Siva Krishna
BAN USERxor may not work all cases..let us take 'abb' and 'acc'. Accoding to xor logic they are anagrams but they are not actually..
 Siva Krishna August 08, 2013sorry its (f(x,c)*x)
 Siva Krishna February 11, 2013but every time it is giving error only
 Siva Krishna November 10, 2012Actually its correct in post order 'y' should come after 'x' and 'z'.
 Siva Krishna August 30, 2012All the numbers are negative case is the only special case,you can check for all the numbers are negative or not if negative return small negative integer otherwise proceed with kandens algo
 Siva Krishna August 26, 2012its not correct according to you 100 is made as just 105 but it to be 155
 Siva Krishna July 27, 2012is the array is sorted or not?
 Siva Krishna July 10, 2012here the output is '0' or any garbage value,why because a[k][j] means a[99][j] means any garbage value which may be greater than 'k' or less than 'k' so "a[k][j]>k" may be true or false.(we may get output '0' in some cases because some values of array are not intialised ,by default they are '0').
 Siva Krishna July 07, 2012No I've executed it on gcc by replacing double to float it is giving only "Not Equal : i=0.8" as output,plz check once again
 Siva Krishna July 06, 2012output :
Equal : i=0.7
Equal : i=0.8
why because here i==0.7 "0.7" is taken as double and 'i' is also double so it is true,but if we take 'float i' then it is false.
6 1
 Siva Krishna July 05, 2012I think it is equal to two iterations
 Siva Krishna July 04, 2012Already given that more than one array can't be used.
 Siva Krishna June 30, 2012The third option in addition to 2 is also right
struct node* m = n;
free( n );
n = m>next;
Sorry for wrong answer.The xor operation here with chars is not correct.
 Siva Krishna May 30, 2012i=0;
while(i<=n/2)
{
if(a[i]==a[ni])
i++;
else
return 0;
}
if(i>n/2)
return 1;

Siva Krishna
May 21, 2012 i checked it by placing %c it is displaying some symbol.
 Siva Krishna April 21, 2012it is always " 1 " as you shift right any number of times because the sign bit is propagating so always there are all bits having value '1' .so ans is "1"
 Siva Krishna April 21, 2012it is always " 1 " as you shift any number of times because the sign bit is propagating so always there are all bits having value '1' .so ans is "1"
 Siva Krishna April 21, 2012First take the first string starting from first char
and find the occurrence of that char in second string and make it null.
Repeat the same process for all chars in first string.Finally if second string is empty then both are anagrams otherwise not.
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Start a counter that increments every second(you have to handle overflow case). When a process enters system, attach the current value of counter to process and call it as timestamp value of a process. Based on the timestamp valued perform scheduling.
 Siva Krishna June 09, 2014