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# Replace two substrings (of a string) with each other

• Difficulty Level : Expert
• Last Updated : 15 Jun, 2021

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``using` `namespace` `std;` `// Function to return the resultant string``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.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``        ``{` `            ``// 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;``        ``}``    ``}` `    ``// Return the updated string``    ``return` `S;``}` `// Driver code``int` `main()``{``    ``string S = ``"aab"``;``    ``string A = ``"aa"``;``    ``string B = ``"bb"``;` `    ``cout << (updateString(S, A, B)) << endl;``}` `// This code is contributed by``// Surendra_Gangwar`

## Java

 `// Java implementation of the approach``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, 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

 `

## Javascript

 ``
Output:
`bbb`

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