31. Next Permutation 🌟🌟

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such an arrangement is not possible, it must rearrange it as the lowest possible order (i.e., sorted in ascending order).

The replacement must be in place and use only constant extra memory.

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.

O(N) Time solution.

• INTUITION:- If we Observe the dictionary of order(permeation order) we can find that there is always Triangle like structure.
• For better understanding here is striver’s video
• 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

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
        }
    }
};

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());
    }
};