Given three strings ‘str’, ‘oldW’ and ‘newW’. The task is find all occurrences of the word ‘oldW’ and replace then with word ‘newW’. Examples:
Input : str[] = "xxforxx xx for xx",
oldW[] = "xx",
newW[] = "geeks"
Output : geeksforgeeks geeks for geeks
The idea is to traverse the original string and count the number of times old word occurs in the string. Now make a new string of sufficient size so that new word can be replaced. Now copy original string to new string with replacement of word.
Implementation:
C
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char * replaceWord( const char * s, const char * oldW,
const char * newW)
{
char * result;
int i, cnt = 0;
int newWlen = strlen (newW);
int oldWlen = strlen (oldW);
for (i = 0; s[i] != '\0' ; i++) {
if ( strstr (&s[i], oldW) == &s[i]) {
cnt++;
i += oldWlen - 1;
}
}
result = ( char *) malloc (i + cnt * (newWlen - oldWlen) + 1);
i = 0;
while (*s) {
if ( strstr (s, oldW) == s) {
strcpy (&result[i], newW);
i += newWlen;
s += oldWlen;
}
else
result[i++] = *s++;
}
result[i] = '\0' ;
return result;
}
int main()
{
char str[] = "xxforxx xx for xx" ;
char c[] = "xx" ;
char d[] = "Geeks" ;
char * result = NULL;
printf ( "Old string: %s\n" , str);
result = replaceWord(str, c, d);
printf ( "New String: %s\n" , result);
free (result);
return 0;
}
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Output:
Old string: xxforxx xx for xx
New String: GeeksforGeeks Geeks for Geeks
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 2: This method involves the in-place update of the string. It is more efficient as it uses only extra space for the new characters to be inserted.
Implementation:
C
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void replaceWord( char * str, char * oldWord, char * newWord)
{
char *pos, temp[1000];
int index = 0;
int owlen;
owlen = strlen (oldWord);
while ((pos = strstr (str, oldWord)) != NULL) {
strcpy (temp, str);
index = pos - str;
str[index] = '\0' ;
strcat (str, newWord);
strcat (str, temp + index + owlen);
}
}
int main()
{
char str[1000], oldWord[100], newWord[100];
printf ( "Enter the string: " );
gets (str);
printf ( "Enter the word to be replaced: " );
gets (oldWord);
printf ( "Replace with: " );
gets (newWord);
replaceWord(str, oldWord, newWord);
printf ( "\nModified string: %s" , str);
return 0;
}
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Input:
1
xxforxx xx for xx
xx
geeks
Output:
geeksforgeeks geeks for geeks
Time Complexity: O(n)
Auxiliary Space: O(1)