Tome Voyage Resolver 4 — 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 resolver value under given operational constraints. The algorithm should add the sum of the first K numbers or the sum of the last K numbers depending on the problem's requirements.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Voyage Resolver 4"
WHY DOES IT MATTER?
The sliding‑window/two‑pointer pattern transforms problems that appear to need nested loops into linear‑time solutions, which is essential for handling massive data streams and real‑time analytics where latency matters.
OPTIMIZATION CHALLENGE
The key insight is that the aggregate of a window can be derived from the previous window by a constant‑time adjustment, eliminating the need to recompute from scratch.
REAL-WORLD CONNECTION
Think of a network router that needs to monitor the total traffic over the most recent 5‑second interval. Instead of recounting every packet each second, it updates a running total as packets enter and leave the time window—exactly the sliding‑window principle.
During an interview, compute the prefix sum and suffix sum in the same loop; this shows you can reason about multiple pointers simultaneously and keep the code concise.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The Two‑Pointer (or sliding‑window) technique excels when a problem asks for the sum or other aggregate over a contiguous sub‑segment of a sequence. A naive solution would recompute the sum for every possible segment, leading to O(N·K) time for a fixed window size K, which quickly becomes prohibitive for N up to 10^5 or more. By maintaining a running total as the window slides one element at a time, we can update the sum in constant time: subtract the element that leaves the window and add the new element that enters. This reduces the overall complexity to linear time while using only O(1) extra space. The same principle applies when we need the sum of the first K elements (a prefix) and the last K elements (a suffix); a single pass can compute both by keeping two pointers—one moving forward from the start and another moving backward from the end—updating two running sums in parallel.
Interview Questions on This Problem
Q1How would you compute the maximum sum obtainable by either the first K or the last K elements of an array in O(N) time?
Initialize two sums: prefixSum for the first K elements and suffixSum for the last K elements. Iterate once: for i from 0 to K‑1 add arr[i] to prefixSum, and for i from N‑1 down to N‑K add arr[i] to suffixSum. The answer is max(prefixSum, suffixSum). This runs in O(N) because K ≤ N and each element is visited at most once.
Q2Explain why a sliding‑window approach is preferable to recomputing sums for each possible K‑length segment when K is fixed.
Recomputing each segment requires O(K) work per segment, leading to O(N·K) total. A sliding window updates the sum in O(1) by removing the leftmost element and adding the new rightmost element, yielding O(N) total. This difference is critical when N is large and K is comparable to N.
Q3In a streaming context where numbers arrive one‑by‑one, how can you maintain the sum of the last K elements without storing the entire stream?
Use a circular buffer (or queue) of size K to hold the most recent K values and a running sum variable. When a new value arrives, subtract the value being evicted (if the buffer is full), add the new value to the sum, and replace the evicted slot. This maintains the last‑K sum in O(1) time and O(K) space.
Examples
Input
[10, 20, 30, 40, 50], 3
Output
180
Explanation: Step-by-step: with input [10, 20, 30, 40, 50] and K = 3, we calculate the sum of the first 3 numbers (10 + 20 + 30 = 60) and the sum of the last 3 numbers (30 + 40 + 50 = 120), then add them together to get the final result: 60 + 120 = 180
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100], 5
Output
550
Explanation: Step-by-step: with input [10, 20, 30, 40, 50, 60, 70, 80, 90, 100] and K = 5, we calculate the sum of the first 5 numbers (10 + 20 + 30 + 40 + 50 = 150) and the sum of the last 5 numbers (60 + 70 + 80 + 90 + 100 = 400), then add them together to get the final result: 150 + 400 = 550
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain two running sums while scanning the array once: one adds the first K elements, the other adds the last K elements using a reverse pointer, achieving O(N) time and O(1) extra space.
Brute Force Approach
Compute the sum of the first K elements and the sum of the last K elements by iterating K times for each, resulting in O(N·K) when K is close to N.
Verified Code Solutions
function solution(nums, k) {
let sumFirstK = 0;
let sumLastK = 0;
for (let i = 0; i < k; i++) {
sumFirstK += nums[i];
sumLastK += nums[nums.length - k + i];
}
return sumFirstK + sumLastK;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int sumFirstK = 0;
int sumLastK = 0;
for (int i = 0; i < k; i++) {
sumFirstK += nums[i];
sumLastK += nums[nums.size() - k + i];
}
return sumFirstK + sumLastK;
}
};class Solution {
public int solution(int[] nums, int k) {
int sumFirstK = 0;
int sumLastK = 0;
for (int i = 0; i < k; i++) {
sumFirstK += nums[i];
sumLastK += nums[nums.length - k + i];
}
return sumFirstK + sumLastK;
}
}def solution(nums, k):
sum_first_k = sum(nums[:k])
sum_last_k = sum(nums[-k:])
return sum_first_k + sum_last_kfunction solution(nums, k) {
let sumFirstK = 0;
let sumLastK = 0;
for (let i = 0; i < k; i++) {
sumFirstK += nums[i];
sumLastK += nums[nums.length - k + i];
}
return sumFirstK + sumLastK;
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.