Medium
Minimum Number of Operations to Reinitialize a Permutation — C++
Full explanation · Time O(sqrt(n)) · Space O(sqrt(n))
// Time: O(sqrt(n))
// Space: O(sqrt(n))
class Solution {
public:
int reinitializePermutation(int n) {
return 1 + discrete_log(2, n / 2, n - 1); // find min x s.t. 2^x mod (n-1) = n/2, result is x + 1
}
private:
// reference: https://cp-algorithms.com/algebra/discrete-log.html
int discrete_log(int a, int b, int m) {
a %= m, b %= m;
int n = sqrt(m) + 1;
int an = 1;
for (int i = 0; i < n; ++i) {
an = (an * 1ll * a) % m;
}
unordered_map<int, int> vals;
for (int q = 0, curr = b; q <= n; ++q) {
vals[curr] = q;
curr = (curr * 1ll * a) % m;
}
for (int p = 1, curr = 1; p <= n; ++p) {
curr = (curr * 1ll * an) % m;
if (vals.count(curr)) {
return n * p - vals[curr];
}
}
return -1;
}
};
// Time: O(n)
// Space: O(1)
class Solution2 {
public:
int reinitializePermutation(int n) {
if (n == 2) {
return 1;
}
int result = 0;
for (int i = 1; !result || i != 1; i = i * 2 % (n - 1)) { // find cycle length
++result;
}
return result;
}
};
// Time: O(n)
// Space: O(1)
class Solution3 {
public:
int reinitializePermutation(int n) {
int result = 0;
for (int i = 1; !result || i != 1; i = (i % 2 == 0) ? i / 2 : n / 2 + (i - 1) / 2) { // find cycle length
++result;
}
return result;
}
};