# Ambigram Detection- Google Top Interview Questions

### Problem Statement :

```You are given a lowercase alphabet string s and a two-dimensional list of strings pairs.
Each element in pairs contains [a, b] meaning that character a and b are considered equivalent.
We say that if a and b are equivalent and b and c are equivalent, then a and c are also equivalent.
Also, any character a is equivalent to itself.

Return whether s is a palindrome given the equivalence relations.

Constraints

0 ≤ n ≤ 100,000 where n is the length of s

0 ≤ p ≤ 100,000 where p is the length of p

Example 1

Input

s = "acba"

pairs = [

["b", "c"]

]

Output

True

Explanation

Since "b" = "c", we have "acba" = "acca", which is a palindrome.

Example 2

Input

s = "zz"

pairs = []

Output

True

Explanation

"z" is equivalent to itself.

Example 3

Input

s = "ac"

pairs = [

["a", "b"],

["b", "c"]

]

Output

True

Explanation

Since "a" = "c", we have "ac" = "cc", which is a palindrome.```

### Solution :

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

bool solve(string s, vector<vector<string>>& pairs) {
vector<int> parent(26);
iota(parent.begin(), parent.end(), 0);

function<int(int)> find = [&](int x) {
return parent[x] == x ? x : parent[x] = find(parent[x]);
};
auto join = [&](int a, int b) { parent[find(a)] = parent[find(b)]; };

for (auto& p : pairs) {
int a = p[0][0] - 'a';
int b = p[1][0] - 'a';
join(a, b);
}
for (char& c : s) {
int i = c - 'a';
c = find(i) + 'a';
}
int i = 0, j = (int)s.size() - 1;
while (i < j) {
if (s[i] ^ s[j]) return false;
i++, j--;
}
return true;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public boolean solve(String s, String[][] pairs) {
DisjointSetUnion dsu = new DisjointSetUnion(26);
for (String[] p : pairs) {
int a = p[0].charAt(0) - 'a';
int b = p[1].charAt(0) - 'a';
dsu.connect(a, b);
}
for (int i = 0; i < s.length(); i++) {
int j = s.length() - 1 - i;
int a = s.charAt(i) - 'a';
int b = s.charAt(j) - 'a';
if (!dsu.connected(a, b))
return false;
}
return true;
}

static class DisjointSetUnion {
public int[] parent;
public int[] weight;
public int count;

public DisjointSetUnion(int N) {
count = N;
parent = new int[N];
for (int i = 0; i < N; i++) parent[i] = i;
weight = new int[N];
Arrays.fill(weight, 1);
}

//"find"
public int root(int p) {
if (p == parent[p])
return p;
return parent[p] = root(parent[p]);
}

//"union"
public void connect(int p, int q) {
p = root(p);
q = root(q);
if (p == q)
return;
if (weight[p] < weight[q]) {
parent[p] = q;
weight[q] += weight[p];
} else {
parent[q] = p;
weight[p] += weight[q];
}
count--;
}

public boolean connected(int p, int q) {
return root(p) == root(q);
}
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, s, pairs):
parent = {}
for c in s:
parent[c] = c

def find(c):
if c not in parent or parent[c] == c:
return c
parent[c] = find(parent[c])
return parent[c]

def union(a, b):
a = find(a)
b = find(b)
if a != b:
parent[b] = a

for s1, e1 in pairs:
union(s1, e1)

s = [find(x) for x in s]

return s == s[::-1]```
```

## Cycle Detection

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## Tree: Preorder Traversal

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