Sequential Subsequence Sum — Problem Statement & Solution Guide
Problem Description
Given an array or sequence of length $N$ representing numerical values or system metrics, compute the sequential subsequence sum according to the target algorithm rules.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sequential Subsequence Sum"
WHY DOES IT MATTER?
Sliding‑window patterns appear in rate‑limiting, moving‑average calculations, real‑time analytics, and any scenario where you need a summary of the most recent K events without re‑scanning the entire history.
OPTIMIZATION CHALLENGE
The key insight is that adjacent windows overlap by K‑1 elements, so you can reuse the previous window’s sum by subtracting the leftmost element and adding the new rightmost element—eliminating the need for a full recomputation.
REAL-WORLD CONNECTION
Think of a network router that tracks the total bytes transmitted over the last 5 seconds to enforce bandwidth caps. As each packet arrives, the router adds its size and discards the size of the packet that fell out of the 5‑second window, mirroring the queue‑based window sum.
When coding, keep the running sum in a separate variable and update it before you push the new element or pop the old one; this avoids off‑by‑one bugs and makes the code easier to read.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The Sequential Subsequence Sum problem asks for the sum of every contiguous subsequence (or "window") of a fixed length K within an array of N numbers. A naïve solution would recompute each window’s sum from scratch, leading to O(N·K) time, which quickly becomes prohibitive when N and K are large (e.g., N=10^6, K=10^5). The optimal paradigm leverages the sliding‑window technique: as the window moves one step to the right, we subtract the element that exits the window and add the new element that enters. This incremental update reduces the per‑step work to O(1), yielding an overall O(N) algorithm. The technique is naturally expressed with a queue (or deque) that stores the current window’s elements, allowing O(1) push‑back and pop‑front operations while maintaining the running sum.
Interview Questions on This Problem
Q1How would you compute the sum of all contiguous subarrays of size K in O(N) time?
Initialize the sum of the first K elements, then slide the window across the array: for each step, subtract the element leaving the window and add the new element entering. Store each window sum in a result list. This uses a simple loop and constant‑time updates, achieving O(N) time and O(1) extra space.
Q2Can you adapt the sliding‑window sum to handle a stream of numbers where the window size is dynamic (e.g., based on a threshold rather than a fixed K)?
Yes. Use a deque to store elements and maintain a running sum. While the sum exceeds the threshold, pop elements from the front and adjust the sum. When a new element arrives, push it to the back and add to the sum. This yields an O(N) solution for the entire stream because each element is inserted and removed at most once.
Q3Why might a candidate choose a prefix‑sum array over a sliding window for this problem, and what are the trade‑offs?
A prefix‑sum array allows O(1) query of any subarray sum after O(N) preprocessing, which is useful if many arbitrary‑range queries follow. However, it uses O(N) extra space and does not support dynamic updates (e.g., streaming data) as efficiently as the sliding‑window queue, which works in O(1) space and handles online inputs.
Examples
Input
[6, 7, 8, 9]
Output
30
Explanation: Step-by-step: The input array is [6, 7, 8, 9]. We iterate through the array and sum all the elements. The sum of the array is 6 + 7 + 8 + 9 = 30.
Input
[2, 4]
Output
6
Explanation: Step-by-step: The input array is [2, 4]. We iterate through the array and sum all the elements. The sum of the array is 2 + 4 = 6.
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
Maintain a running sum and update it by subtracting the element leaving the window and adding the new element entering, achieving O(N) time and O(1) extra space.
Brute Force Approach
Re‑calculate the sum for each window by iterating over its K elements, leading to O(N·K) time.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
sum = 0
for num in nums:
sum += num
return sumfunction solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}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.