Array_Leetcode_Next_Permutation_31.cpp

/*31. Next Permutation

A permutation of an array of integers is an arrangement of its members into a sequence or linear order.

For example, for arr = [1,2,3], the following are all the permutations of arr: [1,2,3], [1,3,2], [2, 1, 3], [2, 3, 1], [3,1,2], [3,2,1].
The next permutation of an array of integers is the next lexicographically greater permutation of its integer. More formally, if all the permutations of the array are sorted in one container according to their lexicographical order, then the next permutation of that array is the permutation that follows it in the sorted container. If such arrangement is not possible, the array must be rearranged as the lowest possible order (i.e., sorted in ascending order).

For example, the next permutation of arr = [1,2,3] is [1,3,2].
Similarly, the next permutation of arr = [2,3,1] is [3,1,2].
While the next permutation of arr = [3,2,1] is [1,2,3] because [3,2,1] does not have a lexicographical larger rearrangement.
Given an array of integers nums, find the next permutation of nums.

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

Example 1:

Input: nums = [1,2,3]
Output: [1,3,2]
Example 2:

Input: nums = [3,2,1]
Output: [1,2,3]
Example 3:

Input: nums = [1,1,5]
Output: [1,5,1]
 

Constraints:

1 <= nums.length <= 100
0 <= nums[i] <= 100*/


class Solution {
public:
    void nextPermutation(vector<int>& nums) {
//optimal O(3N) <----
      int n = nums.size();
      int j =-1; //agar koi  bhi element chota nahi mila uske agle se to matlab reverse karna hai bs array ko (LAST PERMUTATION)
      
//edge case like array length of 1 and 2 will simple trigger j =-1 and simply reverse the whole array and return void.
      for(int i=n-2;i>=0;i--)  //runtime error se bachane ke liye
      {
        if(nums[i]<nums[i+1])
        {
           j = i; 
           cout<<j;
          break;
        }
        
      }

      if(j==-1) 
      {
        
       reverse(nums.begin(),nums.end());

        return ;
      }

      
      //kyunki is question ke hissab se permutation me piche se pehla bada ki nums[j] se bada hoga aur minimum hoga j index ke baad
      for(int i =n-1;i>j;i--) 
      {
          if(nums[i] > nums[j])
          {
            swap(nums[i],nums[j]);
            break;
          }  
      }
      

      reverse(nums.begin() + j+1 , nums.end());
        
    }
};