CareerCup

for swapping variable
num1 = num1 ^ num2 ;

num2 = num1 ^ num2 ;
num1 = num1 ^ num2 ;

for swapping variable

num1 = num1 - num2 ;
num2 = num2 + num1 ;
num1 = num2 - num1;

reversing string
here n is length of string
for(i = 0 ; i < n/2 ; i++)
{
str[i] = str[i] - str[n-1-i];
str[n-1-i] = str[n-1-i] + str[i] ;
str[i] = str[n-1-i] - str[i];

}

Have two pointers a,one to the first character of the string and b one to the last. Swap both and increment a and decrement b. Recursively do this till till both are equal.

int a=2 b=3;

a=a+b;
b=a-b;
a=a-b;

void reverse(char *str, int len)
{
while(--len<=0)
{
*str ^=str[len];
str[len]^=*str;
*str=str[len];
str++;
}
}

void reverse(char *str, int len)
{
while(--len<=0)
{
*str ^=str[len];
str[len]^=*str;
*str=str[len];
str++;
}
}

You can NOT swap two numbers without using temporary variable,
above method may work for small number, consider two big numbers 65535 and 65534
your sum a=a+b overflows
i think even XOR wont work if u take negetive numbers

String reverseMe = "reverse me!";
for (int i = 0; i < reverseMe.length(); i++) {
    reverseMe = reverseMe.substring(1, reverseMe.length() - i)
        + reverseMe.substring(0, 1)
        + reverseMe.substring(reverseMe.length() - i, reverseMe.length());
 }
 System.out.println(reverseMe);

source - stackoverflow stivlo

void(char *p)
{
int i,n;
n=strlen(p);
for(i=0;i<n/2;i++)
{
p[i]=p[i]^p[n-1-i];
p[n-1-i]=p[i]^p[n-1-i];
p[i]=p[i]^p[n-1-i];
}
}
int main()
{
char str[]="string";
void(reverse(str);
puts(str);
}

what will be space complexity of this function:

String reverseMe = "reverse me!";
for (int i = 0; i < reverseMe.length(); i++) {
reverseMe = reverseMe.substring(1, reverseMe.length() - i)
+ reverseMe.substring(0, 1)
+ reverseMe.substring(reverseMe.length() - i, reverseMe.length());
}
System.out.println(reverseMe);


will it be O(n^2)? orO(n^3)

class Rev{
public static void main(String[] args)
{
String s="abcd";
int len=s.length();
for(int i=0;i<len*2;i++)
{
s=s.charAt(i)+s;
i++;
}
s=s.substring(0,len);
System.out.println(s);
}
}