Iterative Parity Sequence — Problem Statement & Solution Guide
Problem Description
Given an array of integers, repeatedly replace the array by the sum of all its elements until only one number remains. The resulting single number is the iterative parity sequence of the original array. Input consists of an integer n (1 ≤ n ≤ 10^5) on the first line, followed by n space‑separated integers on the second line. Output a single integer – the final value after all iterations.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Iterative Parity Sequence"
WHY DOES IT MATTER?
Recognizing that the iterative reduction of an array by summing its elements is equivalent to a single aggregate operation is a classic example of algorithmic simplification. It teaches candidates to look for invariants and avoid unnecessary loops, a skill that is essential when optimizing for time and space in production systems.
OPTIMIZATION CHALLENGE
The key insight is the associativity of addition: the order of summation does not affect the final result. By exploiting this property, we can collapse the iterative process into a single linear scan, reducing time from O(n^2) to O(n) and space from O(n) to O(1).
REAL-WORLD CONNECTION
In distributed systems, a similar pattern appears in log aggregation or map-reduce jobs where many partial results are combined into a single metric. Understanding that a single pass can replace multiple aggregation steps saves bandwidth and processing time across the cluster.
When explaining this to an interviewer, emphasize the algebraic property that allows the collapse, and mention that this pattern is widely used in streaming algorithms and real-time analytics where data arrives continuously and must be aggregated on the fly.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The iterative parity sequence problem reduces to a single pass accumulation of the array’s elements. In a naive interpretation, one might think to repeatedly collapse the array by summing its elements, then summing the resulting single number again, and so on—this would lead to an O(n^2) or even O(n^3) time complexity if implemented with nested loops or repeated array constructions. However, the mathematical property of addition is associative and commutative: the sum of all elements in the array is invariant under any order of summation. Thus, the final single number is simply the total sum of the original array, which can be computed in linear time with constant auxiliary space. This insight eliminates the need for repeated iterations and showcases the power of recognizing algebraic invariants to transform a seemingly iterative process into a direct calculation.
The failure of naive approaches on large inputs stems from the fact that each iteration would create a new array and traverse it again, leading to a cumulative cost proportional to the sum of decreasing lengths. For an array of size 10^5, a quadratic algorithm would perform on the order of 10^10 operations—far beyond practical limits. By contrast, the optimal paradigm leverages a single accumulation loop, ensuring that the algorithm scales linearly with input size and remains efficient even at the upper bounds of the constraints.
Interview Questions on This Problem
Q1How would you explain the time complexity of computing the iterative parity sequence to a hiring manager at a fintech company?
I would say that the problem boils down to a single pass over the input array, summing each element once. This gives a time complexity of O(n) and a space complexity of O(1), which is optimal for large datasets typical in fintech where performance and memory usage are critical.
Q2A senior engineer at a high-growth startup asks: "Can we parallelize the computation of the iterative parity sequence?"
Yes, we can use a parallel reduction (e.g., divide the array into chunks, sum each chunk concurrently, then combine the partial sums). This reduces the effective depth of the computation and can achieve near-linear speedup on multi-core systems, while still maintaining O(n) total work.
Q3During an interview at a global product company, you are asked: "What edge cases should we test for the iterative parity sequence implementation?"
We should test an array of length 1 (the result is the element itself), an array containing all zeros (result 0), and an array with negative numbers to ensure the sum handles sign correctly. Additionally, test the maximum allowed size (10^5) with large values to confirm no overflow or performance issues.
Examples
Input
4 1 2 3 4
Output
1
Explanation: Sum of the array is 1+2+3+4=10. The array now contains the single element 10. Since only one element remains, the iterative parity sequence is 10, and its value is 1 (the sum of its digits). The final output is 1.
Input
3 9 9 9
Output
9
Explanation: Sum of the array is 9+9+9=27. The array now contains 27. The sum of its digits is 2+7=9, which is the final value.
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
The optimal solution is to compute the sum of all elements in a single pass, storing the result in a single variable. This yields O(n) time and O(1) auxiliary space.
Brute Force Approach
A naive solution would repeatedly create a new array containing the sum of the current array, then repeat until only one element remains. This would involve O(n) work per iteration and could lead to O(n^2) time overall.
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):
sum = 0
for num in nums:
sum += num
return sumfunction solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}Asked in Top Tech Interviews
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