All the permutations of a word when arranged in a dictionary, the order of words so obtained is called lexicographical order. in Simple words, it is the one that has all its elements sorted in ascending order, and the largest has all its elements sorted in descending order.
lexicographically is nothing but the greater permutation of it.
For reference: a b c d e f g h i j k l m n o p q r s t u v w x y z
For example, lexicographically next permutation of “gfg” is “ggf” and the 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’.
Example:
CPP
#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;
}
|
Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(1), as constant extra space is used.
The same program can also be implemented without using STL. Below is the code snippet for the same.
The Idea is Based on the Following Facts:
- A sequence sorted in descending order does not have the next permutation. For example “edcba” does not have the next permutation.
- For a sequence that is not sorted in descending order for example “abedc”, we can follow these steps.
- a) Traverse from the right and find the first item that is not following the ascending order. For example in “abedc”, the character ‘b’ does not follow the ascending order.
- b) Swap the found character with the closest greater (or smallest greater) element on the right side of it. In the case of “abedc”, we have ‘c’ as the closest greater element. After swapping ‘b’ and ‘c’, the string becomes “acedb”.
- c) After swapping, reverse the string after the position of the character found in step a. After reversing the substring “edb” of “acedb”, we get “acbde” which is the required next permutation.
Optimizations in steps b) and c)
a) Since the sequence is sorted in decreasing order, we can use binary search to find the closest greater element.
c) Since the sequence is already sorted in decreasing order (even after swapping as we swapped with the closest greater), we can get the sequence sorted (in increasing order) after reversing it.
CPP
#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 ;
}
}
int main()
{
string s = { "gfg" };
bool val = nextpermutation(s);
if (val == false )
cout << "No Word Possible" << endl;
else
cout << s << endl;
return 0;
}
|
Java
import java.io.*;
public class Main {
public static char [] nextPermutation(String s) {
char [] arr = s.toCharArray();
int n = arr.length;
int i = n - 2 ;
while (i >= 0 && arr[i] >= arr[i+ 1 ]) {
i--;
}
if (i < 0 ) {
String st = "No next Permutation possible" ;
char [] ar = st.toCharArray();
return ar;
}
int j = n - 1 ;
while (j >= 0 && arr[j] <= arr[i]) {
j--;
}
swap(arr, i, j);
rev(arr, i+ 1 , n- 1 );
return arr;
}
private static void swap( char [] arr, int i, int j) {
char temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
private static void rev( char [] arr, int start, int end) {
while (start < end) {
swap(arr, start, end);
start++;
end--;
}
}
public static void main(String[] args)
{
String str2 = "gfg" ;
System.out.println(nextPermutation(str2));
}
}
|
Python3
def next_permutation(s: str ) - > str :
arr = list (s)
n = len (arr)
i = n - 2
while i > = 0 and arr[i] > = arr[i + 1 ]:
i - = 1
if i < 0 :
return "No next Permutation possible"
j = n - 1
while j > = 0 and arr[j] < = arr[i]:
j - = 1
arr[i], arr[j] = arr[j], arr[i]
rev(arr, i + 1 , n - 1 )
return ''.join(arr)
def rev(arr: list , start: int , end: int ) - > None :
while start < end:
swap(arr, start, end)
start + = 1
end - = 1
def swap(arr: list , i: int , j: int ) - > None :
temp = arr[i]
arr[i] = arr[j]
arr[j] = temp
if __name__ = = '__main__' :
s = "gfg"
print (next_permutation(s))
|
C#
using System;
class GFG {
public static char [] NextPermutation( string s) {
char [] arr = s.ToCharArray();
int n = arr.Length;
int i = n - 2;
while (i >= 0 && arr[i] >= arr[i+1]) {
i--;
}
if (i < 0) {
string st = "No next Permutation possible" ;
char [] ar = st.ToCharArray();
return ar;
}
int j = n - 1;
while (j >= 0 && arr[j] <= arr[i]) {
j--;
}
Swap(arr, i, j);
Reverse(arr, i+1, n-1);
return arr;
}
private static void Swap( char [] arr, int i, int j) {
char temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
private static void Reverse( char [] arr, int start, int end) {
while (start < end) {
Swap(arr, start, end);
start++;
end--;
}
}
public static void Main( string [] args) {
string str2 = "gfg" ;
Console.WriteLine(NextPermutation(str2));
}
}
|
Javascript
function next_permutation(s) {
let arr = Array.from(s);
let n = arr.length;
let i = n - 2;
while (i >= 0 && arr[i] >= arr[i+1]) {
i--;
}
if (i < 0) {
return "No next Permutation possible" ;
}
let j = n - 1;
while (j >= 0 && arr[j] <= arr[i]) {
j--;
}
let temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
rev(arr, i+1, n-1);
return arr.join( '' );
}
function rev(arr, start, end) {
while (start < end) {
swap(arr, start, end);
start++;
end--;
}
}
function swap(arr, i, j) {
let temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
let s = "gfg" ;
console.log(next_permutation(s));
|
Time Complexity:
- In the worst case, the first step of next_permutation takes O(n) time.
- The binary search takes O(log n) time.
- The reverse takes O(n) time.
Overall time complexity is O(n). Where n is the length of the string.
Auxiliary Space is O(1), because no extra space is used.
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Last Updated :
06 Apr, 2023
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