Shifted Target Index — Problem Statement & Solution Guide
Problem Description
Given an integer array nums of length N, compute the sum of all its elements. The array may contain negative values, and the result should be represented as a 64‑bit signed integer. The input consists of two lines: the first line contains the integer N (1 ≤ N ≤ 10^5), and the second line contains N space‑separated integers, each in the range [-10^9, 10^9]. The output is a single integer representing the total sum of the array elements.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Shifted Target Index"
WHY DOES IT MATTER?
The "one‑pass linear scan" pattern is essential because it guarantees the lowest possible time complexity for problems that require aggregating values across an array. It also minimizes memory usage, which is critical in environments with limited resources or when processing data streams.
OPTIMIZATION CHALLENGE
The key insight is that you can compute the sum in a single traversal while using a 64‑bit accumulator, eliminating the need for auxiliary data structures or multiple passes.
REAL-WORLD CONNECTION
In real‑world engineering, this pattern is analogous to processing sensor data in real time—each new reading is immediately added to a running total without storing the entire history, enabling low‑latency analytics on edge devices.
When presenting this solution in an interview, emphasize the importance of choosing the correct data type early on and mention that the algorithm’s simplicity often leads to fewer bugs compared to more complex approaches.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to computing the sum of an integer array of up to 10^5 elements, each possibly as large as 10^9 in magnitude. A naive approach that iterates through the array once and accumulates the sum is already optimal in terms of asymptotic complexity: it runs in linear time O(N) and uses constant extra space O(1). However, the subtlety lies in the data type used for the accumulator. Since the sum of 10^5 numbers each up to 10^9 can reach 10^14, a 32‑bit signed integer would overflow, leading to incorrect results. Therefore, the algorithm must employ a 64‑bit signed integer (e.g., long long in C++/Java, long in Java, or int64 in Python) to safely hold intermediate and final sums.
In many interview settings, candidates may overlook the need for 64‑bit arithmetic or may attempt to use a more complex data structure such as a prefix sum array or a segment tree, which would add unnecessary overhead. The optimal paradigm is a single-pass linear scan with a 64‑bit accumulator, which guarantees correctness, simplicity, and the lowest possible time and space footprints. This pattern is a classic example of the "one‑pass linear scan" technique that appears frequently in coding interviews, especially when dealing with streaming data or large input sizes.
Interview Questions on This Problem
Q1How would you handle potential integer overflow when summing a large array of integers in a language like C++?
I would use a 64‑bit signed integer type such as long long to store the accumulator. Additionally, I would ensure that the input parsing also uses a 64‑bit type to avoid truncation during reading. If the language provides arbitrary‑precision integers (e.g., BigInteger in Java), I could use that for absolute safety, but for the given constraints a 64‑bit type is sufficient.
Q2In a distributed system that processes log entries, why is a single‑pass sum operation preferable over building a prefix sum array?
A single‑pass sum uses constant memory and minimal CPU cycles, which is critical in high‑throughput systems where latency matters. Building a prefix sum array would require O(N) additional memory and an extra pass to compute the array, increasing both memory usage and processing time. For real‑time analytics, the one‑pass approach is the most efficient.
Q3What edge cases would you test for this problem to ensure robustness?
I would test the following: 1) The smallest input size N=1 with a negative number to confirm handling of negatives. 2) The maximum input size N=10^5 with all values at the upper bound 10^9 to test for overflow. 3) A mix of positive and negative numbers that sum to zero to verify that the accumulator correctly handles cancellation.
Examples
Input
5 1 2 3 4 5
Output
15
Explanation: Add the numbers: 1+2+3+4+5 = 15. The sum is printed as the output.
Input
4 -1 -2 -3 -4
Output
-10
Explanation: Sum the negative numbers: (-1)+(-2)+(-3)+(-4) = -10. The result is output.
Input
6 1000000000 -1000000000 500 500 -500 -500
Output
0
Explanation: The large positive and negative values cancel each other: 1000000000 + (-1000000000) + 500 + 500 + (-500) + (-500) = 0. The final sum is 0.
Constraints
- 1 <= N <= 10^5
- -10^9 <= nums[i] <= 10^9
- The absolute value of the sum will not exceed 10^14
Optimal Approach & Strategy
Use a 64‑bit accumulator and iterate through the input once, adding each number to the accumulator as it is read. This avoids overflow and uses constant extra space.
Brute Force Approach
Read all numbers into an array and then iterate over the array, adding each element to a running total. This is a straightforward linear scan but may risk overflow if a 32‑bit integer is used.
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):
return sum(nums)function solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}Asked in Top Tech Interviews
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