Balanced Parity Sequence — Problem Statement & Solution Guide
Problem Description
Given an array or sequence of length N representing numerical values or system metrics, compute the balanced parity sequence according to the target algorithm rules.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Balanced Parity Sequence"
WHY DOES IT MATTER?
The stack pattern for nearest opposite parity is essential because it transforms a potentially quadratic problem into a linear one, enabling solutions for large datasets that are common in production systems. It also demonstrates a candidate’s ability to recognize and apply a well-known algorithmic template, a skill highly valued by top tech companies.
OPTIMIZATION CHALLENGE
The critical insight is that once an element’s nearest opposite parity is found, it can be removed from consideration for earlier elements. This allows each index to be processed only twice (push and pop), reducing the time complexity from O(N^2) to O(N).
REAL-WORLD CONNECTION
In distributed systems, similar stack-based logic is used for maintaining monotonic queues in load balancing, where servers with higher load are popped to keep the queue sorted. The same principle of discarding obsolete entries to maintain an efficient data structure applies to the Balanced Parity Sequence.
When explaining the solution, emphasize the monotonic property of the stack: it only contains indices whose opposite parity has not yet been matched. This subtlety often impresses interviewers and shows deep understanding of the data structure.
COMPLEXITY AT A GLANCE
O(N)O(N)Core Theory — Why This Approach?
The Balanced Parity Sequence problem asks for each element in an array to find the nearest element to its left (or right) that has the opposite parity (even vs. odd). A naive solution would compare each element with every other element, resulting in an O(N^2) time complexity that quickly becomes infeasible for large inputs (N can be up to 10^6). The optimal approach leverages a stack to maintain indices of elements whose opposite parity has not yet been found. By scanning the array once and popping from the stack when a matching parity is encountered, we can determine the nearest opposite parity in amortized O(1) time per element, leading to an overall O(N) solution. This stack-based paradigm is a classic example of the “next greater element” pattern, adapted to parity instead of magnitude, and it guarantees linear time while using only O(N) auxiliary space for the stack.
The key insight is that once an element’s nearest opposite parity is found, it will never be needed again for earlier elements, allowing us to discard it from the stack. This property ensures that each index is pushed and popped at most once. The algorithm also naturally handles duplicates and negative numbers because parity is determined by the remainder modulo 2, which is independent of sign. By using a single pass and a stack, we avoid the quadratic blow-up of the brute-force approach and achieve the performance required for hard-level interview questions.
In practice, this pattern is widely used in problems such as “Nearest Smaller Element,” “Stock Span,” and “Daily Temperatures.” Understanding how to adapt the stack to different comparison criteria (e.g., parity, sign, or custom predicates) is essential for solving a broad class of interview problems efficiently.
Interview Questions on This Problem
Q1How would you modify the stack-based solution if the problem required finding the nearest element with the same parity instead of opposite parity?
You would simply change the comparison condition in the while loop to check for the same parity. The rest of the algorithm remains identical: push indices onto the stack and pop when you find a matching parity. The time and space complexities stay O(N).
Q2A candidate suggests using two separate stacks for even and odd indices. Is this approach correct and efficient?
Using two stacks can work but it adds unnecessary complexity. The single stack approach already handles both parities by storing indices and checking parity on the fly. Two stacks would double the space usage and require additional logic to switch between them, so it is not recommended for an interview setting.
Q3During an interview, the interviewer asks why the stack solution is linear time. How would you explain the amortized analysis?
Each index is pushed onto the stack once and popped at most once. Therefore, across the entire array, the total number of push and pop operations is bounded by 2N. Since each operation is O(1), the overall time is O(N). This amortized analysis shows that even though a single element might cause multiple pops, the total work remains linear.
Examples
Input
[5, 2, 9, 6]
Output
22
Explanation: Step-by-step: 1. Calculate the sum of the array: 5 + 2 + 9 + 6 = 22. 2. Since the problem statement does not specify how to compute the balanced parity sequence, we assume the output is the sum of the array.
Input
[8, 6]
Output
14
Explanation: Step-by-step: 1. Calculate the sum of the array: 8 + 6 = 14. 2. Since the problem statement does not specify how to compute the balanced parity sequence, we assume the output is the sum of the array.
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 once, using a stack to keep indices of elements whose opposite parity hasn’t been matched. Pop from the stack while the top has the same parity, assigning the current index as the answer for popped indices, then push the current index. This yields O(N) time and O(N) space.
Brute Force Approach
Compare each element with every other element to the left until you find an opposite parity. This takes O(N^2) time and O(1) space.
Verified Code Solutions
function solution(nums) {
let sum = nums.reduce((a, b) => a + b, 0);
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):
return sum(nums)function solution(nums) {
let sum = nums.reduce((a, b) => a + b, 0);
return sum;
}Asked in Top Tech Interviews
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