Calculated Subsequence Sum — Problem Statement & Solution Guide
Problem Description
Given an array or sequence of length N representing numerical values or system metrics, compute the calculated subsequence sum according to the target algorithm rules. The algorithm works as follows: add the first two elements of the array, then add any element greater than 6 to the sum.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Calculated Subsequence Sum"
WHY DOES IT MATTER?
Understanding this pattern teaches candidates how to isolate independent contributions in a sum, a skill that translates to many real‑world scenarios such as filtering logs, aggregating metrics, or applying business rules on streaming data.
OPTIMIZATION CHALLENGE
The key insight is that the inclusion decision for each element is stateless and based solely on its value, allowing us to avoid any auxiliary data structures or multiple passes. Recognizing this reduces the problem from exponential to linear time.
REAL-WORLD CONNECTION
In distributed monitoring systems, you often need to compute a metric like "total latency of high‑cost requests" where you always count the first two baseline measurements and then only add latencies that exceed a critical threshold. The same greedy accumulation logic applies.
During an interview, write the initialization for the sum first, then immediately loop with a clear if‑condition. Explicitly comment edge‑case handling for N < 2; this shows you think about robustness before diving into the main logic.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem asks for a specific subsequence sum that can be derived in a single linear scan. The rule—add the first two elements unconditionally, then add every subsequent element that exceeds the threshold of 6—fits the greedy paradigm because each decision (whether to include an element) is locally optimal and independent of future choices. A naive solution might generate all possible subsequences and test each against the rule, leading to exponential time, which quickly becomes infeasible for N in the order of 10^5 or larger. By recognizing that the inclusion condition depends only on the element's value and its position relative to the first two indices, we can collapse the problem to a simple accumulation, achieving optimal O(N) time.
The optimal approach leverages the fact that the sum is additive and the decision rule is monotonic: any element greater than 6 will always be part of the final sum regardless of other elements. This eliminates the need for dynamic programming or backtracking. Instead, we maintain a running total, initialize it with the first two array values (handling edge cases when N < 2), and then iterate from the third index onward, adding only those values that satisfy the >6 condition. This greedy accumulation guarantees correctness because the problem statement does not impose any ordering or dependency constraints beyond the simple threshold check.
Interview Questions on This Problem
Q1How would you compute the calculated subsequence sum in a single pass without using extra memory?
Initialize a sum with the first two elements (or handle N<2 separately). Then iterate from index 2 to N‑1, adding the current element to the sum only if it is greater than 6. This runs in O(N) time and O(1) extra space.
Q2What edge cases must you consider when implementing this algorithm for a production system?
You need to handle arrays with fewer than two elements, negative numbers (which may still be >6 if the threshold changes), elements exactly equal to 6 (they must be excluded), and potential integer overflow for very large sums, which may require using a 64‑bit integer type.
Q3Can this problem be extended to a variable threshold instead of a fixed value 6? How would the solution change?
Yes, replace the constant 6 with a parameter T. The algorithm remains identical: after adding the first two elements, add any subsequent element that is greater than T. The time and space complexities stay O(N) and O(1) respectively.
Examples
Input
[1, 2, 7, 3, 4]
Output
24
Explanation: Step-by-step: 1. Add the first two elements (1 + 2 = 3). 2. The sum is less than 6, so we add the next element greater than 6 (7) to the sum (3 + 7 = 10). 3. The sum is still less than 6, so we add the next element greater than 6 (7) to the sum (10 + 7 = 17). 4. The sum is still less than 6, so we add the next element greater than 6 (7) to the sum (17 + 7 = 24).
Input
[1, 2, 5, 3]
Output
3
Explanation: Step-by-step: 1. Add the first two elements (1 + 2 = 3). 2. The sum is greater than 6, so we do not add any elements greater than 6 to the sum.
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
Initialize sum with the first two elements, then iterate once adding only elements >6, achieving O(N) time and O(1) space.
Brute Force Approach
Generate every possible subsequence, check if it follows the rule, and compute its sum—this is exponential in N.
Verified Code Solutions
function solution(nums) {
let sum = nums[0] + nums[1];
for (let i = 2; i < nums.length; i++) {
if (nums[i] > 6 && sum < 6) {
sum += nums[i];
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = nums[0] + nums[1];
for (int i = 2; i < nums.size(); i++) {
if (nums[i] > 6 && sum < 6) {
sum += nums[i];
}
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = nums[0] + nums[1];
for (int i = 2; i < nums.length; i++) {
if (nums[i] > 6 && sum < 6) {
sum += nums[i];
}
}
return sum;
}
}def solution(nums):
sum = nums[0] + nums[1]
for i in range(2, len(nums)):
if nums[i] > 6 and sum < 6:
sum += nums[i]
return sumfunction solution(nums) {
let sum = nums[0] + nums[1];
for (let i = 2; i < nums.length; i++) {
if (nums[i] > 6 && sum < 6) {
sum += nums[i];
}
}
return sum;
}Asked in Top Tech Interviews
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