Answer Posted / om

Input: str1 (size M), str2 (size N)
Output:- Union_string str3.

1. Take a auxillary interger array of size 128 and
initialize all its's enteries to zero.(becoz All printable
chacter has value in the range [0,127]. so 128)
int Aux[128]={0}; //O(128) TIME

2. set Couter = strlen(str1) + strlen(str2) +1 ;//extra one
for '\0' character..

3. now scan the first string str1 and put 1 on the index
value equal to str1[i] in auxillary array.If there were
already 1,then Counter--;
Aux[str1[i]]=1; //O(M) TIME

4. now scan the second string str2 and put 1 on the index
value equal to str2[i] in auxillary array.Similar to step 3.
If there were already 1,then Counter--;
Aux[str2[i]=1; //O(N) TIME

5. Now alloacte a memory of size equal to "Counter" for our
resulting union string str3.
char *str3=(char*)malloc(Counter);

6. now again scan the str1 followed by str2. and if any 1 is
found in Aux[str1[i]] or Aux[str2[i]] make it 0. and put
that character to union string str3.

int k=0;
for(int i=0; str1[i]!='\0' ;i++) //O(M)
if(Aux[str1[i]] ==1)
{
str3[k++]=str1[i];
Aux[str1[i]] =0;
}

for(int i=0; str2[i]!='\0' ;i++) //0(N)
if(Aux[str2[i]] ==1)
{
str3[k++]=str2[i];
Aux[str2[i]] =0;
}
str3[k]=\0';


and return str3.


// SO TOTAL O(MAX(M,N)) TIME COMPLEXITY ALGORITHM USING
CONSTANT SPACE OF SIZE 128.

count = 0; for (i = 1;i < = 10; i++);count = count + i; Value of count after execution of the above statements will be a) 0 b) 11 c) 55 d) array

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