Answer / gayathri sundar

#include<stdio.h>
#include<string.h>
main(int argc, char *argv[])
{
char *string = argv[1];
int len = strlen(string);
int i = 0;
int j = len - 1;
printf("string before is %s\n", string);
printf("len is %d\n", len);
while(i <= j)
{
*(string+i) += *(string+j);
*(string+j) = *(string+i) - *(string+j);
*(string+i) = *(string+i) - *(string+j);
i++;
j--;
if(len % 2)
if(i == j) break;
}
printf("reversed string is %s\n", string);
}

Answer / manish pathak

//********this is perfect answer***************

#include<stdio.h>
#include<conio.h>
void main()
{
int i,j,n,num,rem,l;
char s[]="abcde";
l=sizeof(s)-1;
for(i=0,j=l-1;i<=(l-1)/2;i++)//save space n/2
{

s[i]=s[i]+s[j-i];
s[j-i]=s[i]-s[j-i];
s[i]=s[i]-s[j-i];

}

s[l]='\0';
printf("%s",s);
getch();
}

Answer / a

It's all O(n). You're finding the length of the string,
which itself is an O(n) operation.
So, O(n + n/2) = O(n).

Answer / mm chen

#include <iostream>
#include <string>
using std::string;
using std::cout;
using std::endl;
using std::cin;

int main()
{
string s("abcdefghijklmnopqrstuvwxyz");
string::size_type s_size = s.size();

for (string::size_type x = 0; x != s_size; x++){

if ( (s_size -x -1) > x ){

s[x] ^= s[s_size - x -1];
s[s_size - x -1] ^= s[x];
s[x] ^= s[s_size - x -1];

}else{
break;
}

}

cout << s << endl;
}

Answer / anurag

int i=0,len=strlen(str);
int j=len/2;len--;
while(i<j)
{
*(str+i)^=*(str+len)^=*(str+i)^=*(str+len);
len--;i++;
}

Answer / atul kabra

#include<stdio.h>

void reverse(char *);

void main()
{
char str[]="Hello";

reverse(str);
printf("Reverse String is %s",str);

}

void reverse(char *p)
{

char *q=p;
while(*++q!='\0');
q--;

while(p<q)
{
*p=*p+*q;
*q=*p-*q;
*p=*p-*q;
p++;
q--;
}

}

Answer / suraj bhan gupta

#inclue<stdio.h>
#include<string.h>
main(){
char a[10],i;
int len=1;
printf(" Enter string ");
fgets(a,9,stdin);

len = strlen(a);


for(i=0 ; i<(len/2) ; i++){
a[i]=a[i]+a[len-2];
a[len-2]=a[i]-a[len-2];
a[i]=a[i]-a[len-2];
len--;
}
printf("\n Reverse string with n/2 complexity
%s",a);
return 0;
}

Answer / snehal

#inclue<stdio.h>
#include<string.h>
main(){
char a[10],i;
int len=1;
printf(" Enter string ");
fgets(a,9,stdin);

len = strlen(a);


for(i=0 ; i<=(len/2) ; i++){
a[i]=a[i]+a[len-1];
a[len-1]=a[i]-a[len-1];
a[i]=a[i]-a[len-1];
len--;
}
printf("\n Reverse string with n/2 complexity
%s",a);
return 0;
}

Answer / raghuram

#include<iostream.h>

#include<string.h>//complexity-n/2

int main()
{
int i,l,l1;
char str[100];
cout<<"enter string:";
gets(str);
l=strlen(str);
if(l%2==0)
l1=(l/2-1);
else
l1=l/2;
for(i=0;i<=l1;i++)/*swap elements from 2 ends till u reach
middle part of array*/
{
char t=str[i];
str[i]=str[l-i-1];
str[l-i-1]=t;
}
str[l]=0;
cout<<"\n\nreversed string is:"<<str;
getch();
return 0;
}

Answer / prof.muthu (9962940220)

refer c book!
or
contact me

Write a single line c expression to delete a,b,c from aabbcc

2 Answers   Microsoft,

Design an implement of the inputs functions for event mode

0 Answers   Wipro,

program to find the roots of a quadratic equation

14 Answers   College School Exams Tests, Engineering, HP, IIIT, Infosys, Rajiv Gandhi University of Knowledge Technologies RGUKT, SSC,

struct point { int x; int y; }; struct point origin,*pp; main() { pp=&origin; printf("origin is(%d%d)\n",(*pp).x,(*pp).y); printf("origin is (%d%d)\n",pp->x,pp->y); }

1 Answers  

void main() { if(~0 == (unsigned int)-1) printf(“You can answer this if you know how values are represented in memory”); }

1 Answers  

main( ) { int a[ ] = {10,20,30,40,50},j,*p; for(j=0; j<5; j++) { printf(“%d” ,*a); a++; } p = a; for(j=0; j<5; j++) { printf(“%d ” ,*p); p++; } }

1 Answers  

You are given any character string. Find the number of sets of vowels that come in the order of aeiou in the given string. For eg., let the given string be DIPLOMATIC. The answer returned must be "The number of sets is 2" and "The sets are "IO and AI". Vowels that form a singleton set must be neglected. Try to post the program executable in gcc or g++ or in java.

3 Answers  

#define max 5 #define int arr1[max] main() { typedef char arr2[max]; arr1 list={0,1,2,3,4}; arr2 name="name"; printf("%d %s",list[0],name); }

1 Answers  

1 o 1 1 0 1 0 1 0 1 1 0 1 0 1 how to design this function format in c-language ?

2 Answers  

String reverse with time complexity of n/2 with out using temporary variable.

10 Answers   NetApp, Symantec,

Write a program that find and print how many odd numbers in a binary tree

1 Answers  

void main () { int x = 10; printf ("x = %d, y = %d", x,--x++); } a. 10, 10 b. 10, 9 c. 10, 11 d. none of the above

2 Answers   HCL,