Interested to Buy Any Domain ? << Click Here >> for more details...
Write out a function that prints out all the permutations of
a string.
For example, abc would give you abc, acb, bac, bca, cab,
cba. You can assume that all the characters will be unique.
- 5 Answers
- 23057 Views
- IITR, Microsoft, Nike, I also Faced
- E-Mail Answers
Answers were Sorted based on User's Feedback
Answer / fallen angel
use plain recursion
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
void swap(char* src, char* dst)
{
char ch = *dst;
*dst = *src;
*src = ch;
}
/* permute [set[begin], set[end]) */
int permute(char* set, int begin, int end)
{
int i;
int range = end - begin;
if (range == 1) {
printf("set: %s\n", set);
} else {
for(i=0; i<range; i++) {
swap(&set[begin], &set[begin+i]);
permute(set, begin+1, end);
swap(&set[begin], &set[begin+i]);
/* set back */
}
}
return 0;
}
int main()
{
char str[] = "abcd";
permute(str, 0, strlen(str));
return 0;
}
| Is This Answer Correct ? | 28 Yes | 9 No |
Answer / patrick
//I think it can be implemented this way rather
simply..check it..
main(){
char a[15];
int len,fiblen,i,j,count=0,div,on,to;
printf("enter the string\n");
scanf("%s",a);
fiblen=fib(strlen(a));
div=fiblen/(strlen(a));
for(i=0;i<strlen(a);i++){
for(j=0;j<(strlen(a)-1);j++){
for(on=1;on<(strlen(a)-1);on++){
to=on+1;
swap(&a[on],&a[to]);
count++;//no: of anagram
printf("(%d) %s\n",count,a);
}
}
swap(&a[0],&a[i+1]);
}
}
swap(char *a,char *b){
char temp=*a;
*a=*b;
*b=temp;
}
int fib(int a){
if(a==1)
return(1);
else
return(a*fib(a-1));
}
| Is This Answer Correct ? | 4 Yes | 2 No |
Answer / raghuram.a
/*guys..I have implemented Jhonson trotter algorithm..u can
print permutations of 123..n. implement the same for strings!*/
#include<iostream.h>
#include<conio.h>
int min(int a[10],int n)
{
int i,m=1;
for(i=2;i<=n;i++)
{
if(a[m]>a[i])
m=i;
}
return m;
}
void swap(int &a,int &b)
{
int t;
t=a;
a=b;
b=t;
}
int main()
{
int i,j,k,n,flag=0,l,m;
int d[100],a[100];
clrscr();
cout<<"\n\nenter n:\n\n";
cin>>n;
for(i=1;i<=n;i++)
a[i]=i;
for(i=1;i<=n;i++)
d[i]=i-1;
cout<<"\n\npermutations generated for integer 12...n
are:\n\n";
for(i=1;i<=n;i++) //display the given no.
cout<<a[i];
cout<<"\t";
do
{
flag=0;
k=min(a,n);
for(i=1;i<=n;i++)
{
if((a[i]>a[k])&&(d[i]!=0)&&(a[d[i]]<a[i])) //find
mobile integer.
k=i;
}
if(a[k]==1)
break;
l=d[k]; //copy of direction of mobile integer.
m=a[k]; //copy of mobile integer.
if(d[k]==k+1&&d[k+1]==k) //swap directions.
{
if(d[k]==n)
{ d[k+1]=0;
d[k]=k-1;
}
else if(d[k+1]==1)
{
d[k]=0;
d[k+1]=k+2;
}
else
{ d[k]=k-1;
d[k+1]=k+2;
}
}
else if(d[k]==k-1&&d[k-1]==k)
{
d[k]=k+1;
d[k-1]=k-2;
}
/*cout<<d[1]<<d[2]<<d[3]<<d[4];
cout<<"\t";
if u want to know directions of integers.
*/
swap(a[k],a[l]); //swap mobile integer and integer it is
pointing to.
for(i=1;i<=n;i++)
cout<<a[i]; //display the number.
cout<<"\t";
for(i=1;i<=n;i++) //reverse directions of integers
greater than
//mobile integer.
{
if(a[i]>m)
{
if(d[i]==0&&i==n)
d[i]=i-1;
else if(d[i]<i)
d[i]=i+1;
else if(d[i]>i)
d[i]=i-1;
}
}
for(i=1;i<n;i++) //check whether a mobile integer exist or not.
{
if(a[i]<a[i+1])
{
if(d[i+1]!=0)
flag=1;
}
else if(a[i]>a[i+1])
{
if(d[i]!=0)
flag=1;
}
}
}
while(flag==1); //if no mobile
integer(flag=0)terminate the program.
getch();
return 0;
}
| Is This Answer Correct ? | 4 Yes | 3 No |
/*guys..I have implemented Jhonson trotter algorithm..u can
print permutations of 123..n. implement the same for strings!*/
#include<iostream.h>
#include<conio.h>
int min(int a[10],int n)
{
int i,m=1;
for(i=2;i<=n;i++)
{
if(a[m]>a[i])
m=i;
}
return m;
}
void swap(int &a,int &b)
{
int t;
t=a;
a=b;
b=t;
}
int main()
{
int i,j,k,n,flag=0,l,m;
int d[100],a[100];
clrscr();
cout<<"\n\nenter n:\n\n";
cin>>n;
for(i=1;i<=n;i++)
a[i]=i;
for(i=1;i<=n;i++)
d[i]=i-1;
cout<<"\n\npermutations generated for integer 12...n
are:\n\n";
for(i=1;i<=n;i++) //display the given no.
cout<<a[i];
cout<<"\t";
do
{
flag=0;
k=min(a,n);
for(i=1;i<=n;i++)
{
if((a[i]>a[k])&&(d[i]!=0)&&(a[d[i]]<a[i])) //find
mobile integer.
k=i;
}
if(a[k]==1)
break;
l=d[k]; //copy of direction of mobile integer.
m=a[k]; //copy of mobile integer.
if(d[k]==k+1&&d[k+1]==k) //swap directions.
{
if(d[k]==n)
{ d[k+1]=0;
d[k]=k-1;
}
else if(d[k+1]==1)
{
d[k]=0;
d[k+1]=k+2;
}
else
{ d[k]=k-1;
d[k+1]=k+2;
}
}
else if(d[k]==k-1&&d[k-1]==k)
{
d[k]=k+1;
d[k-1]=k-2;
}
/*cout<<d[1]<<d[2]<<d[3]<<d[4];
cout<<"\t";
if u want to know directions of integers.
*/
swap(a[k],a[l]); //swap mobile integer and integer it is
pointing to.
for(i=1;i<=n;i++)
cout<<a[i]; //display the number.
cout<<"\t";
for(i=1;i<=n;i++) //reverse directions of integers
greater than
//mobile integer.
{
if(a[i]>m)
{
if(d[i]==0&&i==n)
d[i]=i-1;
else if(d[i]<i)
d[i]=i+1;
else if(d[i]>i)
d[i]=i-1;
}
}
for(i=1;i<n;i++) //check whether a mobile integer exist or not.
{
if(a[i]<a[i+1])
{
if(d[i+1]!=0)
flag=1;
}
else if(a[i]>a[i+1])
{
if(d[i]!=0)
flag=1;
}
}
}
while(flag==1); //if no mobile
integer(flag=0)terminate the program.
getch();
return 0;
}
| Is This Answer Correct ? | 5 Yes | 5 No |
Answer / shikha
#include <iostream>
using namespace std;
void anagram(char x[],string temp){
int size = 0;
while(x[size] != '\0'){
size++;
}
// cout << size << " " << x << endl;
if(size > 0){
for(int i=0; i<size ; i++){
string temp1 = temp + x[i];
char y[size];
for(int j=0,k=0; j<size-1 ; j++,k++){
if(k != i) y[j] = x[k];
else j--;
}
y[size-1] = '\0';
anagram(y,temp1);
}
}
else{
cout << temp << endl;
temp = "";
}
}
int main(){
char name[] = "abc";
anagram(name, "");
return 0;
}
| Is This Answer Correct ? | 1 Yes | 2 No |
how to programme using switch statements and fuctions, a programme that will output two even numbers, two odd numbers and two prime numbers of the users chioce.
0 Answers Mbarara University of Science and Technology,
Give a oneline C expression to test whether a number is a power of 2?
25 Answers EA Electronic Arts, Google, Motorola,
I need your help, i need a Turbo C code for this problem.. hope u'll help me guys.? Your program will have a 3x3 array. The user will input the sum of each row and each column. Then the user will input 3 values and store them anywhere, or any location or index, temporarily in the array. Your program will supply the remaining six (6) values and determine the exact location of each value in the array. Example: Input: Sum of row 1: 6 Sum of row 2: 15 Sum of row 3: 24 Sum of column 1: 12 Sum of column 2: 15 Sum of column 3: 18 Value 1: 3 Value 2: 5 Value 3: 6 Output: Sum of Row 1 2 3 6 4 5 6 15 7 8 9 24 Sum of Column 12 15 18 Note: Your program will not necessary sort the walues in the array Thanks..
which function is used to clear the buffer stream on gcc? for example: I wrote following code on gcc #include<stdio.h> int main(void) { char ch; int a,b; printf("\nenter two numbers:\t"); scanf("%d%d",&a,&b); printf("enter number is %d and %d",a,b); printf("\nentercharacter:\t"); scanf("%c",&ch); printf("enter character is %c",ch); return 0; } in above progarm ch could not be scan. why?plz tell me solution.
write a program to count the number the same (letter/character foreg: 's') in a given sentence.
void main() { int i=5; printf("%d",i+++++i); }
How do I write a program to print proper subset of given string . Eg :input: abc output:{},{a},{b},{c},{a,b},{a,c},{b,c}, {a,b,c}.I desperately need this program please mail me to saravana6m@gmail.com
main() { int k=1; printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE"); }
Write a program that find and print how many odd numbers in a binary tree
main() { unsigned int i=10; while(i-->=0) printf("%u ",i); }
void main() { int i; char a[]="\0"; if(printf("%s\n",a)) printf("Ok here \n"); else printf("Forget it\n"); }
#include<stdio.h> main() { int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} }; int *p,*q; p=&a[2][2][2]; *q=***a; printf("%d----%d",*p,*q); }
C Code 
