31. Next Permutation (Medium)

Brute Force

• We first generate all the permutations and store them in permutations vector.
• Then we find the given vector in the permutations vector.if we found then we return its next vector as and.
• If given the last vector return the first vector from the permutations vector.
• TC: O(N!*N) - Because there are N! orders and N is the length of every array.
• SC: O(N!) - To store the all permutations, there are N! permutations.

Inbuilt next_permutation

• we can also solve the problem in-place using in-built next_permutation(a.being(),a.end()) function in c++.
• But in an interview this is not expected.

Code

class Solution {
public:
    void nextPermutation(vector<int>& nums){
        next_permutation(nums.begin(), nums.end());
    }
};

O(N) Time solution.

• INTUITION:- If we Observe the dictionary of order(permeation order) we can find that there is always Triangle like structure.
• ALGORITHM:- 1. From back find the largest index k such that nums[k] < nums[k + 1]. If no such index exists, just reverse nums and done. 2.From back find the largest index l > k such that nums[k] < nums[l]. 3.Swap nums[k] and nums[l]. 4.Reverse the sub-array nums[k + 1:].
• REFERENCE:- C++ from Wikipedia(leetcode discussion)

Code

class Solution {
public:
    void nextPermutation(vector<int>& nums)
    {
        int n = nums.size(), k, l;
        for (k = n - 2; k >= 0; k--) { // Step 1
            if (nums[k] < nums[k + 1]) {
                break;
            }
        }
        if (k < 0) {
            reverse(nums.begin(), nums.end());
        } else {
            for (l = n - 1; l > k; l--) { // Step 2
                if (nums[l] > nums[k]) {
                    break;
                }
            }
            swap(nums[k], nums[l]); //Step 3
            reverse(nums.begin() + k + 1, nums.end()); //Step 4
        }
    }
};