LeetCode #1806 — MEDIUM

Minimum Number of Operations to Reinitialize a Permutation

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

The Problem

Problem Statement

You are given an even integer n. You initially have a permutation perm of size n where perm[i] == i (0-indexed).

In one operation, you will create a new array arr, and for each i:

If i % 2 == 0, then arr[i] = perm[i / 2].
If i % 2 == 1, then arr[i] = perm[n / 2 + (i - 1) / 2].

You will then assign arr to perm.

Return the minimum non-zero number of operations you need to perform on perm to return the permutation to its initial value.

Example 1:

Input: n = 2
Output: 1
Explanation: perm = [0,1] initially.
After the 1^st operation, perm = [0,1]
So it takes only 1 operation.

Example 2:

Input: n = 4
Output: 2
Explanation: perm = [0,1,2,3] initially.
After the 1^st operation, perm = [0,2,1,3]
After the 2^nd operation, perm = [0,1,2,3]
So it takes only 2 operations.

Example 3:

Input: n = 6
Output: 4

Constraints:

2 <= n <= 1000
n is even.

Step 01

Brute Force Baseline

Problem summary: You are given an even integer n. You initially have a permutation perm of size n where perm[i] == i (0-indexed). In one operation, you will create a new array arr, and for each i: If i % 2 == 0, then arr[i] = perm[i / 2]. If i % 2 == 1, then arr[i] = perm[n / 2 + (i - 1) / 2]. You will then assign arr to perm. Return the minimum non-zero number of operations you need to perform on perm to return the permutation to its initial value.

Baseline thinking

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

Pattern signal: Array · Math

Example 1

Example 2

Example 3

Step 02

Core Insight

What unlocks the optimal approach

It is safe to assume the number of operations isn't more than n
The number is small enough to apply a brute force solution.

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 #1806: Minimum Number of Operations to Reinitialize a Permutation
class Solution {
    public int reinitializePermutation(int n) {
        int ans = 0;
        for (int i = 1;;) {
            ++ans;
            if (i < (n >> 1)) {
                i <<= 1;
            } else {
                i = (i - (n >> 1)) << 1 | 1;
            }
            if (i == 1) {
                return ans;
            }
        }
    }
}

// Accepted solution for LeetCode #1806: Minimum Number of Operations to Reinitialize a Permutation
func reinitializePermutation(n int) (ans int) {
	for i := 1; ; {
		ans++
		if i < (n >> 1) {
			i <<= 1
		} else {
			i = (i-(n>>1))<<1 | 1
		}
		if i == 1 {
			return ans
		}
	}
}

# Accepted solution for LeetCode #1806: Minimum Number of Operations to Reinitialize a Permutation
class Solution:
    def reinitializePermutation(self, n: int) -> int:
        ans, i = 0, 1
        while 1:
            ans += 1
            if i < n >> 1:
                i <<= 1
            else:
                i = (i - (n >> 1)) << 1 | 1
            if i == 1:
                return ans

// Accepted solution for LeetCode #1806: Minimum Number of Operations to Reinitialize a Permutation
struct Solution;

impl Solution {
    fn reinitialize_permutation(n: i32) -> i32 {
        if n == 2 {
            return 1;
        }
        let mut res = 0;
        let mut i = 1;
        while res == 0 || i != 1 {
            if i % 2 == 0 {
                i /= 2;
            } else {
                i = (n - 1 + i) / 2;
            }
            res += 1;
        }
        res
    }
}

#[test]
fn test() {
    let n = 2;
    let res = 1;
    assert_eq!(Solution::reinitialize_permutation(n), res);
    let n = 4;
    let res = 2;
    assert_eq!(Solution::reinitialize_permutation(n), res);
    let n = 6;
    let res = 4;
    assert_eq!(Solution::reinitialize_permutation(n), res);
}

// Accepted solution for LeetCode #1806: Minimum Number of Operations to Reinitialize a Permutation
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1806: Minimum Number of Operations to Reinitialize a Permutation
// class Solution {
//     public int reinitializePermutation(int n) {
//         int ans = 0;
//         for (int i = 1;;) {
//             ++ans;
//             if (i < (n >> 1)) {
//                 i <<= 1;
//             } else {
//                 i = (i - (n >> 1)) << 1 | 1;
//             }
//             if (i == 1) {
//                 return ans;
//             }
//         }
//     }
// }

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

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.