Segmented Matrix Traversal — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a sequence of numerical metrics to compute a specific aggregate value. Given an array nums of length N, you must identify all elements located at even indices (0-based indexing) and sum their values. The result represents the total weight of the primary data points in the sequence.
The input consists of a single array of integers. The output is a single integer representing the sum of the elements at indices 0, 2, 4, ..., up to the largest even index less than N. If the array is empty, the sum is 0. Note that although the title references matrix traversal and the topic is Heap, this specific problem variant focuses on linear index-based aggregation as a foundational step in segmented processing algorithms.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Segmented Matrix Traversal"
WHY DOES IT MATTER?
Strided iteration forms the core of lower-level tensor operations, signal processing, downsampling, and matrix manipulation libraries like NumPy and BLAS.
OPTIMIZATION CHALLENGE
Eliminating branch mispredictions by stepping directly through indices with i += 2 rather than processing every index and using branching logic (if (i % 2 == 0)).
REAL-WORLD CONNECTION
In digital audio signal processing, stereo audio streams interleave left and right channels into a single 1D array. Extracting or processing a single audio channel requires summing or filtering elements at even or odd stride intervals.
When interviewing, explicitly state your assumptions about integer overflow risks and highlight how direct stride increments avoid conditional branch instructions at the assembly level.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
Processing sequence data through strided memory access is a fundamental pattern in algorithm design, vector computing, and real-time numerical telemetry analysis. When dealing with metric streams, data points are often segmented into interleaved channels or periodic temporal buckets. A naive approach might process the entire sequence sequentially while applying conditional branch evaluations (e.g., checking if the index modulo two equals zero) at every element. On high-throughput streaming systems, this introduces instruction overhead and branch prediction penalties inside tight loops.
To achieve maximum hardware throughput, direct pointer arithmetic or loop striding with a step size of 2 is preferred over exhaustive iteration with branching. This strided traversal leverages hardware CPU cache prefetching by maintaining predictable, monotonic memory reference patterns. Memory controllers fetch contiguous cache lines, and reading every second element still achieves high L1/L2 cache hit rates compared to random or pointer-chasing data access patterns.
From a complexity standpoint, strided linear iteration achieves O(N) temporal efficiency and O(1) spatial footprint. When extended to large-scale distributed streaming datasets (such as processing partitioned time-series buckets), segmenting the data allows parallel accumulation across chunk boundaries without requiring temporary memory allocation or heap-based buffer overhead.
Interview Questions on This Problem
Q1How does strided array access (incrementing by 2) affect CPU cache performance and instruction execution compared to inspecting every index with a modulo check?
Striding by 2 removes conditional branching instructions (i % 2 == 0) from the compiled assembly loop, reducing branch misprediction risks. While memory cache lines still load contiguous blocks of 64 bytes (fetching both even and odd elements into L1 cache), eliminating branch logic reduces clock cycles per iteration. Spatial locality remains high because adjacent target elements reside within the same prefetched cache line.
Q2If `nums` is a continuous streaming pipeline of billions of metric records split across multiple workers, how would you aggregate even-index elements in a distributed environment?
Each partition worker receives a segment along with its global starting offset index. If a partition starts at an even global index, the worker aggregates starting from index 0 with a step of 2; if it starts at an odd global index, it begins at local index 1 with a step of 2. Finally, a central reducer sums the local accumulator aggregates from all partition workers.
Q3How would you modify this algorithm if you were asked to maintain the top K maximum values present at even indices in real time?
You would combine the strided index traversal with a min-heap of maximum size K. As you step through even indices with i += 2, push nums[i] into the min-heap. If the heap size exceeds K, pop the smallest element. This maintains the top K maximum even-indexed elements with a time complexity of O(N log K) and auxiliary space of O(K).
Examples
Input
nums = [1, 2, 3, 4, 5]
Output
9
Explanation: Indices: 0, 1, 2, 3, 4. Even indices are 0, 2, 4. Elements at these indices are 1, 3, 5. Sum = 1 + 3 + 5 = 9.
Input
nums = [10, 20, 30, 40]
Output
40
Explanation: Indices: 0, 1, 2, 3. Even indices are 0, 2. Elements at these indices are 10, 30. Sum = 10 + 30 = 40.
Input
nums = [7]
Output
7
Explanation: Index 0 is even. Element is 7. Sum = 7.
Input
nums = [-5, 10, -15, 20, -25]
Output
-45
Explanation: Even indices: 0, 2, 4. Elements: -5, -15, -25. Sum = -5 + (-15) + (-25) = -45.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The sum of all even-indexed elements fits within a 64-bit integer.
Optimal Approach & Strategy
Initialize a running sum variable to zero and step through the array starting at index 0, incrementing the index pointer by 2 on each iteration. Accumulate each value directly into the running sum until reaching the end of the array.
Brute Force Approach
Iterate through every index from 0 to N-1 and use the modulo operator (i % 2 == 0) to check if the current index is even. If the condition evaluates to true, add the element at that index to a running accumulator.
Verified Code Solutions
/**
* @param {number[]} nums
* @return {number}
*/
var solve = function(nums) {
let sum = 0;
for (let i = 0; i < nums.length; i += 2) {
sum += nums[i];
}
return sum;
};class Solution {
public:
int solve(vector<int>& nums) {
int sum = 0;
for (int i = 0; i < nums.size(); i += 2) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int solve(int[] nums) {
int sum = 0;
for (int i = 0; i < nums.length; i += 2) {
sum += nums[i];
}
return sum;
}
}class Solution:
def solve(self, nums: List[int]) -> int:
return sum(nums[i] for i in range(0, len(nums), 2))/**
* @param {number[]} nums
* @return {number}
*/
var solve = function(nums) {
let sum = 0;
for (let i = 0; i < nums.length; i += 2) {
sum += nums[i];
}
return sum;
};Asked in Top Tech Interviews
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