Given 3 strings S, A and B. The task is to replace every sub-string of S equal to A with B and every sub-string of S equal to B with A. It is possible that two or more sub-strings matching A or B overlap. To avoid confusion about this situation, you should find the leftmost sub-string that matches A or B, replace it, and then continue with the rest of the string.
For example, when matching A = “aa” with S = “aaa”, A[0, 1] will be given preference over A[1, 2].
Note that A and B will have the same length and A != B.
Examples:
Input: S = “aab”, A = “aa”, B = “bb”
Output: bbb
We match the first two characters with A and replacing it with B we get bbb.
Then we continue the algorithm starting at index 3 and we don’t find any more matches.Input: S = “aabbaabb”, A = “aa”, B = “bb”
Output: bbaabbaa
We replace all the occurrences of “aa” with “bb” and “bb” with “aa”, so the resultant string is “bbaabbaa”.
Approach: Go through every possible sub-string from S of length len(A). if any sub-string matches A or B then update the string as required and print the updated string in the end.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include<bits/stdc++.h>using namespace std;// Function to return the resultant stringstring updateString(string S, string A, string B){ int l = A.length(); // Iterate through all positions i for (int i = 0; i + l <= S.length(); i++) { // Current sub-string of // length = len(A) = len(B) string curr = S.substr(i, i + l); // If current sub-string gets // equal to A or B if (curr == A) { // Update S after replacing A string new_string = ""; new_string += S.substr(0, i) + B + S.substr(i + l, S.length()); S = new_string; i += l - 1; } else if(curr == B) { // Update S after replacing B string new_string = ""; new_string += S.substr(0, i) + A + S.substr(i + l, S.length()); S = new_string; i += l - 1; } else { //do nothing } } // Return the updated string return S;}// Driver codeint main(){ string S = "aaxb"; string A = "aa"; string B = "bb"; cout << (updateString(S, A, B)) << endl;}// This code is contributed by// Surendra_Gangwar |
Java
// Java implementation of the approachclass GFG { // Function to return the resultant string static String updateString(String S, String A, String B) { int l = A.length(); // Iterate through all positions i for (int i = 0; i + l <= S.length(); i++) { // Current sub-string of length = len(A) = len(B) String curr = S.substring(i, i + l); // If current sub-string gets equal to A or B if (curr.equals(A)) { // Update S after replacing A String new_string = S.substring(0, i) + B + S.substring(i + l, S.length()); S = new_string; i += l - 1; } else { // Update S after replacing B String new_string = S.substring(0, i) + A + S.substring(i + l, S.length()); S = new_string; i += l - 1; } } // Return the updated string return S; } // Driver code public static void main(String[] args) { String S = "aab"; String A = "aa"; String B = "bb"; System.out.println(updateString(S, A, B)); }} |
Python3
# Python3 implementation of the approach # Function to return the resultant string def updateString(S, A, B): l = len(A) # Iterate through all positions i i = 0 while i + l <= len(S): # Current sub-string of # length = len(A) = len(B) curr = S[i:i+l] # If current sub-string gets # equal to A or B if curr == A: # Update S after replacing A new_string = S[0:i] + B + S[i + l:len(S)] S = new_string i += l - 1 else: # Update S after replacing B new_string = S[0:i] + A + S[i + l:len(S)] S = new_string i += l - 1 i += 1 # Return the updated string return S # Driver code if __name__ == "__main__": S = "aab" A = "aa" B = "bb" print(updateString(S, A, B)) # This code is contributed by Rituraj Jain |
C#
// C# implementation of the approach using System;class GFG { // Function to return the resultant string static string updateString(string S, string A, string B) { int l = A.Length; // Iterate through all positions i for (int i = 0; i + l <= S.Length; i++) { // Current sub-string of length = len(A) = len(B) string curr = S.Substring(i, l); // If current sub-string gets equal to A or B if (curr.Equals(A)) { // Update S after replacing A string new_string = S.Substring(0, i) + B + S.Substring(i + l); S = new_string; i += l - 1; } else { // Update S after replacing B string new_string = S.Substring(0, i) + A + S.Substring(i + l); S = new_string; i += l - 1; } } // Return the updated string return S; } // Driver code public static void Main() { string S = "aab"; string A = "aa"; string B = "bb"; Console.WriteLine(updateString(S, A, B)); } } // This code is contributed by Ryuga |
PHP
<?php// PHP implementation of the approach// Function to return the resultant stringfunction updateString($S, $A, $B){ $l = strlen($A); // Iterate through all positions i for ($i = 0; $i + $l <= strlen($S); $i++) { // Current sub-string of length = len(A) = len(B) $curr = substr($S, $i, $i + $l); // If current sub-string gets equal to A or B if (strcmp($curr, $A) == 0) { // Update S after replacing A $new_string = substr($S, 0, $i) . $B . substr($S, $i + $l, strlen($S)); $S = $new_string; $i += $l - 1; } else { // Update S after replacing B $new_string = substr($S, 0, $i) . $A . substr($S, $i + $l, strlen($S)); $S = $new_string; $i += $l - 1; } } // Return the updated string return $S;}// Driver code$S = "aab";$A = "aa";$B = "bb";echo(updateString($S, $A, $B));// This code is contributed by Code_Mech. |
Javascript
<script>// Javascript implementation of the approach // Function to return the resultant string function updateString(S, A, B) { let l = A.length; // Iterate through all positions i for(let i = 0; i + l <= S.length; i++) { // Current sub-string of length = len(A) = len(B) let curr = S.substring(i, i + l); // If current sub-string gets equal to A or B if (curr == A) { // Update S after replacing A let new_string = S.substring(0, i) + B + S.substring(i + l); S = new_string; i += l - 1; } else { // Update S after replacing B let new_string = S.substring(0, i) + A + S.substring(i + l); S = new_string; i += l - 1; } } // Return the updated string return S; }// Driver codelet S = "aab"; let A = "aa"; let B = "bb"; document.write(updateString(S, A, B)); // This code is contributed by mukesh07</script> |
Output
bbxb
Complexity Analysis:
- Time Complexity: O(n*n), as substr function is being used inside the loop
- Auxiliary Space: O(n)
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