# String Equivalence Relations - Amazon Top Interview Questions

### Problem Statement :

```You are given three lowercase alphabet strings a, b and target. Strings a and b have the same length and are defined to be equivalent: a[i] = b[i]. For example, if a = "abc" and b = "xyz", then "a" = "x", "b" = "y" and "c" = "z".

Also, we can make the following kinds of inferences for characters:

c = c
a = b implies b = a
a = b and b = c implies a = c

Return the smallest lexicographically equivalent string for target.

Constraints

n ≤ 1,000 where n is the length of a and b
m ≤ 1,000 where m is the length of target

Example 1

Input

a = "axc"
b = "xdz"
target = "ddxz"

Output

"aaac"

Explanation

We know that "a" = "x" and "x" = "d", so "a" = "d". So we can replace the "d"s and "x" with "a"s. Then we can directly replace "z" with "c".

Example 2

Input

a = "abc"
b = "def"

target = "xyz"

Output

"xyz"

Explanation

There's no inferences we can make here.```

### Solution :

```                        ```Solution in C++ :

int findpapa(vector<char> &papa, int c) {
if (papa[c] != c) {
papa[c] = findpapa(papa, papa[c]);
}
return papa[c];
}

string solve(string a, string b, string target) {
vector<char> papa(26);
for (int i = 0; i < 26; i++) {
papa[i] = i;
}
for (int i = 0; i < a.size(); i++) {
// cout<<a[i]-'a'<<" "<<b[i]-'a'<<endl;
int p1 = findpapa(papa, a[i] - 'a');
int p2 = findpapa(papa, b[i] - 'a');
// cout<<p1<<" "<<p2<<endl;
if (p1 < p2) {
papa[p2] = p1;
} else {
papa[p1] = p2;
}
// cout<<"main nhi hagga"<<endl;
}
for (char &c : target) {
int p1 = findpapa(papa, c - 'a');
c = ('a' + p1);
}
return target;
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, a, b, target):
parent = {}

def find(x):
if x not in parent:
return x
if parent[x] != x:
parent[x] = find(parent[x])
return parent[x]

def union(x, y):
x0 = parent[x]
y0 = parent[y]
if x0 == y0:
return
if y0 < x0:
x0, y0 = y0, x0
parent[y0] = x0

for c1, c2 in zip(a, b):
if c1 not in parent:
parent[c1] = c1
if c2 not in parent:
parent[c2] = c2
union(c1, c2)

return "".join(find(t) for t in target)```
```

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