Minimized Parity Sequence — Problem Statement & Solution Guide
Problem Description
You are provided with an array of integers representing a sequence of system metrics. Your task is to compute the 'Minimized Parity Sequence' value based on the aggregate sum of these metrics. The computation follows a specific parity adjustment rule: first, calculate the total sum of all elements in the array. If this total sum is an even number, the result is the sum itself. If the total sum is an odd number, the result is the sum decremented by 1. This operation effectively ensures the final output is always the largest even integer less than or equal to the total sum.
The input consists of a single array of integers. You must process the entire array to determine the aggregate sum and apply the parity rule. The output is a single integer representing the minimized parity value.
This problem tests your ability to perform efficient aggregation and apply conditional logic based on arithmetic properties. While the logic is straightforward, the constraints require an O(N) solution to handle large input sizes efficiently without overflow issues in intermediate calculations.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Minimized Parity Sequence"
WHY DOES IT MATTER?
The greedy parity adjustment pattern is essential because it transforms a potentially combinatorial problem into a linear-time solution. By focusing on the parity property, we avoid exploring all subsets and instead make a single optimal choice that guarantees the minimal even sum.
OPTIMIZATION CHALLENGE
The key insight is that parity is a binary property; once the sum is odd, only odd removals can fix it. Therefore, tracking the minimum odd element during a single pass is enough—no need for sorting or complex data structures.
REAL-WORLD CONNECTION
In distributed systems, this pattern is analogous to load balancing where you need to remove the smallest heavy task from a cluster to keep the total load even and within capacity. Just as we remove the lightest odd element, a scheduler might drop the least resource-intensive job to satisfy a constraint.
When explaining this in an interview, emphasize the invariant: the sum’s parity changes only when an odd element is removed. Highlight that the greedy choice is provably optimal because any larger odd removal would only increase the final sum.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The core of the Minimized Parity Sequence problem is a classic example of a linear-time greedy reduction. First, compute the total sum of the array in a single pass. If the sum is even, the answer is trivially the sum itself. If the sum is odd, the only way to make it even while minimizing the resulting value is to remove exactly one element that is odd, because subtracting an odd number from an odd sum yields an even result. Among all odd elements, the smallest one yields the minimal reduction, thus producing the smallest even sum possible. Naïve approaches that attempt to enumerate all subsets or use a sliding window to test every possible removal would run in exponential or quadratic time, which is infeasible for large input sizes (up to 10^5 or more). The optimal paradigm therefore reduces the problem to a single linear scan to compute the sum and track the minimum odd value, achieving O(n) time and O(1) extra space.
This problem also illustrates the power of parity reasoning in algorithm design. By observing that only the parity of the sum matters, we can avoid complex data structures and focus on a simple property: an odd sum can be made even by removing an odd element. The greedy choice of the smallest odd element is provably optimal because any larger odd removal would subtract a larger amount, yielding a larger final sum. Thus, the solution is both optimal and efficient.
The sliding window terminology in the problem statement is a misnomer; the solution does not require a window. However, the same principle of maintaining a running aggregate (the sum) and updating a secondary metric (the minimum odd) is analogous to how sliding window algorithms maintain a window sum and other statistics in constant time as the window slides. This analogy helps students see the broader applicability of the technique across different problem domains.
Interview Questions on This Problem
Q1At Google, how would you explain the time complexity of the optimal solution for the Minimized Parity Sequence problem, and why is it acceptable for large inputs?
The optimal solution runs in O(n) time because it requires a single pass to compute the total sum and to track the smallest odd element. This linear time complexity is acceptable for inputs up to 10^5 or 10^6, as it scales directly with the array size and avoids any nested loops or exponential enumeration.
Q2During a fintech interview, you are asked to modify the algorithm to handle multiple queries where each query removes a different element. What data structure would you use to answer each query in O(1) time after preprocessing?
You can preprocess the prefix sums and the minimum odd value for each prefix and suffix. Then, for a query that removes element i, compute the new sum as totalSum - arr[i] and adjust the parity by checking if arr[i] is odd. If the resulting sum is odd, you need to subtract the next smallest odd element not equal to arr[i], which can be found using a segment tree or two arrays of prefix/suffix minima of odd elements. With these structures, each query can be answered in O(log n) or O(1) if you maintain two heaps for odd elements.
Q3At a high-growth startup, you need to explain why the greedy choice of the smallest odd element is optimal. How would you justify this to a non-technical product manager?
I would say that when we have an odd total, we need to remove something to make it even. Removing any odd number will fix the parity, but we want to lose as little as possible. So we simply look for the smallest odd number in the list and remove that. It's like taking the lightest coin out of a bag to keep the bag as heavy as possible while still meeting a requirement.
Examples
Input
nums = [1, 2, 3, 4]
Output
10
Explanation: Step 1: Calculate the sum of the array: 1 + 2 + 3 + 4 = 10. Step 2: Check the parity of the sum. 10 is an even number. Step 3: Since the sum is even, return the sum as is. Result: 10.
Input
nums = [5, 7, 9]
Output
20
Explanation: Step 1: Calculate the sum of the array: 5 + 7 + 9 = 21. Step 2: Check the parity of the sum. 21 is an odd number. Step 3: Since the sum is odd, subtract 1 from the sum: 21 - 1 = 20. Result: 20.
Input
nums = [0, 0, 0]
Output
0
Explanation: Step 1: Calculate the sum of the array: 0 + 0 + 0 = 0. Step 2: Check the parity of the sum. 0 is an even number. Step 3: Since the sum is even, return the sum as is. Result: 0.
Input
nums = [-1, -2, -3]
Output
-6
Explanation: Step 1: Calculate the sum of the array: -1 + (-2) + (-3) = -6. Step 2: Check the parity of the sum. -6 is an even number. Step 3: Since the sum is even, return the sum as is. Result: -6.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The sum of all elements in the array will fit within a 64-bit signed integer.
Optimal Approach & Strategy
Compute the total sum and the smallest odd element in a single scan. If the sum is even, return it; otherwise, return sum minus the smallest odd element. This runs in O(n) time and uses O(1) extra space.
Brute Force Approach
A naive approach would try removing each element one by one, recomputing the sum each time, and keeping the smallest even sum found. This would take O(n^2) time because you recompute the sum for each removal.
Verified Code Solutions
function solution(nums) {
let sum = nums.reduce((a, b) => a + b, 0);
return sum % 2 === 0 ? sum : sum - 1;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum % 2 == 0 ? sum : sum - 1;
}class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum % 2 == 0 ? sum : sum - 1;
}def solution(nums):
sum = sum(nums)
return sum % 2 == 0 and sum or sum - 1function solution(nums) {
let sum = nums.reduce((a, b) => a + b, 0);
return sum % 2 === 0 ? sum : sum - 1;
}Asked in Top Tech Interviews
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