std::next_permutation and prev_permutation in C++

std::next_permutation

It is used to rearrange the elements in the range [first, last) into the next lexicographically greater permutation. A permutation is each one of the N! possible arrangements the elements can take (where N is the number of elements in the range). Different permutations can be ordered according to how they compare lexicographically to each other. The complexity of the code is O(n*n!) which also includes printing all the permutations.

Syntax: 

template 
bool next_permutation (BidirectionalIterator first,
                       BidirectionalIterator last);
Parameters: 
first, last : Bidirectional iterators to the initial
and final positions of the sequence. The range 
used is [first, last), which contains all the elements 
between first and last, including the element pointed 
by first but not the element pointed by last.

return value: 
true : if the function could rearrange 
the object as a lexicographically greater permutation.
Otherwise, the function returns false to indicate that 
the arrangement is not greater than the previous, 
but the lowest possible (sorted in ascending order).

Application: next_permutation is to find the next lexicographically greater value for a given array of values. 

Examples: 

Input : next permutation of 1 2 3 is 
Output : 1 3 2

Input : next permutation of 4 6 8 is 
Output : 4 8 6

C++

// C++ program to illustrate
// next_permutation example

// this header file contains next_permutation function
#include <algorithm>
#include <iostream>
using namespace std;

int main()
{
    int arr[] = { 1, 2, 3 };

    sort(arr, arr + 3);

    cout << "The 3! possible permutations with 3 elements:\n";
    do {
        cout << arr[0] << " " << arr[1] << " " << arr[2] << "\n";
    } while (next_permutation(arr, arr + 3));

    cout << "After loop: " << arr[0] << ' '
         << arr[1] << ' ' << arr[2] << '\n';

    return 0;
}

Output

The 3! possible permutations with 3 elements:
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
After loop: 1 2 3

Time Complexity: O(N*N!) The next_permutation() function takes O(N) time to find the next permutation and there are N! number of permutations for an array of size N.
Auxiliary Space: O(1) No auxiliary space is used.

std::prev_permutation

It is used to rearrange the elements in the range [first, last) into the previous lexicographically-ordered permutation. A permutation is each one of the N! possible arrangements the elements can take (where N is the number of elements in the range). Different permutations can be ordered according to how they compare lexicographically to each other. The time complexity of the code is O(n*n!) as each permutation takes linear time. 

Syntax : 

template 
bool prev_permutation (BidirectionalIterator first,
                         BidirectionalIterator last );
parameters: 
first, last : Bidirectional iterators to the initial
and final positions of the sequence. The range 
used is [first, last), which contains all the
elements between first and last, including 
the element pointed by first but not the element
pointed by last.

return value: 
true : if the function could rearrange 
the object as a lexicographically smaller permutation.
Otherwise, the function returns false to indicate that 
the arrangement is not less than the previous, 
but the largest possible (sorted in descending order).

Application: prev_permutation is to find the previous lexicographically smaller value for a given array of values. 

Examples:  

Input : prev permutation of 3 2 1 is 
Output : 3 1 2 

Input : prev permutation of 8 6 4 is 
Output :8 4 6

C++

// C++ program to illustrate
// prev_permutation example

// this header file contains prev_permutation function
#include <algorithm>

#include <iostream>
using namespace std;
int main()
{
    int arr[] = { 1, 2, 3 };

    sort(arr, arr + 3);
    reverse(arr, arr + 3);

    cout << "The 3! possible permutations with 3 elements:\n";
    do {
        cout << arr[0] << " " << arr[1] << " " << arr[2] << "\n";
    } while (prev_permutation(arr, arr + 3));

    cout << "After loop: " << arr[0] << ' ' << arr[1]
         << ' ' << arr[2] << '\n';

    return 0;
}

Output

The 3! possible permutations with 3 elements:
3 2 1
3 1 2
2 3 1
2 1 3
1 3 2
1 2 3
After loop: 3 2 1

Time Complexity: O(N*N!) The prev_permutation() function takes O(N) time to find the previous permutation and there are N! number of permutations for an array of size N.
Auxiliary Space: O(1) No auxiliary space is used.