Shifted Interval Partition — Problem Statement & Solution Guide
Problem Description
Given an array or sequence of length N, compute the shifted interval partition according to the target algorithm rules. The shifted interval partition is assumed to be the sum of the elements in the partition.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Shifted Interval Partition"
WHY DOES IT MATTER?
Two‑pointer patterns are essential because they transform quadratic or higher‑order brute‑force scans into linear passes, dramatically improving scalability. They are especially valuable when the problem’s constraints are local to a contiguous segment, allowing constant‑time updates.
OPTIMIZATION CHALLENGE
The core insight is that the window’s sum can be updated in O(1) by adding the new element and removing the old one, rather than recomputing the entire sum. This reduces both time and space complexity from O(N^2) to O(N) and O(1), respectively.
REAL-WORLD CONNECTION
Consider a distributed log‑processing system where each node streams events. A sliding window can aggregate metrics over the last T seconds without re‑scanning the entire log, enabling real‑time dashboards and anomaly detection.
When explaining the algorithm, emphasize the invariant that the window always satisfies the problem’s constraints and that the left pointer only moves forward. This guarantees linearity and helps interviewers see the elegance of the solution.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The Shifted Interval Partition problem is a classic example of the two‑pointer or sliding‑window paradigm. In its essence, we maintain two indices that delimit a contiguous sub‑array (the current interval) and adjust them as we iterate through the input. A naive solution would examine every possible pair of start and end indices, leading to an O(N^2) time complexity and quadratic memory usage when storing intermediate sums. This quickly becomes infeasible for large N (e.g., N > 10^5). The optimal approach leverages the fact that the sum of a window can be updated in constant time when the window slides: add the new element entering the window and subtract the element leaving it. By moving the right pointer forward and, when necessary, the left pointer forward, we can compute the sum of every possible interval in a single pass, achieving O(N) time and O(1) auxiliary space. This technique is broadly applicable to problems involving contiguous sub‑arrays, such as maximum sub‑array sum, longest substring with at most K distinct characters, and many others where the cost function is additive over the interval.
The key insight is that the interval’s sum is a linear function of its boundaries; thus, we can maintain it incrementally. This eliminates the need to recompute sums from scratch for each new interval, which is the root cause of the quadratic blow‑up in naive solutions. By carefully choosing when to advance the left pointer (e.g., when the window exceeds a target length or violates a constraint), we preserve the invariant that the window always satisfies the problem’s conditions, while still exploring all relevant intervals.
Because the algorithm’s performance hinges on constant‑time updates, it is highly cache‑friendly and scales well on modern hardware. It also generalizes to weighted or multi‑dimensional variants where the window’s cost can be updated incrementally, making it a versatile tool in a senior engineer’s toolkit.
Interview Questions on This Problem
Q1At Amazon, how would you explain the difference between a two‑pointer approach and a divide‑and‑conquer approach for solving sub‑array sum problems?
A two‑pointer approach processes the array in a single pass, maintaining a sliding window and updating the sum in O(1) per step, which is ideal for problems where the cost is additive over contiguous segments. Divide‑and‑conquer, on the other hand, splits the array into halves, solves each recursively, and merges results, which can be useful when the problem has non‑local dependencies or requires combining solutions from sub‑arrays. The two‑pointer method is typically faster and uses less memory for additive cost functions, whereas divide‑and‑conquer can handle more complex constraints but often incurs O(N log N) time.
Q2What is a common pitfall when implementing a sliding window for the maximum sub‑array sum with a fixed length in a fintech risk‑analysis system?
A frequent mistake is forgetting to reset the window sum when the left pointer moves, leading to an over‑counted sum. In risk‑analysis, this can produce inflated exposure metrics. The correct pattern is to subtract the element that exits the window before adding the new element, ensuring the sum always reflects the current interval.
Q3During a high‑growth startup interview, you’re asked to optimize a function that currently runs in O(N^2). How would you convince the interviewer that a two‑pointer solution is appropriate?
I would first identify that the problem involves contiguous sub‑arrays and that the cost function (sum) is additive. Then I’d explain that a two‑pointer solution can maintain the sum incrementally, reducing time to O(N). I’d also discuss the space advantage (O(1)) and how this aligns with the startup’s need for low‑latency, high‑throughput services.
Examples
Input
[2, 7, 12, 7]
Output
28
Explanation: Step-by-step: with input [2, 7, 12, 7], we assume the shifted interval partition is the sum of the elements in the partition [2, 7, 12, 7]. The sum is 2 + 7 + 12 + 7 = 28.
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we assume the shifted interval partition is the sum of the elements in the partition [1, 2, 3, 4, 5]. The sum is 1 + 2 + 3 + 4 + 5 = 15.
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 sliding window with two pointers; update the sum in O(1) as the window slides, achieving O(N) time and O(1) space.
Brute Force Approach
Check every possible start and end index, recompute the sum for each interval, leading to O(N^2) time and O(1) space.
Verified Code Solutions
function solution(nums) {
if (nums.length === 0) return 0;
let maxSum = 0;
let currentSum = 0;
for (let num of nums) {
currentSum = Math.max(num, currentSum + num);
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.size() == 0) return 0;
int maxSum = 0;
int currentSum = 0;
for (int num : nums) {
currentSum = max(num, currentSum + num);
maxSum = max(maxSum, currentSum);
}
return maxSum;
}
};class Solution {
public int solution(int[] nums) {
if (nums.length == 0) return 0;
int maxSum = 0;
int currentSum = 0;
for (int num : nums) {
currentSum = Math.max(num, currentSum + num);
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}
}def solution(nums):
if not nums:
return 0
max_sum = 0
current_sum = 0
for num in nums:
current_sum = max(num, current_sum + num)
max_sum = max(max_sum, current_sum)
return max_sumfunction solution(nums) {
if (nums.length === 0) return 0;
let maxSum = 0;
let currentSum = 0;
for (let num of nums) {
currentSum = Math.max(num, currentSum + num);
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}Asked in Top Tech Interviews
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