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);
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)
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
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