# Replace all occurrences of string AB with C without using extra space

Given a string str that may contain one more occurrences of “AB”. Replace all occurrences of “AB” with “C” in str.

Examples:

```Input  : str = "helloABworld"
Output : str = "helloCworld"

Input  : str = "fghABsdfABysu"
Output : str = "fghCsdfCysu"
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to find all occurrences of “AB”. For every occurrence, replace it with C and more all characters one position back.

```// C++ program to replace all occurrences of "AB"
// with "C"
#include <bits/stdc++.h>

void translate(char* str)
{
if (str[0] == '\0')
return;

// Start traversing from second chracter
for (int i=1; str[i] != '\0'; i++)
{
// If previous character is 'A' and
// current character is 'B"
if (str[i-1]=='A' && str[i]=='B')
{
// Replace previous character with
// 'C' and move all subsequent
// characters one position back
str[i-1] = 'C';
for (int j=i; str[j]!='\0'; j++)
str[j] = str[j+1];
}
}
return;
}

// Driver code
int main()
{
char str[] = "helloABworldABGfG";
translate(str);
printf("The modified string is :\n");
printf("%s", str);
}
```

Output:

```The modified string is :
helloCworldCGfG
```

Time Complexity : O(n2)
Auxiliary Space : O(1)

An efficient solution is to keep track of two indexes, one for modified string (i in below code) and other for original string (j in below code). If we find “AB” at current index j, we increment j by 2 and i by 1. Otherwise we increment both and copy character from j to i.

Below is implementation of above idea.

```// Efficient C++ program to replace all occurrences
// of "AB" with "C"
#include <bits/stdc++.h>

void translate(char* str)
{
int len = strlen(str);
if (len < 2)
return;

int i = 0;  // Index in modified string
int j = 0; // Index in original string

// Traverse string
while (j < len-1)
{
// Replace occurrence of "AB" with "C"
if (str[j] == 'A' && str[j+1] == 'B')
{
// Increment j by 2
j = j + 2;
str[i++] = 'C';
continue;
}
str[i++] = str[j++];
}

if (j == len-1)
str[i++] = str[j];

// add a null character to terminate string
str[i] = '\0';
}

// Driver code
int main()
{
char str[] = "helloABworldABGfG";
translate(str);
printf("The modified string is :\n");
printf("%s", str);
}
```

Output:

```The modified string is :
helloCworldCGfG
```

Time Complexity : O(n)
Auxiliary Space : O(1)

This article is contributed by Roshni Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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