Generate all possible strings formed by replacing letters with given respective symbols

Given a string S consisting of N characters and an array M[] of pairs of characters such that any character M[i][0] may be changed with the character M[i][1] within the string S, the duty is to generate all of the doable strings shaped by changing some characters of the string with their respective symbols within the array M[].

Examples:

Enter: S = “aBc”, M = {‘a’ : ‘$’, ‘B’ : ‘#’, ‘c’ : ‘^’}
Output:
aBc
aB^
a#c
a#^
$Bc
$B^
$#c
$#^

Enter: S = “a”, M={‘a’ : ‘$’}
Output:
a
$

Strategy: The given downside may be solved by utilizing Backtracking to generate all doable strings by changing every character with the mapped character within the array M[]. Comply with the steps beneath to resolve the issue:

Retailer all of the pairs of the mapped characters within the array M[] in a map, say Map.
Outline a recursive perform say generateLetters(S, P), the place S is the modified string and P is the index of the present character:
- Test for the bottom case i.e., if index P equals to the N then print the string S and return.
- Don’t change the present character and recursively name for the perform, generateLetters(S, P + 1).
- Now, change the character, S[P] with the respective image within the map M and name for the perform, generateLetters(S, P+1).
After finishing the above steps, name for the perform generateLetters(S, 0) to print all doable strings.

Under is the implementation of the above method:

C++

#embrace <bits/stdc++.h>

utilizing namespace std;

void generateLetters(string S, int P,

unordered_map<char, char> M)

{

if (P == S.measurement()) {

cout << S << "n";

return;

}

generateLetters(S, P + 1, M);

S[P] = M[S[P]];

generateLetters(S, P + 1, M);

return;

}

int foremost()

{

string S = "aBc";

unordered_map<char, char> M;

M['a'] = '$';

M['B'] = '#';

M['c'] = '^';

M['d'] = '&';

M['1'] = '*';

M['2'] = '!';

M['E'] = '@';

generateLetters(S, 0, M);

return 0;

}

Output:


aBc
aB^
a#c
a#^
$Bc
$B^
$#c
$#^

Time Complexity: O(N*2^N)
Auxiliary House: O(1)

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