triples and compute each sum, keeping the maximum, which is O(n³).
Verified Code Solutions
function solution(nums) {
let maxSum = -Infinity;
for (let i = 0; i <= nums.length - 3; i++) {
for (let j = i + 1; j <= nums.length - 2; j++) {
for (let k = j + 1; k <= nums.length - 1; k++) {
maxSum = Math.max(maxSum, nums[i] + nums[j] + nums[k]);
}
}
}
return maxSum;
}class Solution {
public:
int solution(vector<int> nums) {
int max_sum = INT_MIN;
for (int i = 0; i <= nums.size() - 3; i++) {
for (int j = i + 1; j <= nums.size() - 2; j++) {
for (int k = j + 1; k <= nums.size() - 1; k++) {
max_sum = max(max_sum, nums[i] + nums[j] + nums[k]);
}
}
}
return max_sum;
}
};class Solution {
public int solution(int[] nums) {
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i <= nums.length - 3; i++) {
for (int j = i + 1; j <= nums.length - 2; j++) {
for (int k = j + 1; k <= nums.length - 1; k++) {
maxSum = Math.max(maxSum, nums[i] + nums[j] + nums[k]);
}
}
}
return maxSum;
}
}def solution(nums):
max_sum = float('-inf')
for i in range(len(nums) - 2):
for j in range(i + 1, len(nums) - 1):
for k in range(j + 1, len(nums)):
max_sum = max(max_sum, nums[i] + nums[j] + nums[k])
return max_sumfunction solution(nums) {
let maxSum = -Infinity;
for (let i = 0; i <= nums.length - 3; i++) {
for (let j = i + 1; j <= nums.length - 2; j++) {
for (let k = j + 1; k <= nums.length - 1; k++) {
maxSum = Math.max(maxSum, nums[i] + nums[j] + nums[k]);
}
}
}
return maxSum;
}Asked in Top Tech Interviews
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