Optimal Parity Sequence — Problem Statement & Solution Guide

ArraysHardReversal

TimeO(N)

SpaceO(1)

Problem Description

You are given an integer array nums of length N. For each position i (0‑based) define its parity value as 0 if nums[i] is even and 1 if nums[i] is odd. A parity sequence is any permutation of these parity values obtained by reversing any contiguous sub‑array of nums any number of times. Your task is to transform the original parity sequence into the lexicographically smallest possible sequence using the allowed reversals and output that final sequence. The output should be a string of 0s and 1s without spaces, where the i‑th character corresponds to the parity of the element that ends up at position i after all operations.

DSA Pattern Breakdown

"Optimal Parity Sequence"

hard

WHY DOES IT MATTER?

The counting‑and‑reconstruct pattern is essential for problems where the target order depends only on the frequency of elements, not their identities. It guarantees linear time and constant extra space, which is critical for large datasets and real‑time systems.

OPTIMIZATION CHALLENGE

The key insight is that the reversal operation can bring any element to any position, so the only constraint is the multiset of parities. Recognizing that all 0’s are identical allows us to ignore relative order and use a simple counter.

REAL-WORLD CONNECTION

Imagine a warehouse sorting packages by weight class (light vs heavy). Instead of moving each package individually, you simply count how many light packages there are and place them all at the front of the aisle. This reduces handling time and avoids unnecessary shuffling.

When explaining this to an interviewer, emphasize that the reversal operation is a universal move for binary data, and that the optimal solution is to count and rebuild, not to simulate reversals.

COMPLEXITY AT A GLANCE

⏱ Time:O(N)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem reduces to a classic sorting‑by‑parity task. Each element’s parity is either 0 (even) or 1 (odd). Reversing any contiguous sub‑array is a powerful operation that can move any element to any position while preserving the multiset of parities. For a binary sequence, the lexicographically smallest arrangement is simply all 0’s followed by all 1’s. A naive approach would attempt to generate every possible sequence obtainable by a series of reversals, which is exponential in N and infeasible for large inputs. The optimal paradigm is a greedy counting strategy: count the number of 0’s (or 1’s) and construct the target sequence in linear time. This leverages the fact that all 0’s are indistinguishable and can be placed at the front without affecting the relative order of 1’s, which is irrelevant for lexicographic minimality.

Interview Questions on This Problem

Q1How would you transform a binary array into its lexicographically smallest form using only subarray reversals, and why is this operation sufficient?

Because a subarray reversal can bring any element to any position, we can greedily move each 0 to the front. Since all 0’s are identical, the final sequence is simply all 0’s followed by all 1’s. This is optimal because any other arrangement would have a 1 before a 0, making it lexicographically larger.

Q2In a distributed system, you need to reorder a stream of even/odd flags to minimize latency. Which algorithmic pattern from this problem would you apply and why?

I would apply the counting‑and‑reconstruct pattern. By streaming the flags and counting zeros, I can immediately output the minimal sequence without storing the entire stream, which is ideal for low‑latency, high‑throughput environments.

Q3During a coding interview at a fintech startup, the interviewer asks: "Can you explain why a two‑pointer approach would not be necessary here?" How would you respond?

A two‑pointer approach is unnecessary because the parity values are binary and indistinguishable within each group. Counting zeros gives us the exact number needed for the front of the array, eliminating the need to swap elements pairwise.

Examples

Example 1

Input

[3, 8, 5, 2, 7]

Output

Explanation: Step‑by‑step: start with [3,8,5,2,7] → parity 10101. Reverse 0‑3 → [2,5,8,3,7] → parity 01011. Reverse 1‑2 → [2,8,5,3,7] → parity 00111. Reverse 3‑4 → [2,8,5,7,3] → parity 00101, which is lexicographically minimal.

Example 2

Input

[4, 1, 6, 9, 2, 5]

Output

Explanation: Parity of the original array is [0,1,0,1,0,1]. By repeatedly reversing sub‑arrays that start at an even element and end at the next odd element, we can push all evens to the left. One possible sequence: reverse 1‑3 → [4,9,6,1,2,5] → parity [0,1,0,1,0,1]; reverse 3‑5 → [4,9,6,5,2,1] → parity [0,1,0,1,0,1]; reverse 1‑4 → [4,2,6,5,9,1] → parity [0,0,0,1,1,1]. The resulting parity string "000111" is the smallest possible.

Example 3

Input

[11, 13, 15]

Output

Explanation: All numbers are odd, so every parity value is 1. Any reversal leaves the parity sequence unchanged. Hence the only possible output is "111".

Constraints

1 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
The algorithm must run in O(N log N) time or better
Only in‑place reversals of contiguous sub‑arrays are allowed

Optimal Approach & Strategy

Count the number of 0’s in the parity array and output that many 0’s followed by the remaining 1’s. This runs in linear time and uses constant extra space.

Brute Force Approach

Generate all possible sequences by applying every combination of subarray reversals, then pick the lexicographically smallest. This explores an exponential number of states and is infeasible for large N.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N)

function solution(nums) {
   if (nums.length === 0) return 0;
   if (nums.length === 1) return nums[0];
   let reversed = nums.slice().reverse();
   return reversed.reduce((a, b) => a + b, 0);
}

class Solution {
   public:
   int solution(vector<int>& nums) {
       if (nums.size() == 0) return 0;
       if (nums.size() == 1) return nums[0];
       vector<int> reversed(nums.size());
       for (int i = 0; i < nums.size(); i++) {
           reversed[i] = nums[nums.size() - 1 - i];
       }
       int sum = 0;
       for (int num : reversed) {
           sum += num;
       }
       return sum;
   }
};

class Solution {
   public int solution(int[] nums) {
       if (nums.length == 0) return 0;
       if (nums.length == 1) return nums[0];
       int[] reversed = new int[nums.length];
       for (int i = 0; i < nums.length; i++) {
           reversed[i] = nums[nums.length - 1 - i];
       }
       int sum = 0;
       for (int num : reversed) {
           sum += num;
       }
       return sum;
   }
}

def solution(nums):
   if len(nums) == 0:
       return 0
   if len(nums) == 1:
       return nums[0]
   reversed = nums[::-1]
   return sum(reversed)

function solution(nums) {
   if (nums.length === 0) return 0;
   if (nums.length === 1) return nums[0];
   let reversed = nums.slice().reverse();
   return reversed.reduce((a, b) => a + b, 0);
}

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Optimal Parity Sequence Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Arrays

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Optimal Parity Sequence — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor