# Maximize the Number of Equivalent Pairs After Swaps - Google Top Interview Questions

### Problem Statement :

```You are given a list of integers of the same length A and B.

You are also given a two-dimensional list of integers C where each element is of the form [i, j] which means that you can swap A[i] and A[j] as many times as you want.

Return the maximum number of pairs where A[i] = B[i] after the swapping.

Constraints

n ≤ 100,000 where n is the length of A and B

m ≤ 100,000 where m is the length of C

Example 1

Input

A = [1, 2, 3, 4]

B = [2, 1, 4, 3]

C = [

[0, 1],

[2, 3]

]

Output

4

Explanation

We can swap A[0] with A[1] then A[2] with A[3].```

### Solution :

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

class UnionFind {
private:
vector<int> parents, rank;

public:
UnionFind(int n) {
parents.resize(n);
rank.resize(n);
for (int i = 0; i < n; i++) {
parents[i] = i;
rank[i] = 1;
}
}

int find(int node) {
int root = node;

while (root != parents[root]) {
root = parents[root];
}

// Path compression
while (node != root) {
int temp = parents[node];
parents[node] = root;
node = temp;
}

return root;
}

void unify(int a, int b) {
int rootA = find(a);
int rootB = find(b);

if (rootA == rootB) return;

// Union by rank
if (rank[rootA] > rank[rootB]) {
parents[rootB] = rootA;
} else if (rank[rootB] > rank[rootA]) {
parents[rootA] = rootB;
} else {
parents[rootB] = rootA;
rank[rootA]++;
}
}

vector<int> get_parents_array() {
return parents;
}
};

// Time and Space: O(N)
int solve(vector<int>& A, vector<int>& B, vector<vector<int>>& C) {
int n = A.size();
UnionFind union_find(n);

for (vector<int>& edge : C) {
union_find.unify(edge[0], edge[1]);  // Do unions to form groups
}

vector<int> parents = union_find.get_parents_array();
unordered_map<int, vector<int>> grp_map;

for (int i = 0; i < n; i++) {
int parent = union_find.find(i);
grp_map[parent].push_back(i);  // Map parents to list of indices in their group
}

int count = 0;

for (auto& grp : grp_map) {  // For each group
vector<int>& indices = grp.second;
unordered_map<int, int> value_map;

for (int idx : indices) {  // Map values found
value_map[A[idx]]++;
}

for (int idx : indices) {  // For same indices check how many matched values are found
if (--value_map[B[idx]] >= 0) {
count++;
}
}
}

return count;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
class DisjointSet {
int node;
DisjointSet parent;
public DisjointSet(int val) {
this.node = val;
this.parent = this;
}
}

private Map<Integer, DisjointSet> map = new HashMap();
private Map<Integer, List<Integer>> swappableMap = new HashMap();
public int solve(int[] A, int[] B, int[][] C) {
int count = 0;
if (A.length == 0 || B.length == 0)
return 0;
for (int i = 0; i < A.length; i++) map.put(i, new DisjointSet(i));

for (int[] arr : C) {
int idx1 = arr[0];
int idx2 = arr[1];
union(idx1, idx2);
}

for (int i = 0; i < A.length; i++) {
DisjointSet set = map.get(i);
DisjointSet par = find(set);
}

for (int key : swappableMap.keySet()) {
List<Integer> list = swappableMap.get(key);

Map<Integer, Integer> freq1 = new HashMap();
Map<Integer, Integer> freq2 = new HashMap();

for (int i = 0; i < list.size(); i++) {
int idx = list.get(i);
freq1.put(A[idx], freq1.getOrDefault(A[idx], 0) + 1);
freq2.put(B[idx], freq2.getOrDefault(B[idx], 0) + 1);
}
for (int num : freq1.keySet()) {
count += (Math.min(freq1.get(num), freq2.getOrDefault(num, 0)));
}
}
return count;
}

private void union(int idx1, int idx2) {
DisjointSet set1 = map.get(idx1);
DisjointSet set2 = map.get(idx2);

DisjointSet f1 = find(set1);
DisjointSet f2 = find(set2);

if (f1.node == f2.node)
return;
f1.parent = f2;
}

private DisjointSet find(DisjointSet set) {
if (set.parent == set)
return set;
return set.parent = find(set.parent);
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, A, B, edges):
N = len(A)
graph = [[] for _ in range(N)]
for u, v in edges:
graph[u].append(v)
graph[v].append(u)

ans = 0
seen = [False] * N
for u in range(N):
if not seen[u]:
queue = [u]
seen[u] = True
for node in queue:
for nei in graph[node]:
if not seen[nei]:
queue.append(nei)
seen[nei] = True

count = Counter(B[i] for i in queue)
for i in queue:
if count[A[i]]:
count[A[i]] -= 1
ans += 1

return ans```
```

## Kundu and Tree

Kundu is true tree lover. Tree is a connected graph having N vertices and N-1 edges. Today when he got a tree, he colored each edge with one of either red(r) or black(b) color. He is interested in knowing how many triplets(a,b,c) of vertices are there , such that, there is atleast one edge having red color on all the three paths i.e. from vertex a to b, vertex b to c and vertex c to a . Note that

## Super Maximum Cost Queries

Victoria has a tree, T , consisting of N nodes numbered from 1 to N. Each edge from node Ui to Vi in tree T has an integer weight, Wi. Let's define the cost, C, of a path from some node X to some other node Y as the maximum weight ( W ) for any edge in the unique path from node X to Y node . Victoria wants your help processing Q queries on tree T, where each query contains 2 integers, L and

## Contacts

We're going to make our own Contacts application! The application must perform two types of operations: 1 . add name, where name is a string denoting a contact name. This must store name as a new contact in the application. find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting partial with and print the co

## No Prefix Set

There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio

## Cube Summation

You are given a 3-D Matrix in which each block contains 0 initially. The first block is defined by the coordinate (1,1,1) and the last block is defined by the coordinate (N,N,N). There are two types of queries. UPDATE x y z W updates the value of block (x,y,z) to W. QUERY x1 y1 z1 x2 y2 z2 calculates the sum of the value of blocks whose x coordinate is between x1 and x2 (inclusive), y coor

## Direct Connections

Enter-View ( EV ) is a linear, street-like country. By linear, we mean all the cities of the country are placed on a single straight line - the x -axis. Thus every city's position can be defined by a single coordinate, xi, the distance from the left borderline of the country. You can treat all cities as single points. Unfortunately, the dictator of telecommunication of EV (Mr. S. Treat Jr.) do