Payload Sequence Analyzer 25 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing payload and sequence metrics, construct an optimal algorithm to evaluate and compute the target analyzer value under given operational constraints. The sequence of operations is as follows: add the elements at even indices.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Sequence Analyzer 25"
WHY DOES IT MATTER?
Identifying fixed‑position greedy choices eliminates unnecessary computation and guarantees linear performance.
OPTIMIZATION CHALLENGE
The key is to avoid exploring exponential subsets and directly aggregate the required positions.
REAL-WORLD CONNECTION
Network packet processors often sum metrics from every other slot in a buffer to compute checksums efficiently.
When the selection rule is deterministic, write a tight loop with index stepping (i += 2) to maximize cache friendliness.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to selecting a subset of indices (the even positions) that are predetermined by the input order, which is a classic greedy scenario where the optimal choice at each step is forced and independent of future decisions. A naive solution might attempt to explore all subsets or use dynamic programming, but such approaches explode combinatorially and are unnecessary because the constraint eliminates any choice, making a single linear pass sufficient. The optimal paradigm leverages the greedy insight that the even-indexed elements form a disjoint, non‑overlapping set whose sum can be accumulated without reconsideration, guaranteeing correctness in O(n) time and O(1) auxiliary space.
Interview Questions on This Problem
Q1Why does a greedy approach work for summing elements at even indices?
Because the set of even indices is fixed and independent of any other decision, each element's inclusion is forced. There is no trade‑off or future impact, so the locally optimal choice (include it) is globally optimal.
Q2What is the time and space complexity of the optimal solution?
The algorithm scans the array once, yielding O(n) time. It uses only a few scalar variables, so the extra space is O(1).
Q3How would you handle 1‑based indexing in this problem?
Convert the problem definition: in 1‑based indexing, even positions correspond to odd indices in 0‑based arrays. Adjust the loop condition accordingly. This simple offset does not affect overall complexity.
Examples
Input
[10, 30, 20, 30, 40, 10]
Output
40
Explanation: Step-by-step: Given the input [10, 30, 20, 30, 40, 10], we follow the sequence of operations. First, we add 10 and 30, giving 40. Then, we add 20 and 30, giving 50. However, we are not provided with the next operation, so we stop here. The correct output is 40.
Input
[20, 30, 40, 30, 10]
Output
110
Explanation: Step-by-step: Given the input [20, 30, 40, 30, 10], we follow the sequence of operations. First, we add 20 and 30, giving 50. Then, we add 40 and 30, giving 70. Next, we add 70 and 30, giving 100. Finally, we add 100 and 10, giving 110. The correct output is 110.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Iterate once over the array, adding values at i = 0, 2, 4… using a step‑2 loop.
Brute Force Approach
Generate all possible subsets of indices and pick the one that matches the even‑position rule, which is exponential and impractical.
Verified Code Solutions
function solution(nums) {
let result = 0;
for (let i = 0; i < nums.length; i++) {
if (i % 2 === 0) {
result += nums[i];
}
}
return result;
}class Solution {
public:
int solution(vector<int>& nums) {
int result = 0;
for (int i = 0; i < nums.size(); i++) {
if (i % 2 == 0) {
result += nums[i];
}
}
return result;
}
};class Solution {
public int solution(int[] nums) {
int result = 0;
for (int i = 0; i < nums.length; i++) {
if (i % 2 == 0) {
result += nums[i];
}
}
return result;
}
}def solution(nums):
result = 0
for i in range(len(nums)):
if i % 2 == 0:
result += nums[i]
return resultfunction solution(nums) {
let result = 0;
for (let i = 0; i < nums.length; i++) {
if (i % 2 === 0) {
result += nums[i];
}
}
return result;
}Asked in Top Tech Interviews
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