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)
                    


View More Similar Problems

Reverse a linked list

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio

View Solution →

Compare two linked lists

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

View Solution →

Merge two sorted linked lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C

View Solution →

Get Node Value

This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t

View Solution →

Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -

View Solution →

Cycle Detection

A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer

View Solution →