In-place replace multiple occurrences of a pattern

Given a string and a pattern, replace multiple occurrences of a pattern by character ‘X’. The conversion should be in-place and solution should replace multiple consecutive (and non-overlapping) occurrences of a pattern by a single ‘X’.

String – GeeksForGeeks
Pattern – Geeks
Output: XforX
 
String – GeeksGeeks
Pattern – Geeks
Output: X

String – aaaa
Pattern – aa
Output: X

String – aaaaa
Pattern – aa
Output: Xa

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

The idea is to maintain two index i and j for in-place replacement. Index i always points to next character in the output string. Index j traverses the string and searches for one or more pattern match. If a match is found, we put character ‘X’ at index i and increment index i by 1 and index j by length of the pattern. Index i is increment only once if we find multiple consecutive occurrences of the pattern. If the pattern is not found, we copy current character at index j to index i and increment both i and j by 1. Since pattern length is always more than equal to 1 and replacement is only 1 character long, we would never overwrite unprocessed characters i.e j >= i is invariant.

// C++ program to in-place replace multiple 
// occurrences of a pattern by character ‘X’ 
#include <bits/stdc++.h> 
using namespace std; 
  
// returns true if pattern is prefix of str 
bool compare(char* str, char* pattern) 
{ 
    for (int i = 0; pattern[i]; i++) 
        if (str[i] != pattern[i]) 
            return false; 
    return true; 
} 
  
// Function to in-place replace multiple 
// occurrences of a pattern by character ‘X’ 
void replacePattern(char* str, char* pattern) 
{ 
    // If pattern is null or empty string, 
    // nothing needs to be done 
    if (pattern == NULL) 
        return; 
  
    int len = strlen(pattern); 
    if (len == 0) 
        return; 
  
    int i = 0, j = 0; 
    int count; 
  
    // for each character 
    while (str[j]) { 
        count = 0; 
  
        // compare str[j..j+len] with pattern 
        while (compare(str + j, pattern)) { 
            // increment j by length of pattern 
            j = j + len; 
            count++; 
        } 
  
        // If single or multiple occurrences of pattern 
        // is found, replace it by character 'X' 
        if (count > 0) 
            str[i++] = 'X'; 
  
        // copy character at current position j 
        // to position i and increment i and j 
        if (str[j]) 
            str[i++] = str[j++]; 
    } 
  
    // add a null character to terminate string 
    str[i] = '\0'; 
} 
  
// Driver code 
int main() 
{ 
    char str[] = "GeeksforGeeks"; 
    char pattern[] = "Geeks"; 
  
    replacePattern(str, pattern); 
    cout << str; 
  
    return 0; 
} 

Output:

XforX

The time complexity of above algorithm is O(n*m), where n is length of string and m is length of the pattern.

Implementation using STL

The idea of this implementation is to use the STL in-built functions to search for pattern string in main string and then erasing it from the main string

// C++ program to in-place replace multiple 
// occurrences of a pattern by character ‘X’ 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to in-place replace multiple 
// occurrences of a pattern by character ‘X’ 
void replacePattern(string str, string pattern) 
{ 
  
    // making an iterator for string str 
    string::iterator it = str.begin(); 
    // run this loop until iterator reaches end of string 
    while (it != str.end()) { 
        // searching the first index in string str where 
        // the first occurrence of string pattern occurs 
        it = search(str.begin(), str.end(), pattern.begin(), pattern.end()); 
        // checking if iterator is not pointing to end of the 
        // string str 
        if (it != str.end()) { 
            // erasing the full pattern string from that iterator 
            // position in string str 
            str.erase(it, it + pattern.size()); 
            // inserting 'X' at that iterator posiion 
            str.insert(it, 'X'); 
        } 
    } 
  
    // this loop removes consecutive 'X' in string s 
    // Example: GeeksGeeksforGeeks was changed to 'XXforX' 
    // running this loop will change it to 'XforX' 
    for (int i = 0; i < str.size() - 1; i++) { 
        if (str[i] == 'X' && str[i + 1] == 'X') { 
            // removing 'X' at posiion i in string str 
            str.erase(str.begin() + i); 
            i--; // i-- because one character was deleted 
            // so repositioning i 
        } 
    } 
    cout << str; 
} 
  
// Driver code 
int main() 
{ 
    string str = "GeeksforGeeks"; 
    string pattern = "Geeks"; 
  
    replacePattern(str, pattern); 
  
    return 0; 
} 

Output:

XforX

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

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Recommended Posts: