Medium

Minimum Number of Operations to Reinitialize a PermutationC++

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;
    }
};