The set [1, 2, 3, ..., n] contains a total of n! unique permutations.
By listing and labeling all of the permutations in order, we get the following sequence for n = 3:
"123"
"132"
"213"
"231"
"312"
"321"
Given n and k, return the kth permutation sequence.
- Generate all the possible permutation.
- Sort the permutations.
- Return the kth permutation.
- TC: O(N! * N) + O(N! Log N!), O(N!) - generate every possible permutation , another O(N) time is to make a deep copy and store every sequence in the data structure. Also, O(N! Log N!) time required to sort the data structure
- SC: O(N!)
- Time Complexity: O(N) * O(N) = O(N^2)
- Space Complexity: O(N)
class Solution {
public:
string getPermutation(int n, int k)
{
int fact = 1;
vector<int> numbers;
for (int i = 1; i < n; i++) {
fact = fact * i;
numbers.push_back(i);
}
numbers.push_back(n);
string ans = "";
k--;
while (true) {
ans += to_string(numbers[k / fact]);
numbers.erase(numbers.begin() + k / fact);
if (numbers.size() == 0) break;
k %= fact;
fact /= numbers.size();
}
return ans;
}
};