Write a C Program to Print All Permutations of a Given String _ GeeksforGeeks

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Write a C program to print all permutations of a given string

A permutation, also called an arrangement number or order, is a rearrangement of the elements of an

ordered list S into a one-to-one correspondence with S itself. A string of length n has n! permutation.

Source: Mathword(http://mathworld.wolfram.com/Permutation.html)

Below are the permutations of string ABC.

ABC, ACB, BAC, BCA, CAB, CBA
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Here is a solution using backtracking.

# include

/* Function to swap values at two pointers */

void swap (char *x, char *y){

char temp;

temp = *x;

*x = *y;

*y = temp;

}

/* Function to print permutations of string

This function takes three parameters:

1. String

2. Starting index of the string

3. Ending index of the string. */void permute(char *a, int i, int n)

{

int j;

if (i == n)

printf("%s\n", a);

else

{

for (j = i; j





}

Output:

ABC

ACB

BAC

BCA

CBA

CAB

Algorithm Paradigm: Backtracking

Time Complexity: O(n*n!)

Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same

problem.

Related Tpoics:

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Find the first non-repeating character from a stream of characters

Dynamic Programming | Set 33 (Find if a string is interleaved of two other strings)

Remove b and ac from a given stringDynamic Programming | Set 29 (Longest Common Substring)

Write your own atoi()

String matching where one string contains wildcard characters

Count words in a given string

97Like Tweet 1 8

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Kai

The solution doesn't work if the string contains duplicate charaters like 'AAB' , obviously thiscase should be handled by the permutation algo.

Pratham

I wrote the code for variable length string but the program gives segmentation fault.

#include

void swap(char *i,char *j)

{

char t;t=*i;

*i=*j;

*j=t;

}

void perm(char *str,int i,int n)

{

int j;

if(i==n)

{printf("%s",str);

}

else

{

for(j=i;j





int n;

printf("Enter the string");

scanf("%s",string);

n=strlen(string);

perm(string,0,n);

return 0;

}

please tell me where i went wrong

Javed

in the main() funcion you need to pass 1 less than strlen(string) i.e the maximum actu

index in the string i.e. n-1 to the function perm().

Pratham

I tried it javed but still it is giving me the same segmentation fault .

I am using dev C++ ide

SHANTANU AGRAWAL

http://ideone.com/WPCW48

Ankit

Can someone tell me that when I use char *temp in the swap function why doesn't my progra

work?

sonu

When length of string is unknown ,i.e. Char array can be of any size , Pass by referen

is used .

Vignesh A

The code works fine with out Back Tracking.. Please explain the logic

void permute(char *a, int i, int n)

{

int j;

if (i == n)

printf("%s\n", a);

else
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{

for (j = i; j





if(a+i == a+j && i != j) continue;

// rest of the code as given

}

mragrid

What if i want the algorithm to stop if it reaches a limit ? for example if we have 3628800 (10!

permutations and i want it to stop when we have generated 1000 permutations ? and what timcomplexity we will have that way ?

Jingguo Yao

Lexicographic permutation generation algorithm handles duplicated characters. See page 31

of The Art of Computer Programming, Volume 4A, Chapter 7.2.1.2.

importsys

MIN_INT = -sys.maxint - 1deflexicographic_order_generation(A):

A[0:0] = [MIN_INT]

A.sort()

n = len(A) - 1

whileTrue:

print("".join([str(x) forx inA[1:]]))

j = n - 1

whileA[j] >= A[j + 1]:

j -= 1 ifj == 0:

return

l = n

whileA[j] >= A[l]:

l -= 1

A[j], A[l] = A[l], A[j]

suffix = A[j+1:]

suffix.reverse()

A = A[:(j + 1)] + suffix

A = [1, 2, 3]

lexicographic_order_generation(A)

A = list("AABAAC")

lexicographic_order_generation(A)
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mathsmaths

thanks

DevilGeek

Just like a tower of Hanoi.

atiq

//Repeatation will not harm my code....like AABA ... will give result with no repeatation

http://ideone.com/acjCxb

Happy coding

Sriharsha g.r.v well done and thanq

Sriharsha g.r.v

it will not work for repeated character strings i.e "abaaa" it prints many repeated strings

Gaurav Ramesh

you can change the swap condition .. to only swap when character at i and j are notsame..

so your condition inside for will look like:

for(..){

if(a+i == a+j && i != j) continue;

// rest of the code as given..

}

akshita

will you help me in understanding this code plz

Gaurav Ramesh

Hi Akshita,

If you've understood the main code, the change I've suggested is just t
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prevent swapping of the letters when both letters at i and j are same !

thus avoiding the repeated permutations..

derek

not correct,

for example 'aabb'

Gaurav Ramesh

Hey Derek, I don't see why this shouldn't work for your input ! when i a

are both pointing to a or b, they shouldn't be swapped and just continu

Guest

should be

if((*(char *)(a+i) == *(char *)(a+j)) && (i != j)) continue;

Sameer

http://codingrecipies.blogspot...

DarkProtocol

Without recursion is greatly appreciated!!! Thanks in advance.

GuestPost

//check this

//http://www.cplusplus.com/refer...

Puzzled

Can someone explain the time complexity

Neha Garg

i am unable to understand this code ..plz explain it step wise

Guy

the idea is that for every N length string we can calculate the permutation of that string
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the permutations of the rest of the string and then the permutations of the string starts

with the second letter + all the permutations of the rest of the string and so on until we

reach the strings which starts with the last letter of the string + all the permutations of

the rest of the string .

following that logic the recursion tree grows.

the swap simply moves to the start the letter we wanted to start with and is used to

make all the changes "in place" thus not to waste extra space.(if space is not a problem then this could also be done by creating an array of strings

and concatenating the first letter to the already generated string each time instead of t

swap).

Bharath G M

http://www.youtube.com/watch?v...Good one..

all izz well

iam not able to understand this program ....

can anyone trace and explian it plz ..........

Bharath G M

http://www.youtube.com/watch?v..... Good one

Nhan

Time Complexity is O(n!), not O(n*n!)

First for-loop is n times, second for-loop in the first recursive call is (n-1) times and so on.

Hence O(n!), or more tightly _Theta_(n!) as both lower and upper bounds are (n!)

Silent

what is the need of backtrack step here?? answer is still correct without this step..

Marsha Donna

the need 4 backtracking is to always work with the original string..in the second iterati

after a and b are swapped the string becomes bac..when the function is recurively

called..the permutation of bac keepin b fixed are Bac and Bca...if we leave the string a

bca..in the next iteration when we have to swap a and c(in original string abc) to form

cba it would not be possible correctly unless we revert Bca back to bac...then it would

go back change bac to the original string abc after which a and c are swapped to cba
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the next iteration...see the figure above

Pranjal

Its correct for the ABC case --- Try it with 4 characters and you will see the importanc

of backtracking !!!

Guest

// TreeToDLL.cpp : Defines the entry point for the console application.

#include "stdafx.h"

#include

usingnamespacestd;

structNode

{

intdata; Node *left;

Node *right;

};

classTree

{

staticintl;

staticintflip;

staticintleafdepth;

staticNode * prev;staticNode * previous;

staticintcount;

public:

Node *head;

intmaxdepth;

Tree()

{

head = NULL;

maxdepth = 0;

}

voidInorder(Node *node)

{

if(node)

{

Inorder(node->left);

std::cout





}

voidTreetoDLL(Node *node)

{

if(node)

{

TreetoDLL(node->left);

if(previous)

{

previous->right = node;

node->left = previous;

}

if(count == 0)

{

head = node;

count = count + 1;

}

previous = node;

TreetoDLL(node->right);

}

}

voidPrintDLL()

{

Node *tmp = head;

while(tmp)

{

std::coutright;

}

}

voidInsertNode(Node *node,intdata)

{

Node * tmp = node;

if(head == NULL)

{

head = newNode;

head->data = data;

head->left = NULL;

head->right = NULL;

return;

}

elseif(tmp)

{

prev = tmp;

if data > tm ->data





{

l=0;

tmp = tmp -> right;

InsertNode(tmp,data);

}

else

{

l=1;

InsertNode(tmp->left,data);

}

}

if(l == 1)

{

prev->left = newNode;

prev->left->data = data;

prev->left->left = NULL;

prev->left->right = NULL;

l = -1;

return;

}

elseif(l == 0)

{

prev->right = newNode;

prev->right->data = data;

prev->right->left = NULL;

prev->right->right = NULL;

l = -1;

return;

}

}

};

intTree::l =0;

intTree::flip =0;

intTree::leafdepth =-1;

intTree::count = 0;

Node * Tree::prev = NULL;

Node * Tree::previous = NULL;

intmain(intargc, char* argv[])

{

Tree t;

t.InsertNode(t.head,10);














t.InsertNode(t.head,11);t.InsertNode(t.head,1);



t.TreetoDLL(t.head);

t.PrintDLL();

return0;

}

raghvendra

#include

#include

#include

using namespace std;

char str[100];

void permutation(int n)

{

char x;

if(n





}

}

intmain()

{

cin>>str;

cout





}

Indra Kumar Gurjar

/* withour recursion */.

#include

#includestruct node

{

char *str;

int curr;

struct node* next;.

};

typedef struct node node;.

node * new_node(char *str, int curr, node* next).

{

char *temp;.

temp=(char*)malloc(strlen(str)+1);

node* ptr=(node*)malloc(sizeof(node));.

ptr->str=temp;

strcpy(ptr->str, str);

Indra Kumar Gurjar

/* withour recursion */.

#include

#include

struct node

{

char *str;

int curr;

struct node* next;.

};

typedef struct node node;.

node * new_node(char *str, int curr, node* next).

{

char *temp;.

temp=(char*)malloc(strlen(str)+1);

node* ptr=(node*)malloc(sizeof(node));.

-> =





strcpy(ptr->str, str);

Irfanali117

publicclasspermute {

staticvoidpermute(intlevel, String permuted,

booleanused[], String original) {

intlength = original.length();

if(level == length) {

System.out.println(permuted);

} else{

for(inti = 0; i < length; i++) {

if(!used[i]) {

used[i] = true;

permute(level + 1, permuted + original.charAt(i),

used, original);

used[i] = false;

}

Raj Kishor

[sourcecode language="C++"]

//permutation of a string

#include

#include

using namespace std;

void permute(string ,string , int);

int main()

{

string str="",dats;

coutdats;

cout





Load more comments

void permute(string str,string dats, int j)

Kamal Zuhairi Zamli

is this vb...no c version of the code ?

vishal

//In C#....

public static void permutation(string strPrefix, string str)

{

int n = str.Length;

if(n == 1)

Console.WriteLine(strPrefix + str);

for (int i = 0; i < str.Length; i++)

{

string prefix = strPrefix + str.Substring(i, 1);

permutation(prefix, str.Substring(0,i) + str.Substring(i+1,n-i-1));

}

}

rajashekar007

What if there are duplicates in the string?
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String String

String Java
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Write a C Program to Print All Permutations of a Given String _ GeeksforGeeks

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