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.


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

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

Example 1


A = [1, 2, 3, 4]

B = [2, 1, 4, 3]

C = [

    [0, 1],

    [2, 3]





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

Solution :


                        Solution in C++ :

class UnionFind {
    vector<int> parents, rank;

    UnionFind(int 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;

    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

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

    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);
            swappableMap.computeIfAbsent(par.node, k -> new ArrayList()).add(i);

        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)
        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:

        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]:
                            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

View More Similar Problems

Unique Colors

You are given an unrooted tree of n nodes numbered from 1 to n . Each node i has a color, ci. Let d( i , j ) be the number of different colors in the path between node i and node j. For each node i, calculate the value of sum, defined as follows: Your task is to print the value of sumi for each node 1 <= i <= n. Input Format The first line contains a single integer, n, denoti

View Solution →

Fibonacci Numbers Tree

Shashank loves trees and math. He has a rooted tree, T , consisting of N nodes uniquely labeled with integers in the inclusive range [1 , N ]. The node labeled as 1 is the root node of tree , and each node in is associated with some positive integer value (all values are initially ). Let's define Fk as the Kth Fibonacci number. Shashank wants to perform 22 types of operations over his tree, T

View Solution →

Pair Sums

Given an array, we define its value to be the value obtained by following these instructions: Write down all pairs of numbers from this array. Compute the product of each pair. Find the sum of all the products. For example, for a given array, for a given array [7,2 ,-1 ,2 ] Note that ( 7 , 2 ) is listed twice, one for each occurrence of 2. Given an array of integers, find the largest v

View Solution →

Lazy White Falcon

White Falcon just solved the data structure problem below using heavy-light decomposition. Can you help her find a new solution that doesn't require implementing any fancy techniques? There are 2 types of query operations that can be performed on a tree: 1 u x: Assign x as the value of node u. 2 u v: Print the sum of the node values in the unique path from node u to node v. Given a tree wi

View Solution →

Ticket to Ride

Simon received the board game Ticket to Ride as a birthday present. After playing it with his friends, he decides to come up with a strategy for the game. There are n cities on the map and n - 1 road plans. Each road plan consists of the following: Two cities which can be directly connected by a road. The length of the proposed road. The entire road plan is designed in such a way that if o

View Solution →

Heavy Light White Falcon

Our lazy white falcon finally decided to learn heavy-light decomposition. Her teacher gave an assignment for her to practice this new technique. Please help her by solving this problem. You are given a tree with N nodes and each node's value is initially 0. The problem asks you to operate the following two types of queries: "1 u x" assign x to the value of the node . "2 u v" print the maxim

View Solution →