LeetCode #1734 — MEDIUM

Decode XORed Permutation

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

There is an integer array perm that is a permutation of the first n positive integers, where n is always odd.

It was encoded into another integer array encoded of length n - 1, such that encoded[i] = perm[i] XOR perm[i + 1]. For example, if perm = [1,3,2], then encoded = [2,1].

Given the encoded array, return the original array perm. It is guaranteed that the answer exists and is unique.

Example 1:

Input: encoded = [3,1]
Output: [1,2,3]
Explanation: If perm = [1,2,3], then encoded = [1 XOR 2,2 XOR 3] = [3,1]

Example 2:

Input: encoded = [6,5,4,6]
Output: [2,4,1,5,3]

Constraints:

3 <= n < 10⁵
n is odd.
encoded.length == n - 1

Step 01

Brute Force Baseline

Problem summary: There is an integer array perm that is a permutation of the first n positive integers, where n is always odd. It was encoded into another integer array encoded of length n - 1, such that encoded[i] = perm[i] XOR perm[i + 1]. For example, if perm = [1,3,2], then encoded = [2,1]. Given the encoded array, return the original array perm. It is guaranteed that the answer exists and is unique.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Bit Manipulation

Example 1

[3,1]

Example 2

[6,5,4,6]

Core Insight

What unlocks the optimal approach

Compute the XOR of the numbers between 1 and n, and think about how it can be used. Let it be x.
Think why n is odd.
perm[0] = x XOR encoded[1] XOR encoded[3] XOR encoded[5] ...
perm[i] = perm[i-1] XOR encoded[i-1]

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1734: Decode XORed Permutation
class Solution {
    public int[] decode(int[] encoded) {
        int n = encoded.length + 1;
        int a = 0, b = 0;
        for (int i = 0; i < n - 1; i += 2) {
            a ^= encoded[i];
        }
        for (int i = 1; i <= n; ++i) {
            b ^= i;
        }
        int[] perm = new int[n];
        perm[n - 1] = a ^ b;
        for (int i = n - 2; i >= 0; --i) {
            perm[i] = encoded[i] ^ perm[i + 1];
        }
        return perm;
    }
}

// Accepted solution for LeetCode #1734: Decode XORed Permutation
func decode(encoded []int) []int {
	n := len(encoded) + 1
	a, b := 0, 0
	for i := 0; i < n-1; i += 2 {
		a ^= encoded[i]
	}
	for i := 1; i <= n; i++ {
		b ^= i
	}
	perm := make([]int, n)
	perm[n-1] = a ^ b
	for i := n - 2; i >= 0; i-- {
		perm[i] = encoded[i] ^ perm[i+1]
	}
	return perm
}

# Accepted solution for LeetCode #1734: Decode XORed Permutation
class Solution:
    def decode(self, encoded: List[int]) -> List[int]:
        n = len(encoded) + 1
        a = b = 0
        for i in range(0, n - 1, 2):
            a ^= encoded[i]
        for i in range(1, n + 1):
            b ^= i
        perm = [0] * n
        perm[-1] = a ^ b
        for i in range(n - 2, -1, -1):
            perm[i] = encoded[i] ^ perm[i + 1]
        return perm

// Accepted solution for LeetCode #1734: Decode XORed Permutation
struct Solution;

impl Solution {
    fn decode(encoded: Vec<i32>) -> Vec<i32> {
        let n = encoded.len() + 1;
        let mut first = 0;
        for i in 0..n {
            first ^= (i + 1) as i32;
        }
        for i in 0..n - 1 {
            if i % 2 == 1 {
                first ^= encoded[i];
            }
        }
        let mut res = vec![first];
        for i in 1..n {
            res.push(res[i - 1] ^ encoded[i - 1]);
        }
        res
    }
}

#[test]
fn test() {
    let encoded = vec![3, 1];
    let res = vec![1, 2, 3];
    assert_eq!(Solution::decode(encoded), res);
    let encoded = vec![6, 5, 4, 6];
    let res = vec![2, 4, 1, 5, 3];
    assert_eq!(Solution::decode(encoded), res);
}

// Accepted solution for LeetCode #1734: Decode XORed Permutation
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1734: Decode XORed Permutation
// class Solution {
//     public int[] decode(int[] encoded) {
//         int n = encoded.length + 1;
//         int a = 0, b = 0;
//         for (int i = 0; i < n - 1; i += 2) {
//             a ^= encoded[i];
//         }
//         for (int i = 1; i <= n; ++i) {
//             b ^= i;
//         }
//         int[] perm = new int[n];
//         perm[n - 1] = a ^ b;
//         for (int i = n - 2; i >= 0; --i) {
//             perm[i] = encoded[i] ^ perm[i + 1];
//         }
//         return perm;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.