std::next_permutation and prev_permutation in C++

Last Updated : 20 Jan, 2026

In C++, next_permutation() and prev_permutation() (in <algorithm>) rearrange a container’s elements into the next or previous lexicographical order among all possible permutations.

std::next_permutation()

The std::next_permutation in C++ is used to rearrange the elements of the given range [first, last) to the lexicographical larger permutation if it exists.

Syntax

std::next_permutation(first, last);

Parameters

  • first: Iterator to the first element of the given range.
  • last: Iterator to the theoretical element just after the last element of the given range.

Return Value

  • Returns true if the container could be rearranged to the to the lexicographical larger permutation.
  • Returns false otherwise.
C++ 
#include <bits/stdc++.h>
using namespace std;

int main() {

    vector<int> v = {1, 2, 3};
    do {
        for (auto i: v) cout << i << " ";
        cout << endl;
    } while (next_permutation(v.begin(), v.end()));

    return 0;
}

Output
1 2 3 
1 3 2 
2 1 3 
2 3 1 
3 1 2 
3 2 1

Explanation: The total number of permutations of vector of 3 elements is 3! = 6. We already took the smallest possible permutation as the starting point, so we were able to print all the permutations using next_permutation().

std::prev_permutation()

The std::prev_permutation is used to rearrange the elements of the given range [first, last) in the lexicographical smaller permutation if it exists.

Syntax

std::prev_permutataion(first, last)

Parameters

  • first: Iterator to the first element of the given range.
  • last: Iterator to the theoretical element just after the last element of the given range.

Return Value

  • Returns true if the container could be rearranged to the to the lexicographical smaller permutation.
  • Returns false otherwise.
C++ 
#include <bits/stdc++.h>
using namespace std;

int main() {
  vector<int> v = {2, 1, 3};
    do {
        for (auto i: v) cout << i << " ";
        cout << endl;
    } while (prev_permutation(v.begin(), v.end()));

    return 0;
}

Output
2 1 3 
1 3 2 
1 2 3

Explanation: The total number of permutations of vector of 3 elements is 3! = 6. But we were only able to print 3 permutations because we didn't took the largest permutation as starting point for prev_permutation() function. So, all the permutation greater than the permutation {2, 1, 3} are left out.

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