In C++, how are the different permutations generated by std::next_permutation conceptually ordered?

Introduction / Context:
The C++ standard library provides std::next_permutation and std::prev_permutation to iterate through permutations of a sequence. These algorithms are heavily used in combinatorial problems, interview questions, and competitive programming. The order in which permutations are generated is not arbitrary. The question tests whether you know that std::next_permutation produces permutations in lexicographical order according to the ordering of the elements in the range.

Given Data / Assumptions:

• We are using C++ and the algorithms std::next_permutation and std::prev_permutation from the <algorithm> header.
• The input sequence is stored in a container such as std::vector, std::string, or an array.
• A strict weak ordering for element comparison is provided, usually the default operator<.
• We want to understand the conceptual order of permutations, not internal implementation details.

Concept / Approach:
Lexicographical order generalises dictionary order to sequences of arbitrary elements. Given a comparison operator, one sequence is considered less than another if it has a smaller element in the first position where they differ. std::next_permutation rearranges the sequence into the next permutation in lexicographical order. When the sequence is already at the largest possible lexicographical arrangement, std::next_permutation transforms it back into the smallest arrangement and returns false. This contract means that if you start with a sorted range and repeatedly call std::next_permutation, you will enumerate all permutations in sorted lexicographical order.

Step-by-Step Solution:
Step 1: Sort the sequence in ascending order to obtain the smallest lexicographical permutation. Step 2: Call std::next_permutation to transform the sequence into the next lexicographically larger arrangement. Step 3: Observe that each call produces a sequence that compares greater than the previous one under lexicographical comparison, until the maximum permutation is reached. Step 4: Recognise that this behaviour corresponds exactly to lexicographical ordering of permutations based on element values. Step 5: Choose option a, which states that permutations are generated in lexicographical order when comparing elements in the range.

Verification / Alternative check:
For a simple example, consider the sequence 1 2 3. Starting from 1 2 3 and calling std::next_permutation successively yields 1 3 2, 2 1 3, 2 3 1, 3 1 2, and 3 2 1. These sequences form the lexicographical order of permutations of three elements. When 3 2 1 is reached, the next call resets the sequence to 1 2 3 and returns false. This behaviour confirms the lexicographical nature of the ordering.

Why Other Options Are Wrong:
Option b suggests that permutations are generated by always placing the highest element first, which would not produce all permutations in a consistent order. Option c claims a random order without any defined pattern, which contradicts the deterministic contract of std::next_permutation. Option d refers to memory addresses rather than values, but lexicographical order is defined on values and comparison operations, not on where elements are stored in memory.

Common Pitfalls:
A common mistake is to call std::next_permutation on an unsorted initial sequence and expect to iterate through all permutations. While the function still works, the order will be lexicographical starting from that specific arrangement, not necessarily beginning from the globally smallest permutation. Another pitfall is confusing lexicographical order with numeric or lexicographic sorting of individual elements rather than entire sequences. For reliable generation of all permutations in increasing order, always sort the range first and then repeatedly apply std::next_permutation.

Final Answer:
The permutations generated by std::next_permutation are ordered in lexicographical order based on the comparison of the elements in the range.