Replace two substrings (of a string) with each other
Last Updated :
15 Sep, 2022
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 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 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 code
int main()
{
string S = "aaxb";
string A = "aa";
string B = "bb";
cout << (updateString(S, A, B)) << endl;
}
// This code is contributed by
// Surendra_Gangwar
// 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 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# 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 implementation of the approach
// Function to return the resultant string
function 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.
<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 code
let S = "aab";
let A = "aa";
let B = "bb";
document.write(updateString(S, A, B));
// This code is contributed by mukesh07
</script>
Complexity Analysis:
- Time Complexity: O(n*n), as substr function is being used inside the loop
- Auxiliary Space: O(n)
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