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Verified Parity Sequence Solution

Problem Statement

Given an array of integers, compute the Verified Parity Sequence. This sequence is defined as the cumulative sum of the array elements, where the final value represents the total aggregate of all inputs. The problem requires processing the array in a single pass to determine this aggregate value efficiently.

The input consists of a single array of integers. The output must be a single integer representing the sum of all elements in the array. Although the term 'parity' often refers to even/odd properties, in this specific context, it denotes the verified total sum of the sequence. The solution must handle large arrays and large integer values without overflow, assuming standard 64-bit integer arithmetic.

Your task is to implement a function that takes an array of integers and returns the sum of its elements. The implementation should be optimized for time complexity, ensuring it runs in linear time relative to the size of the input array. Space complexity should be constant, as no additional data structures are required beyond the input array and a single accumulator variable.

Example 1
Input
nums = [1, 2, 3, 4, 5]
Output
15

Explanation: Initialize sum = 0. Iterate through the array: sum = 0 + 1 = 1; sum = 1 + 2 = 3; sum = 3 + 3 = 6; sum = 6 + 4 = 10; sum = 10 + 5 = 15. The final verified parity sequence is 15.

Example 2
Input
nums = [-1, 0, 1, -2, 2]
Output
0

Explanation: Initialize sum = 0. Iterate through the array: sum = 0 + (-1) = -1; sum = -1 + 0 = -1; sum = -1 + 1 = 0; sum = 0 + (-2) = -2; sum = -2 + 2 = 0. The final verified parity sequence is 0.

Example 3
Input
nums = [1000000000, 1000000000, 1000000000]
Output
3000000000

Explanation: Initialize sum = 0. Iterate through the array: sum = 0 + 1000000000 = 1000000000; sum = 1000000000 + 1000000000 = 2000000000; sum = 2000000000 + 1000000000 = 3000000000. The final verified parity sequence is 3000000000.

Example 4
Input
nums = [-5, -10, -15]
Output
-30

Explanation: Initialize sum = 0. Iterate through the array: sum = 0 + (-5) = -5; sum = -5 + (-10) = -15; sum = -15 + (-15) = -30. The final verified parity sequence is -30.

Constraints

  • 1 <= nums.length <= 10^5
  • -10^9 <= nums[i] <= 10^9
  • The sum of all elements will fit within a 64-bit signed integer.
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Verified Parity Sequence — Problem Statement & Solution Guide

HashingHardFrequency Counter
TimeO(n)
|
SpaceO(1)

Problem Description

Given an array of integers, compute the Verified Parity Sequence. This sequence is defined as the cumulative sum of the array elements, where the final value represents the total aggregate of all inputs. The problem requires processing the array in a single pass to determine this aggregate value efficiently.

The input consists of a single array of integers. The output must be a single integer representing the sum of all elements in the array. Although the term 'parity' often refers to even/odd properties, in this specific context, it denotes the verified total sum of the sequence. The solution must handle large arrays and large integer values without overflow, assuming standard 64-bit integer arithmetic.

Your task is to implement a function that takes an array of integers and returns the sum of its elements. The implementation should be optimized for time complexity, ensuring it runs in linear time relative to the size of the input array. Space complexity should be constant, as no additional data structures are required beyond the input array and a single accumulator variable.

DSA Pattern Breakdown

DSA Pattern Breakdown

"Verified Parity Sequence"

hard

WHY DOES IT MATTER?

The single-pass accumulation pattern is essential because it transforms a quadratic-time problem into linear time, making it scalable to massive inputs. It also reduces memory overhead, which is critical in environments with limited resources.

OPTIMIZATION CHALLENGE

The key insight is that you only need to remember the current total; you never need to revisit previous elements. By updating a single accumulator, you eliminate the need for nested loops or auxiliary arrays.

REAL-WORLD CONNECTION

In financial transaction processing, you often need the total balance after each transaction. The same running sum logic is used to update account balances in real time, ensuring consistency and performance.

When presenting this in an interview, emphasize the O(n) time and O(1) space, and mention that the algorithm is trivially parallelizable if needed. Also, highlight the importance of choosing the correct integer type to avoid overflow.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)
💾 Space:O(1)

Core Theory — Why This Approach?

The Verified Parity Sequence problem is essentially a prefix sum computation: you need to accumulate the sum of all elements in a single pass. A naive approach might recompute the sum for each prefix, leading to an O(n^2) time complexity, which quickly becomes infeasible for large arrays (e.g., 10^7 elements). The optimal paradigm uses a single accumulator variable that is updated as you iterate through the array, achieving O(n) time and O(1) auxiliary space. This pattern is a classic example of linear-time reduction, where you transform a potentially expensive nested operation into a simple iterative accumulation. It also highlights the importance of choosing the right data type (e.g., 64‑bit integers) to avoid overflow when the sum can exceed the 32‑bit range.

Interview Questions on This Problem

Q1How would you modify the algorithm if the array elements were streamed in real time and you needed to provide the running total after each new element?

You would maintain the same accumulator but expose it after each iteration. In a streaming context, you simply update the running total and output it, ensuring the algorithm remains O(1) per element and O(1) space overall.

Q2In a distributed system, how could you compute the total sum of a large dataset split across multiple nodes?

Each node computes a local sum of its partition in O(k) time, then the partial sums are aggregated (e.g., via a reduce operation). This reduces the overall time to O(n) while keeping memory usage per node minimal.

Q3What potential pitfalls should you watch for when implementing this algorithm in a language with fixed integer sizes, and how would you mitigate them?

The primary pitfall is integer overflow. Mitigate it by using a 64‑bit integer type (long long in C++/Java, long in Python) or by performing checks before addition. If the language supports arbitrary precision integers, use that for safety.

Examples

Example 1

Input

nums = [1, 2, 3, 4, 5]

Output

15

Explanation: Initialize sum = 0. Iterate through the array: sum = 0 + 1 = 1; sum = 1 + 2 = 3; sum = 3 + 3 = 6; sum = 6 + 4 = 10; sum = 10 + 5 = 15. The final verified parity sequence is 15.

Example 2

Input

nums = [-1, 0, 1, -2, 2]

Output

0

Explanation: Initialize sum = 0. Iterate through the array: sum = 0 + (-1) = -1; sum = -1 + 0 = -1; sum = -1 + 1 = 0; sum = 0 + (-2) = -2; sum = -2 + 2 = 0. The final verified parity sequence is 0.

Example 3

Input

nums = [1000000000, 1000000000, 1000000000]

Output

3000000000

Explanation: Initialize sum = 0. Iterate through the array: sum = 0 + 1000000000 = 1000000000; sum = 1000000000 + 1000000000 = 2000000000; sum = 2000000000 + 1000000000 = 3000000000. The final verified parity sequence is 3000000000.

Example 4

Input

nums = [-5, -10, -15]

Output

-30

Explanation: Initialize sum = 0. Iterate through the array: sum = 0 + (-5) = -5; sum = -5 + (-10) = -15; sum = -15 + (-15) = -30. The final verified parity sequence is -30.

Constraints

  • 1 <= nums.length <= 10^5
  • -10^9 <= nums[i] <= 10^9
  • The sum of all elements will fit within a 64-bit signed integer.

Optimal Approach & Strategy

The optimal solution uses a single accumulator updated in a single loop, achieving O(n) time and O(1) space. Use a 64‑bit integer to handle large sums.

Brute Force Approach

A naive solution might recompute the sum for each prefix, leading to nested loops and O(n^2) time. It would also use additional space to store intermediate sums.

Verified Code Solutions

JavaScript Solution
Time: O(n)
function solution(nums) {
   let sum = 0;
   for (let num of nums) {
       sum += num;
   }
   return sum;
}

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