Shifted Parity Sequence — Problem Statement & Solution Guide
Problem Description
Given an array or sequence of length N representing numerical values or system metrics, compute the sum of the sequence according to the target algorithm rules.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Shifted Parity Sequence"
WHY DOES IT MATTER?
Sliding‑window (two‑pointer) patterns are essential for any problem that asks for a maximal contiguous segment under a simple, monotonic predicate—such as sum limits, distinct counts, or parity constraints. Mastery of this pattern lets engineers turn quadratic brute‑force scans into linear‑time solutions, a frequent expectation in high‑throughput systems.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the parity condition is *local* and *binary*: each element is either compatible with shift 0 or shift 1, never both. This allows us to maintain two independent windows simultaneously, advancing pointers only when a violation occurs, which collapses the naïve O(N²) enumeration to O(N).
REAL-WORLD CONNECTION
Think of a network packet processor that must forward a burst of packets only while they meet a header‑parity rule (e.g., even‑length payloads on even‑numbered slots). The processor slides a window over the incoming stream, dropping the earliest packet as soon as a rule violation appears, ensuring constant‑time per packet and maximal throughput.
When coding the solution, keep a single running sum and two left pointers (one per shift). Update the sum by adding A[r] and, if the window becomes invalid for a shift, subtract A[l] while moving that left pointer. This avoids recomputing sums from scratch and keeps the code clean and bug‑free.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The Shifted Parity Sequence problem asks for the maximum sum of a contiguous sub‑array whose elements align with a global parity shift. For a chosen shift s (0 or 1), an element A[i] is considered “valid” if (A[i] % 2) == ((i + s) % 2). A naïve solution would enumerate every possible sub‑array, verify the parity condition for each element, and compute its sum – an O(N²) time algorithm that quickly becomes infeasible for N up to 10⁵ or higher. The optimal paradigm leverages the two‑pointer (sliding‑window) technique: we maintain a window [l, r) that always satisfies the parity condition for the current shift. When the right pointer encounters an invalid element, we advance the left pointer until the window regains validity. During this process we keep a running sum, updating the answer whenever the window expands. Because each index is visited at most twice (once by the right pointer, once by the left), the overall complexity collapses to linear time, O(N), with O(1) extra space.
Interview Questions on This Problem
Q1How would you modify the two‑pointer solution if the problem asked for the longest sub‑array (by length) instead of the maximum sum?
The same sliding‑window framework applies; instead of tracking a running sum, maintain the window length (r‑l). Whenever the window is valid, compare its length to the best length seen so far. The update step is O(1), so the overall algorithm remains O(N).
Q2Can the Shifted Parity Sequence be solved in a single pass without explicitly trying both parity shifts?
Yes. Observe that for any index i, the parity condition depends only on (A[i] % 2) XOR (i % 2). This value tells you which shift (0 or 1) makes the element valid. By maintaining two parallel windows—one for each shift—you can update both in the same linear scan, effectively handling both possibilities in a single pass.
Q3Why does the two‑pointer technique guarantee optimality for this problem, and could a divide‑and‑conquer approach be competitive?
Two‑pointer guarantees optimality because the constraint (all elements in the window must satisfy a simple parity predicate) is monotonic: extending the window can only break validity, never restore it without moving the left edge. Divide‑and‑conquer would require merging sub‑solutions and would still be O(N log N), which is asymptotically slower than the linear sliding‑window solution.
Examples
Input
[10, 7, 4, 11]
Output
32
Explanation: Step-by-step: with input [10, 7, 4, 11], we sum the elements directly, giving output 10 + 7 + 4 + 11 = 32. This is a simple sum operation.
Input
[8, 6]
Output
14
Explanation: Step-by-step: with input [8, 6], we sum the elements directly, giving output 8 + 6 = 14. This is a simple sum operation.
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
Use two sliding windows (one per shift) with left and right pointers, maintaining a running sum and adjusting the left edge only on parity violations – O(N) time.
Brute Force Approach
Enumerate every sub‑array, check each element’s parity against the chosen shift, and compute its sum – O(N²) time.
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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