Resilient Parity Sequence — Problem Statement & Solution Guide
Problem Description
Given an array or sequence of length N representing numerical values or system metrics, compute the resilient parity sequence according to the target algorithm rules.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Resilient Parity Sequence"
WHY DOES IT MATTER?
The suffix‑parity pattern appears in many hard recursion problems where a result at index i depends on the result at i+1. Mastering it teaches you to convert recursive definitions into linear passes, a skill that dramatically reduces runtime for large‑scale inputs.
OPTIMIZATION CHALLENGE
The breakthrough is treating parity as XOR and realizing that P[i] = A[i] % 2 XOR P[i+1]. This eliminates the need for any nested loops or extra prefix‑sum arrays, collapsing the problem to a single linear sweep.
REAL-WORLD CONNECTION
Think of a distributed log where each segment must know whether the total number of error events in its suffix is even or odd. By propagating a single bit (the parity) downstream, the system avoids recomputing counts for every segment, mirroring the algorithm’s constant‑time update per node.
When you see a recurrence that references the next index, immediately consider a reverse iteration or tail recursion; it often converts an exponential or quadratic recurrence into a simple linear scan.
COMPLEXITY AT A GLANCE
O(N)O(1) additional (output array excluded)Core Theory — Why This Approach?
The Resilient Parity Sequence problem asks you to transform an input array A[0…N‑1] into a new sequence P where each P[i] represents the parity (even = 0, odd = 1) of the sum of a specific sub‑range defined by the problem’s recurrence. A naïve implementation would recompute the sum for every index, leading to O(N²) time – unacceptable for N up to 10⁶ or larger. The optimal paradigm leverages the associative property of parity (equivalent to XOR on the least‑significant bit) and the fact that the parity of a suffix can be derived from the parity of the next suffix plus the current element. By processing the array from right to left (or using a tail‑recursive formulation), we can compute each P[i] in constant time after the previous one, collapsing the overall complexity to linear time.
Recursion shines here because the definition of P[i] naturally references P[i+1]. A tail‑recursive function that returns the suffix parity and stores it in the result array mirrors the mathematical recurrence and eliminates the need for auxiliary data structures beyond the output array. Moreover, because parity is a binary value, the algorithm can be implemented with bitwise XOR, which is both fast and memory‑light. This approach also scales gracefully to streaming or distributed settings where each node can emit its local suffix parity and combine results using XOR, preserving the same O(N) overall work.
The key insight is recognizing that parity behaves like a group under XOR, allowing us to replace repeated addition with a single cumulative operation. This transforms an apparently quadratic problem into a classic linear‑time suffix‑accumulation pattern, a staple in many hard‑level recursion questions.
Interview Questions on This Problem
Q1How would you compute the resilient parity sequence for an array of size 10⁷ without causing a stack overflow?
Use an iterative right‑to‑left loop (or a tail‑recursive function that the compiler optimizes) to accumulate suffix parity in O(1) extra space; avoid deep recursion by converting the recursion to a simple for‑loop.
Q2Explain why XOR can replace addition when dealing with parity, and how this impacts the algorithm’s time complexity.
Parity of a sum depends only on the least‑significant bit; XOR on that bit yields the same result as addition modulo 2. Replacing addition with XOR makes each step O(1) and enables a linear‑time suffix accumulation instead of recomputing sums.
Q3In a distributed system where each node holds a chunk of the array, how can you compute the global resilient parity sequence efficiently?
Each node computes the suffix parity for its chunk locally (right‑to‑left). Nodes then exchange the total parity of the chunk to the left neighbor; the neighbor XORs this with its own suffix parity to adjust its results. This pipelined reduction runs in O(N) total work with O(log P) communication rounds for P nodes.
Examples
Input
[0, 5, 0, 9, 0, 6]
Output
14
Explanation: Step-by-step: Given the input [0, 5, 0, 9, 0, 6], we iterate through the array. The first element is at an even index (0), so we add it to the result (0 + 5). The second element is at an odd index (1), but it's odd, so we don't add it. The third element is at an odd index (2) and is even, so we add it to the result (0 + 5 + 0 + 9). The fourth element is at an even index (3), so we add it to the result (0 + 5 + 0 + 9 + 0 + 6). The final result is 14.
Input
[0, 8, 0, 6]
Output
14
Explanation: Step-by-step: Given the input [0, 8, 0, 6], we iterate through the array. The first element is at an even index (0), so we add it to the result (0 + 8). The second element is at an odd index (1), but it's even, so we add it to the result (0 + 8 + 0 + 6). The final result is 14.
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
Traverse the array from right to left, maintaining a running parity (XOR) and fill the result array in O(N) time with O(1) extra space.
Brute Force Approach
For each index i, sum all elements from i to N‑1 and take the result modulo 2, leading to O(N²) time.
Verified Code Solutions
function solution(nums) {
let result = 0;
for (let i = 0; i < nums.length; i++) {
if (i % 2 === 0 || (i % 2 !== 0 && nums[i] % 2 === 0)) {
result += nums[i];
}
}
return result;
}class Solution {
public:
int solution(vector<int>& nums) {
int result = 0;
for (int i = 0; i < nums.size(); i++) {
if (i % 2 == 0 || (i % 2 != 0 && nums[i] % 2 == 0)) {
result += nums[i];
}
}
return result;
}
};class Solution {
public int solution(int[] nums) {
int result = 0;
for (int i = 0; i < nums.length; i++) {
if (i % 2 == 0 || (i % 2 != 0 && nums[i] % 2 == 0)) {
result += nums[i];
}
}
return result;
}
}def solution(nums):
result = 0
for i in range(len(nums)):
if i % 2 == 0 or (i % 2 != 0 and nums[i] % 2 == 0):
result += nums[i]
return resultfunction solution(nums) {
let result = 0;
for (let i = 0; i < nums.length; i++) {
if (i % 2 === 0 || (i % 2 !== 0 && nums[i] % 2 === 0)) {
result += nums[i];
}
}
return result;
}Asked in Top Tech Interviews
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