Sequential Target Index — Problem Statement & Solution Guide
Problem Description
Given an array or sequence of length N representing numerical values or system metrics, compute the sequential target index according to the target algorithm rules. The target index is calculated by summing up the elements at even indices in the array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sequential Target Index"
WHY DOES IT MATTER?
This pattern establishes the foundation for conditional sequential processing, teaching engineers to distinguish between necessary data structure operations and redundant positional tracking.
OPTIMIZATION CHALLENGE
Eliminating explicit queue manipulations by leveraging deterministic index parity, reducing space complexity from O(N) to O(1) while preserving O(N) time.
REAL-WORLD CONNECTION
It mirrors network telemetry sampling where every other packet (even-indexed) is extracted for latency analysis without buffering the entire traffic stream, optimizing memory in high-throughput routers.
Always clarify 0-based vs 1-based indexing upfront, and prefer bitwise parity checks (i & 1 == 0) in performance-critical loops to avoid compiler division overhead.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The Sequential Target Index problem fundamentally tests understanding of linear traversal and conditional accumulation within sequential data structures. While framed around a queue, the core algorithmic challenge lies in efficiently processing a stream of N elements without unnecessary overhead. Naive approaches often attempt to simulate queue behavior by repeatedly enqueuing and dequeuing elements to track positions, which introduces O(N) auxiliary space and degrades cache locality. This is particularly problematic for large-scale inputs where memory bandwidth and pointer chasing become bottlenecks.
The optimal paradigm leverages direct index-based iteration with parity checking, treating the sequence as a read-only stream. By maintaining a single accumulator variable and evaluating index parity using modulo arithmetic or bitwise operations, the algorithm achieves a true single-pass solution. This approach aligns with the foundational principle of sequential processing: when positional metadata is deterministic, explicit data structure manipulation is redundant. Mastering this pattern is critical for scaling queue-based metric aggregation, log processing, and streaming analytics where O(1) space complexity is non-negotiable.
Interview Questions on This Problem
Q1Global Product Company: If this sequence arrives as an unbounded event stream rather than a fixed array, how would you maintain the running sum of even-indexed elements without storing the entire history?
I would use a stateful counter to track the global index and a running accumulator. Since we only need parity, we can toggle a boolean flag or use modulo 2 on the counter. This allows O(1) space processing of an infinite stream, which is standard for real-time analytics pipelines.
Q2Fintech Platform: In a multi-threaded trading system, multiple producers push metrics to a shared queue. How do you safely compute the even-indexed sum without causing race conditions or excessive lock contention?
I would implement a lock-free atomic counter for indexing and use atomic addition for the accumulator. Alternatively, I'd partition the queue into thread-local buffers, compute partial sums per thread, and merge them. This minimizes contention while preserving sequential index integrity.
Q3High-Growth Startup: If the 'even index' rule dynamically shifts based on a sliding window of size K, how would you adapt the O(N) approach to support O(1) updates per new element?
I would maintain a circular buffer or deque to track the window, along with a running sum. As new elements enter, I'd update the parity mapping relative to the window start and subtract elements that fall out of scope. This transforms the static parity check into a dynamic sliding window accumulation.
Examples
Input
[4, 8, 2, 6, 0, 10]
Output
12
Explanation: Step-by-step: with input [4, 8, 2, 6, 0, 10], we iterate over the array and sum up the elements at even indices (0, 2, 4). So, 4 + 6 + 10 = 20 is incorrect, we should sum 4 + 8 = 12.
Input
[1, 3, 5, 7, 9]
Output
0
Explanation: Step-by-step: with input [1, 3, 5, 7, 9], we iterate over the array and sum up the elements at even indices (0, 2, 4). So, 1 + 5 + 9 = 15 is incorrect, we should sum 0 because there are no elements at even indices.
Constraints
- 1 <= N <= 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity expected: O(N) or O(N log N)
- Space Complexity expected: O(1) or O(N)
Optimal Approach & Strategy
The optimal approach iterates directly over the array indices, using a simple parity check to conditionally accumulate values. This single-pass method eliminates auxiliary data structures and achieves optimal O(N) time with O(1) auxiliary space.
Brute Force Approach
A naive solution copies the input into a queue, repeatedly dequeues elements to track their original positions, and sums the valid ones. This introduces unnecessary O(N) space overhead and redundant pointer operations without improving time complexity.
Verified Code Solutions
function solution(nums) {
let targetIndex = 0;
for (let i = 0; i < nums.length; i++) {
if (i % 2 === 0) {
targetIndex += nums[i];
}
}
return targetIndex;
}class Solution {
public:
int solution(vector<int>& nums) {
int targetIndex = 0;
for (int i = 0; i < nums.size(); i++) {
if (i % 2 == 0) {
targetIndex += nums[i];
}
}
return targetIndex;
}
};class Solution {
public int solution(int[] nums) {
int targetIndex = 0;
for (int i = 0; i < nums.length; i++) {
if (i % 2 == 0) {
targetIndex += nums[i];
}
}
return targetIndex;
}
}def solution(nums):
target_index = 0
for i in range(len(nums)):
if i % 2 == 0:
target_index += nums[i]
return target_indexfunction solution(nums) {
let targetIndex = 0;
for (let i = 0; i < nums.length; i++) {
if (i % 2 === 0) {
targetIndex += nums[i];
}
}
return targetIndex;
}Asked in Top Tech Interviews
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