std::prev_permutation

Defined in header `<algorithm>`
template< class BidirIt > bool prev_permutation( BidirIt first, BidirIt last );	(1)	(constexpr since C++20)
template< class BidirIt, class Compare > bool prev_permutation( BidirIt first, BidirIt last, Compare comp );	(2)	(constexpr since C++20)

Transforms the range [first, last) into the previous permutation. Returns true if such permutation exists, otherwise transforms the range into the last permutation (a descending order) and returns false.

1) The set of all permutations is ordered lexicographically with respect to operator<(until C++20)std::less{}(since C++20).

2) The set of all permutations is ordered lexicographically with respect to comp.

If the type of *first is not Swappable(until C++11)BidirIt is not ValueSwappable(since C++11), the behavior is undefined.

Parameters

first, last	-	the pair of iterators defining the range of elements to permute
comp	-	comparison function object (i.e. an object that satisfies the requirements of Compare) which returns `true` if the first argument is less than the second. The signature of the comparison function should be equivalent to the following: boolcmp(constType1&a,constType2&b); While the signature does not need to have `const&`, the function must not modify the objects passed to it and must be able to accept all values of type (possibly const) `Type1` and `Type2` regardless of value category (thus, `Type1&` is not allowed, nor is `Type1` unless for `Type1` a move is equivalent to a copy(since C++11)). The types Type1 and Type2 must be such that an object of type BidirIt can be dereferenced and then implicitly converted to both of them.
Type requirements
-`BidirIt` must meet the requirements of ValueSwappable and LegacyBidirectionalIterator.

Return value

true if the new permutation precedes the old in lexicographical order. false if the first permutation was reached and the range was reset to the last permutation.

Exceptions

Any exceptions thrown from iterator operations or the element swap.

Complexity

Given \(\scriptsize N\)N as std::distance(first, last):

1,2) At most \(\scriptsize \frac{N}{2}\)N/2 swaps.

Possible implementation

template<class BidirIt>
bool prev_permutation(BidirIt first, BidirIt last)
{
    if (first == last)
        return false;
    BidirIt i = last;
    if (first == --i)
        return false;
    
    while (1)
    {
        BidirIt i1, i2;
        
        i1 = i;
        if (*i1 < *--i)
        {
            i2 = last;
            while (!(*--i2 < *i))
                ;
            std::iter_swap(i, i2);
            std::reverse(i1, last);
            return true;
        }
        
        if (i == first)
        {
            std::reverse(first, last);
            return false;
        }
    }
}

Notes

Averaged over the entire sequence of permutations, typical implementations use about 3 comparisons and 1.5 swaps per call.

Implementations (e.g. MSVC STL) may enable vectorization when the iterator type satisfies LegacyContiguousIterator and swapping its value type calls neither non-trivial special member function nor ADL-found swap.

Example

The following code prints all six permutations of the string "cab" in reverse order.

#include <algorithm>
#include <iostream>
#include <string>

int main()
{
    std::string s = "cab";
    
    do
    {
        std::cout << s << ' ';
    }
    while (std::prev_permutation(s.begin(), s.end()));
    
    std::cout << s << '\n';
}

Output:

cab bca bac acb abc cba

is_permutation (C++11)	determines if a sequence is a permutation of another sequence (function template & algorithm function object)
ranges::is_permutation (C++20)
next_permutation	generates the next greater lexicographic permutation of a range of elements (function template & algorithm function object)
ranges::next_permutation (C++20)
ranges::prev_permutation (C++20)	generates the next smaller lexicographic permutation of a range of elements (algorithm function object)