how to find out the union of two character arrays?
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Answer / prakashdasari
char *str1,*str2,*res;
while(*str1!=NULL)
str++;
while(*str2!=NULL)
{
*str1 == *str2;
str1++;
str2++;
}
*str1 = '\0';
//so total string is in one string ie str1 itself;
//now i will remove duplicates from entire string
for(i=0;str[i]!=NULL;i++)
{
for(j=0;str[j]!=NULL;j++)
{
if(str[i]==str[j])
{
flag = 1;break;
}
else flag = 0;
}
if(flag == 0)
{
*res = str[i];
res++;
}
*res = '\0';
}
now resultant string (res) is union of two character
arrays....
| Is This Answer Correct ? | 6 Yes | 1 No |
Answer / 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.
| Is This Answer Correct ? | 2 Yes | 2 No |
Write a program in c to print 1 121 12321 1234321 123454321
11 Answers ANR, College School Exams Tests, Mu Sigma, Wipro,
#include<stdio.h> main() { int a[3]; int *I; a[0]=100;a[1]=200;a[2]=300; I=a; Printf(“%d\n”, ++*I); Printf(“%d\n”, *++I); Printf(“%d\n”, (*I)--); Printf(“%d\n”, *I); } what is the o/p a. 101,200,200,199 b. 200,201,201,100 c. 101,200,199,199 d. 200,300,200,100
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