27. Next Permutation

Topic :

arrays

Difficulty :

medium

Problem Link :

link

problem statement

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

solution

// find the index_1 such that a[index_1]<a[index_1+1]//
// find index_2 such that a[index_2]>a[index_1] search from the end//  
// swap (a[index_1] ,a[index_2]) //
// reverse (index+1 --> last)//
import java.io.*;
import java.util.*;
class NextPermutation
{ 
    static void reverse (int a[], int ini ,int fin)
    { while (ini<fin)
        { swap(a,ini++,fin--);
        }
    }
    static void swap(int a[],int i,int j)
    { int temp=a[i];
        a[i]=a[j];
        a[j]=temp;
    }
    static void display(int a[])
    { for(int ele : a)
        {System.out.print(ele+" ");}
    }
    static void Nextpermutation(int a[])
    { if(a==null||a.length<=1)
        return ;
        int i=a.length-2; // start from the second last element cause that can be the break point
        while (i>=0 && a[i]>=a[i+1])
          i--;
        
          if(i>=0) // if and only if we find a break point i.e our ind_1
          { // we search for ind_2
             int j=a.length-1; // again start from the end 
             while (a[j]<=a[i]) 
                j--;
                
                swap(a,i,j);
            }
            
            reverse (a,i+1,a.length-1);   // reverse 
        
        
    }
   public static void main(String args[])
   { int a[]={1,2,3};
       Nextpermutation(a);
      display(a);
    }
}