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# std::next_permutation

From cppreference.com
< cpp‎ | algorithm

Defined in header `<algorithm>` |  |
template< class BidirIt > bool next_permutation( BidirIt first, BidirIt last ); | (1) |
template< class BidirIt, class Compare > bool next_permutation( BidirIt first, BidirIt last, Compare comp ); | (2) |

Transforms the range `[first, last)` into the next permutation from the set of all permutations that are lexicographically ordered with respect to `operator<` or `comp`. Returns true if such permutation exists, otherwise transforms the range into the first permutation (as if by `std::sort(first, last)`) and returns false.

### [edit] Parameters

first, last | - | 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: bool cmp(const Type1 &a, const Type2 &b); The signature does not need to have const &, but the function object must not modify the objects passed to it. 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 `BidirectionalIterator`.

### [edit] Return value

true if the new permutation is lexicographically greater than the old. false if the last permutation was reached and the range was reset to the first permutation.

### [edit] Exceptions

Any exceptions thrown from iterator operations or the element swap.

### [edit] Complexity

At most N/2 swaps, where N = std::distance(first, last).

### [edit] Possible implementation

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

### [edit] Example

The following code prints all three permutations of the string "aba"

Run this code

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

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

Output:

    aab
    aba
    baa

### [edit] See also

is_permutation (C++11) | determines if a sequence is a permutation of another sequence (function template) [edit]
prev_permutation | generates the next smaller lexicographic permutation of a range of elements (function template) [edit]

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