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1806. Minimum Number of Operations to Reinitialize a Permutation

Description

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 1st 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 1st operation, perm = [0,2,1,3]
After the 2nd 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.

Solutions

Solution 1: Find Pattern + Simulation

We observe the change pattern of the numbers and find that:

  1. The even-indexed numbers of the new array are the numbers in the first half of the original array in order;
  2. The odd-indexed numbers of the new array are the numbers in the second half of the original array in order.

That is, if the index $i$ of a number in the original array is in the range [0, n >> 1), then the new index of this number is i << 1; otherwise, the new index is (i - (n >> 1)) << 1 | 1.

In addition, the path of number movement is the same in each round of operation. As long as a number (except for numbers $0$ and $n-1$) returns to its original position, the entire sequence will be consistent with the previous one.

Therefore, we choose the number $1$, whose initial index is also $1$. Each time we move the number $1$ to a new position, until the number $1$ returns to its original position, we can get the minimum number of operations.

The time complexity is $O(n)$, and the space complexity is $O(1)$.

  • 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;
            }
        }
    }
}

  • 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;
            }
        }
    }
};

  • 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


  • 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
		}
	}
}

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