# Program to maximize the number of equivalent pairs after swapping in Python

Suppose we have a list of numbers A and list of numbers B of same length. We also have a 2D list of numbers C where each element is of the form [i, j] this indicates we can swap A[i] and A[j] as many times as we want. We have to find the maximum number of pairs where A[i] = B[i] after the swapping.

So, if the input is like A = [5, 6, 7, 8], B = [6, 5, 8, 7], C = [[0, 1],[2, 3]], then the output will be 4, as we can swap A[0] with A[1] then A[2] with A[3].

To solve this, we will follow these steps −

• N := size of A
• graph := a graph by attaching given edges bidirectionally.
• ans := 0
• seen := a list of size N and fill with False
• for u in range 0 to N, do
• if seen[u] is zero, then
• queue := a queue and insert u
• seen[u] := True
• for each node in queue, do
• for each nei in graph[node], do
• if seen[nei] is false, then
• insert nei at the end of queue
• seen[nei] := True
• count := a map that contains count of B[i] elements for all i in queue
• for each i in queue, do
• if count[A[i]] is non-zero, then
• count[A[i]] := count[A[i]] - 1
• ans := ans + 1
• return ans

Let us see the following implementation to get better understanding −

## Example

Live Demo

from collections import Counter
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
ob = Solution()
A = [5, 6, 7, 8]
B = [6, 5, 8, 7]
C = [[0, 1],[2, 3]]
print(ob.solve(A, B, C))

## Input

[5, 6, 7, 8], [6, 5, 8, 7], [[0, 1],[2, 3]]

## Output

4