Lexicographically next permutation in C++

Given a word, find the lexicographically greater permutation of it. For example, lexicographically next permutation of “gfg” is “ggf” and next permutation of “acb” is “bac”.

Note: In some cases, the next lexicographically greater word might not exist, e.g, “aaa” and “edcba”

In C++, there is a specific function that saves us from a lot of code. It’s in the file #include <algorithm>. The function is next_permutation(a.begin(), a.end()). It returns ‘true’ if the function could rearrange the object as a lexicographically greater permutation. Otherwise, the function returns ‘false’.



Let us look at the code snippet here :

filter_none

edit
close

play_arrow

link
brightness_4
code

// Find the next lexicographically
// greater permutation of a word
  
#include <algorithm>
#include <iostream>
  
using namespace std;
  
int main()
{
    string s = { "gfg" };
    bool val
        = next_permutation(s.begin(),
                           s.end());
    if (val == false)
        cout << "No Word Possible"
             << endl;
    else
        cout << s << endl;
    return 0;
}

chevron_right


Output:

ggf

The same program can also be implemented without using STL. Below is the code snippet for the same:

filter_none

edit
close

play_arrow

link
brightness_4
code

// Find the next lexicographically
// greater permutation of a word
  
#include <iostream>
  
using namespace std;
  
void swap(char* a, char* b)
{
    if (*a == *b)
        return;
    *a ^= *b;
    *b ^= *a;
    *a ^= *b;
}
void rev(string& s, int l, int r)
{
    while (l < r)
        swap(&s[l++], &s[r--]);
}
  
int bsearch(string& s, int l, int r, int key)
{
    int index = -1;
    while (l <= r) {
        int mid = l + (r - l) / 2;
        if (s[mid] <= key)
            r = mid - 1;
        else {
            l = mid + 1;
            if (index == -1 || s[index] >= s[mid])
                index = mid;
        }
    }
    return index;
}
  
bool nextpermutation(string& s)
{
    int len = s.length(), i = len - 2;
    while (i >= 0 && s[i] >= s[i + 1])
        --i;
    if (i < 0)
        return false;
    else {
        int index = bsearch(s, i + 1, len - 1, s[i]);
        swap(&s[i], &s[index]);
        rev(s, i + 1, len - 1);
        return true;
    }
}
  
// Driver code
int main()
{
    string s = { "gfg" };
    bool val = nextpermutation(s);
    if (val == false)
        cout << "No Word Possible" << endl;
    else
        cout << s << endl;
    return 0;
}
// This code is contributed by Mysterious Mind

chevron_right


Output:

ggf

Time Complexity:

  1. In the worst case, the first step of next_permutation takes O(n) time.
  2. Binary search takes O(logn) time.
  3. Reverse takes O(n) time.

Overall time complexity is O(n).
Where n is the length of the string.

Reference: http://www.cplusplus.com/reference/algorithm/next_permutation/

This article is contributed by Harshit Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up