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 :



title-img




                        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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