Leetcode 1806: Minimum Number of Operations to Reinitialize a Permutation
You are given an even integer
n and a permutation perm of size n where initially perm[i] = i (0-indexed). In each operation, you create a new array arr based on certain rules and update perm to arr. The goal is to determine the minimum number of operations required to restore perm to its initial state.Problem
Approach
Steps
Complexity
Input: The input consists of an even integer `n` and a permutation `perm` of size `n` where initially `perm[i] = i` (0-indexed).
Example: n = 4
Constraints:
• 2 <= n <= 1000
• n is even.
Output: Return the minimum number of operations required to restore the permutation to its initial state.
Example: Output: 2
Constraints:
• The solution should handle all even integers n within the given range.
Goal: The goal is to find the minimum number of operations required to restore the permutation `perm` to its original state.
Steps:
• Initialize the permutation `perm` as [0, 1, 2, ..., n-1].
• Perform operations to generate a new array `arr` based on the rules described.
• Check if `arr` matches the initial permutation. If not, update `perm` to `arr` and repeat until the match is found.
Goal: The input consists of an even integer `n` and a permutation `perm` where `n` is between 2 and 1000.
Steps:
• n is always even.
• The solution should handle both small and large values of `n` efficiently.
Assumptions:
• The input will always be an even integer n.
• The permutation `perm` is initially [0, 1, 2, ..., n-1].
• Input: n = 4
• Explanation: The initial permutation is `[0, 1, 2, 3]`. After the first operation, the permutation becomes `[0, 2, 1, 3]`. After the second operation, the permutation returns to `[0, 1, 2, 3]`, so it takes 2 operations.
Approach: We will simulate the process of performing the operations and count how many operations it takes to return to the original permutation.
Observations:
• We need to find a way to efficiently check when the permutation returns to the original state.
• Simulating the operation and checking for equality after each step is a straightforward approach to solve the problem.
Steps:
• Initialize the permutation `perm` as [0, 1, 2, ..., n-1].
• Perform operations by creating the array `arr` according to the given rules.
• Compare `arr` with the original permutation and check if they are the same.
• Repeat until the permutation matches the initial state and count the number of operations.
Empty Inputs:
• There will always be an input value for `n` which is even.
Large Inputs:
• For large inputs, the solution should handle the operations efficiently, especially since n can go up to 1000.
Special Values:
• Consider the smallest value of n (n = 2) and ensure the solution works for that edge case.
Constraints:
• The solution should operate efficiently with time complexity proportional to n.
int reinitializePermutation(int n) {
vector<int> prem(n, 0), arr(n, 0);
for(int i = 0; i < n; i++) {
prem[i] = i;
}
int cnt = 1;
while(1) {
for(int i = 0; i < n; i++) {
arr[i] = (i % 2) ? prem[n/2 + (i - 1)/2] : prem[i/2];
}
bool x = false;
for(int i= 0; i < n; i++)
if(arr[i] != i) x = true;
if(!x) break;
prem = arr;
cnt++;
}
return cnt;
}
1 : Function Definition
int reinitializePermutation(int n) {
Define the function `reinitializePermutation` which takes an integer `n` representing the size of the array and returns an integer representing the number of reinitializations required.
2 : Variable Initialization
vector<int> prem(n, 0), arr(n, 0);
Initialize two vectors `prem` and `arr` of size `n` to store the current and new permutation values respectively.
3 : Loop and Recursion
for(int i = 0; i < n; i++) {
Start a loop to iterate through all indices of the vector.
4 : Variable Initialization
prem[i] = i;
Initialize the `prem` array such that each index `i` holds the value `i`.
5 : Variable Initialization
int cnt = 1;
Initialize the counter variable `cnt` to 1, representing the number of reinitializations performed.
6 : Loop and Recursion
while(1) {
Start an infinite loop to continuously apply the reinitialization process until the condition is met.
7 : Loop and Recursion
for(int i = 0; i < n; i++) {
Start a loop to apply the reinitialization rule to each index of the array `arr`.
8 : Mathematical Operations
arr[i] = (i % 2) ? prem[n/2 + (i - 1)/2] : prem[i/2];
Reinitialize each index of the array `arr` according to a specific rule: if `i` is odd, use the second half of `prem`, otherwise use the first half.
9 : Conditionals
bool x = false;
Initialize a boolean variable `x` to track whether the array `arr` is different from the original `prem`.
10 : Loop and Recursion
for(int i= 0; i < n; i++)
Start a loop to compare each index of `arr` to its original value in `prem`.
11 : Conditionals
if(arr[i] != i) x = true;
If any index in `arr` is not equal to its corresponding index `i`, set `x` to true, indicating that reinitialization is still incomplete.
12 : Flow Control
if(!x) break;
If `x` remains false (indicating the array is in its final state), break out of the infinite loop.
13 : Variable Initialization
prem = arr;
Update the `prem` array to match the newly reinitialized `arr` for the next iteration.
14 : Mathematical Operations
cnt++;
Increment the counter `cnt` to reflect the number of reinitialization steps taken.
15 : Return
return cnt;
Return the counter `cnt`, which represents the total number of reinitialization steps required.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: Each operation requires O(n) time for generating the new array and checking for equality. The worst-case scenario is when the solution requires multiple operations to restore the original permutation.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) due to the space required to store the permutation and the intermediate array.
| LeetCode Solutions Library / DSA Sheets / Course Catalog |
|---|
comments powered by Disqus