In-place replace multiple occurrences of a pattern

Difficulty Level : Medium
Last Updated : 15 Jul, 2021

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.

C++

// 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++

// 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 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 review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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