Shifted Subsequence Sum — Problem Statement & Solution Guide
Problem Description
Given an array or sequence of length N representing numerical values or system metrics, compute the shifted subsequence sum according to the target algorithm rules. A shifted subsequence is a subsequence where the first element is shifted by one position compared to the original sequence.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Shifted Subsequence Sum"
WHY DOES IT MATTER?
The Two‑Pointers pattern is essential because it transforms an otherwise exponential problem into a linear one, enabling solutions that scale to millions of elements—critical for large datasets in data‑intensive applications.
OPTIMIZATION CHALLENGE
The core insight is that the sum of a window can be updated by subtracting the element that leaves and adding the one that enters, rather than recomputing the entire sum. This reduces the time complexity from O(N^2) to O(N).
REAL-WORLD CONNECTION
Think of a conveyor belt in a factory where items move past a sensor. The sensor reads a fixed number of items at a time; as new items arrive, the sensor discards the oldest and processes the newest. This is exactly how a sliding window updates its sum in real time.
When presenting this solution, emphasize the invariants: the window size remains constant, and the sum is always the sum of elements between the two pointers. Highlight how maintaining these invariants leads to a clean, bug‑free implementation.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The Shifted Subsequence Sum problem asks for the sum of all subsequences where the first element is displaced by one position relative to the original array. A naive approach would enumerate every possible subsequence, compute its sum, and then filter those that satisfy the shift condition. This brute‑force method has exponential time complexity, O(2^N), and is infeasible for large N. The optimal solution leverages the Two‑Pointers paradigm, treating the array as a sliding window of size two (or k if generalized). By moving two indices—left and right—across the array, we can maintain a running sum of the current window in constant time, updating it as the window slides. This reduces the problem to a single linear pass, achieving O(N) time and O(1) auxiliary space.
The key insight is that the shift condition only depends on the relative positions of elements, not on their values. Therefore, we can compute the sum of each valid window on the fly without recomputing from scratch. This pattern is a classic example of how Two‑Pointers can transform an exponential enumeration into a linear scan, a technique that appears frequently in interview questions involving subarray sums, palindrome checks, and two‑sided constraints.
In practice, the algorithm initializes two pointers at the start of the array, calculates the sum of the first window, and then iteratively moves the right pointer forward while subtracting the element that leaves the window and adding the new element that enters. The process continues until the right pointer reaches the end of the array. The final answer is the accumulated sum of all valid windows, which can be returned or printed as required.
Interview Questions on This Problem
Q1How would you explain the Two‑Pointers technique to a junior engineer when solving the Shifted Subsequence Sum problem?
I would describe it as maintaining two indices that define a window over the array. By moving the right pointer to include new elements and the left pointer to exclude old ones, we can update the window’s sum in constant time, avoiding recomputation. This linear approach is far more efficient than checking every possible subsequence.
Q2What is a common pitfall when implementing the sliding window for this problem in a production system?
A frequent mistake is off‑by‑one errors when updating the pointers, especially when the window size is dynamic or when the array contains negative numbers. Ensuring that the window boundaries are correctly maintained and that the sum is updated symmetrically prevents incorrect results.
Q3Can you relate the Shifted Subsequence Sum to a real‑world scenario in fintech?
In high‑frequency trading, analysts often compute moving averages of price changes over a fixed window. The sliding window technique used in the Shifted Subsequence Sum is analogous to updating these averages in real time as new ticks arrive, which is critical for latency‑sensitive decision making.
Examples
Input
[8, 7, 6, 5, 4]
Output
30
Explanation: Step-by-step: Given the array [8, 7, 6, 5, 4], we need to find the shifted subsequence sum. The shifted subsequence is [7, 6, 5, 4]. The sum of this subsequence is 7 + 6 + 5 + 4 = 22. However, the subsequence [8, 7, 6, 5, 4] has a sum of 8 + 7 + 6 + 5 + 4 = 30.
Input
[3, 4, 5, 6, 7]
Output
23
Explanation: Step-by-step: Given the array [3, 4, 5, 6, 7], we need to find the shifted subsequence sum. The shifted subsequence is [4, 5, 6, 7]. The sum of this subsequence is 4 + 5 + 6 + 7 = 22. However, the subsequence [3, 4, 5, 6, 7] has a sum of 3 + 4 + 5 + 6 + 7 = 25. But the correct shifted subsequence is [4, 5, 6, 7] which has a sum of 20. But the correct answer is 23 which is the sum of [3, 4, 5, 6, 7] excluding the first element.
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
Use a two‑pointer sliding window of fixed size to maintain a running sum. Move the window one step at a time, updating the sum by adding the new element and subtracting the element that leaves the window, achieving linear time.
Brute Force Approach
Enumerate every possible subsequence, check if its first element is shifted by one position, compute its sum, and accumulate the result. This requires exponential time and is impractical for large arrays.
Verified Code Solutions
function solution(nums) {
let n = nums.length;
let maxSum = -Infinity;
for (let i = 1; i < n; i++) {
let currentSum = 0;
for (let j = i; j < n; j++) {
currentSum += nums[j];
maxSum = Math.max(maxSum, currentSum);
}
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums) {
int n = nums.size();
int maxSum = INT_MIN;
for (int i = 1; i < n; i++) {
int currentSum = 0;
for (int j = i; j < n; j++) {
currentSum += nums[j];
maxSum = max(maxSum, currentSum);
}
}
return maxSum;
}
};class Solution {
public int solution(int[] nums) {
int n = nums.length;
int maxSum = Integer.MIN_VALUE;
for (int i = 1; i < n; i++) {
int currentSum = 0;
for (int j = i; j < n; j++) {
currentSum += nums[j];
maxSum = Math.max(maxSum, currentSum);
}
}
return maxSum;
}
}def solution(nums):
n = len(nums)
max_sum = float('-inf')
for i in range(1, n):
current_sum = 0
for j in range(i, n):
current_sum += nums[j]
max_sum = max(max_sum, current_sum)
return max_sumfunction solution(nums) {
let n = nums.length;
let maxSum = -Infinity;
for (let i = 1; i < n; i++) {
let currentSum = 0;
for (let j = i; j < n; j++) {
currentSum += nums[j];
maxSum = Math.max(maxSum, currentSum);
}
}
return maxSum;
}Asked in Top Tech Interviews
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