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Formatted question description: https://leetcode.ca/all/1806.html

1806. Minimum Number of Operations to Reinitialize a Permutation

Level

Medium

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: nums = [0,1] initially.

After the 1st operation, nums = [0,1]

So it takes only 1 operation.

Example 2:

Input: n = 4

Output: 2

Explanation: nums = [0,1,2,3] initially.

After the 1st operation, nums = [0,2,1,3]

After the 2nd operation, nums = [0,1,2,3]

So it takes only 2 operations.

Example 3:

Input: n = 6

Output: 4

Constraints:

  • 2 <= n <= 1000
  • n is even.

Solution

Simply simulate the process. Create two arrays init of perm, each of which has length n, and initialize init[i] = perm[i] = i for 0 <= i < n. Then update perm each time until perm equals init, and return the number of operations.

  • class Solution {
    public int reinitializePermutation(int n) {
        int[] init = new int[n];
        int[] perm = new int[n];
        for (int i = 0; i < n; i++) {
            init[i] = i;
            perm[i] = i;
        }
        int ops = 0;
        while (true) {
            ops++;
            update(n, perm);
            if (Arrays.equals(perm, init))
                break;
        }
        return ops;
    }

    public void update(int n, int[] perm) {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            if (i % 2 == 0)
                arr[i] = perm[i / 2];
            else
                arr[i] = perm[n / 2 + (i - 1) / 2];
        }
        System.arraycopy(arr, 0, perm, 0, n);
    }
}

############

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;
            }
        }
    }
}

  • // OJ: https://leetcode.com/contest/weekly-contest-234/problems/minimum-number-of-operations-to-reinitialize-a-permutation/
// Time: O(logN)
// Space: O(1)
class Solution {
public:
    int reinitializePermutation(int n) {
        int i = 1, cnt = 0;
        while (true) {
            ++cnt;
            if (i < n / 2) {
                i *= 2;
            } else  {
                i = n - i - 1;
                i *= 2;
                i = n - i - 1;
            }
            if (i == 1) break;
        }
        return cnt;
    }
};

  • 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

############

# 1806. Minimum Number of Operations to Reinitialize a Permutation
# https://leetcode.com/problems/minimum-number-of-operations-to-reinitialize-a-permutation/

class Solution:
    def reinitializePermutation(self, n: int) -> int:
        arr = [i for i in range(n)]
        desired = [i for i in range(n)]
        
        count = 0
        
        while count == 0 or arr != desired:
            tmp = [0] * n 
            
            for i in range(n):
                if i % 2 == 0:
                    tmp[i] = arr[i // 2]
                else:
                    tmp[i] = arr[n // 2 + (i-1) // 2]
            
            arr = tmp
            count += 1
            
        return count
        


  • 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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