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
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; } |
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; } |
XforX
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