In-place replace multiple occurrences of a pattern

01 Jul, 2022
Given a string and a pattern, replace multiple occurrences of a pattern by character ‘X’. The conversion should be in-place and the 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
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.
Implementation:
C++
Python3
Javascript
// 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 strbool 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 codeint main(){    char str[] = "GeeksforGeeks";    char pattern[] = "Geeks";     replacePattern(str, pattern);    cout << str;     return 0;}
# Python3 program to in-place replace multiple# occurrences of a pattern by character ‘X’ # returns true if pattern is prefix of Strdef compare(Str, pattern):     if(len(Str) != len(pattern)):        return False     for i in range(len(pattern)):        if (Str[i] != pattern[i]):            return False             return True # Function to in-place replace multiple# occurrences of a pattern by character ‘X’def replacePattern(Str,pattern):     # If pattern is null or empty String,    # nothing needs to be done    if (pattern == ""):        return     Len = len(pattern)     if (Len == 0):        return     i, j = 0, 0     # for each character    while (j < len(Str)):         count = 0         # compare Str[j..j+len] with pattern        while (compare(Str[j:j+Len], pattern)):            # increment j by length of pattern            j = j + Len            count += 1         # If single or multiple occurrences of pattern        # is found, replace it by character 'X'        if (count > 0):            Str = Str[0:i] + 'X' + Str[i+1:]            i += 1                  # copy character at current position j        # to position i and increment i and j        if (j<len(Str)):            Str = Str[0:i] + Str[j] + Str[i+1:]            i += 1            j += 1     # add a null character to terminate String    Str = Str[0:i]    return Str # Driver codeStr = "GeeksforGeeks"pattern = "Geeks" Str = replacePattern(Str, pattern)print(Str) # This code is contributed by shinjanpatra
<script> // JavaScript program to in-place replace multiple// occurrences of a pattern by character ‘X’ // returns true if pattern is prefix of strfunction compare(str, pattern){    for (let i = 0; i<pattern.length; i++)        if (str[i] != pattern[i])            return false;    return true;} // Function to in-place replace multiple// occurrences of a pattern by character ‘X’function replacePattern(str,pattern){    // If pattern is null or empty string,    // nothing needs to be done    if (pattern == "")        return;     let len = pattern.length;    if (len == 0)        return;     let i = 0, j = 0;    let count;     // for each character    while (j<str.length) {         count = 0;         // compare str[j..j+len] with pattern        while (compare(str.substring(j,j+len), 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 = str.substring(0,i) + 'X' + str.substring(i+1,)            i++        }         // copy character at current position j        // to position i and increment i and j        if (j<str.length){            str = str.substring(0,i) + str[j] + str.substring(i+1,)            i++;j++        }    }     // add a null character to terminate string    str = str.substring(0,i);    return str;} // Driver codelet str = "GeeksforGeeks";let pattern = "Geeks"; str = replacePattern(str, pattern);document.write(str,"</br>"); // This code is contributed by shinjanpatra </script>
XforX
Time complexity: O(n*m) where n is length of string and m is length of the pattern.
Auxiliary Space: O(1) 
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++
// 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 position            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 position i in string str            str.erase(str.begin() + i);            i--; // i-- because one character was deleted            // so repositioning i        }    }    cout << str;} // Driver codeint main(){    string str = "GeeksforGeeks";    string pattern = "Geeks";     replacePattern(str, pattern);     return 0;}
XforX
Time Complexity: O(n*m)Auxiliary Space: O(1)
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 review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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