Answer / santhoo035

#include<iostream.h>
#include<string.h>
int main()
{
int flag=0,i=0,j=0;
char ar[4]="abc" ;
char temp[4];
cout<<"Plz enter a string with size #3";
cin>>temp;
for(j=0;ar[j];j++)
{
flag=0;
for(i=0;temp[i];i++)
{
if(ar[j]==temp[i])
flag+=1;
}
if(flag==0||flag>1)
break;
}
if(flag==0||flag>1)
cout<<"Not accepted";
else
cout<<"Accepted";
}
Note:PLz convert "iostream.h" into "stdio.h" and "cout<<"
to "printf"

Answer / ew

The above is broken. If the ar[4] is "aac" rather
than "abc" it fails.

#include "stdio.h"

#define CountOf(a) (sizeof(a)/sizeof(*a))

int main(int argc, _TCHAR* argv[])
{


char data[] = "abc";
char test[CountOf(data)];

printf("Enter %d characters: ", CountOf(data)-1);
scanf("%s", test);

for (int i=0;i < CountOf(data);i++)
{
for (int j=0;j < CountOf(data);j++)
{
if (data[i] == test[j])
{
test[j] = 0;
break;
}
}
}

for (int i=0;i < CountOf(data);i++)
{

if (test[i]) {
printf("\nNot ");
break;
}
}
printf("\nAccepted");

return 1;
}

Answer / rahul shandilya

both of the solutions above given have complexity O(N^2),
i have tried to solve it in O(N).let me inform if there is
any bug or better solution .

#include<iostream>
using namespace std;
int main()
{
int a[130],a1[130];
memset(a,0,sizeof(a));
memset(a1,0,sizeof(a));
char ans[1000];
char str[]="putanystringhere";
for(int i=0;i<sizeof(str)/sizeof(char)-1;i++)
{
a[(int)str[i]]+=1;
}
cout<<"Enter any string of length
"<<sizeof(str)/sizeof(char)-1<<" :";
cin>> ans;
for(int i=0;i<sizeof(str)/sizeof(char)-1;i++)
{
a1[(int)ans[i]]+=1;
}
bool flag=true;
for(int i=0;i<sizeof(str)/sizeof(char)-1;i++)
{
if(a[(int)str[i]]!=a1[(int)str[i]])
{
flag=false;
break;
}
}
if(flag==true)
cout<<"No error";
else
cout<<"Error";
system("pause");
return 0;
}

Answer / sivan

if we ask the user to enter the second string having the
same characters as in the first string, the program
essentailly reduces to finding whether there is a
repetition in the second string. A program for this is
given below.

#include<iostream>
using namespace std;

int main()
{
char string1[10];
char string2[10];
int flag = 0;

cout<<"\n Enter the first string";
cin>>string1;
cout<<"\n Enter the second string with the same
characters as entered in string1";
cin>>string2;
for(int i = 0; i < strlen(string2); i++)
{
for(int j = i+1; j < strlen(string2) ; j++)
{
if(string2[i] == string2[j])
{
flag = 1;
break;
}
}
}

if(flag ==1)
cout<<"\n wrong";
else
cout<<"\n correct";

return 0;
}

Answer / yegullelew

My answer is O(n) - but it is done in C# sorry for C guys
public bool isAccepted(string original, string
toCompare)
{
if (original.Length != toCompare.Length)
return false;
bool[] flags = new bool[128];
for (int i = 0; i < toCompare.Length; i++)
flags[toCompare[i]] = true;
for (int i = 0; i < original.Length; i++)
if (!flags[original[i]])
return false;
return true;
}

union u { struct st { int i : 4; int j : 4; int k : 4; int l; }st; int i; }u; main() { u.i = 100; printf("%d, %d, %d",u.i, u.st.i, u.st.l); } a. 4, 4, 0 b. 0, 0, 0 c. 100, 4, 0 d. 40, 4, 0

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