LeetCode 1806: Minimum Number of Operations to Reinitialize a Permutation
LeetCode 1806 Solution Explanation
Explanation
To solve this problem, we can simulate the operations described in the problem statement. We start with the initial permutation and repeatedly apply the operations until we reach the initial permutation again. We keep track of the number of operations performed and return this count as the minimum number of operations required.
Algorithm:
- Start with the initial permutation perm where perm[i] = i.
- Create a new array arr of size n.
- Perform the operations on perm to generate arr according to the given rules.
- Assign arr to perm.
- Increment the count of operations.
- Repeat steps 3-5 until perm is equal to the initial permutation.
- Return the count of operations.
Time Complexity: O(n) Space Complexity: O(n)
LeetCode 1806 Solutions in Java, C++, Python
class Solution {
public int reinitializePermutation(int n) {
int[] perm = new int[n];
for (int i = 0; i < n; i++) {
perm[i] = i;
}
int[] arr = new int[n];
int count = 0;
while (true) {
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];
}
}
count++;
perm = arr.clone();
boolean isInitialPerm = true;
for (int i = 0; i < n; i++) {
if (perm[i] != i) {
isInitialPerm = false;
break;
}
}
if (isInitialPerm) {
break;
}
}
return count;
}
}Interactive Code Editor for LeetCode 1806
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