# Find same contacts in a list of contacts in Python

PythonServer Side ProgrammingProgramming

Suppose we have a list of contacts holding username, email and phone number in any order, we have to find the same contacts (When same person have many different contacts) and return the same contacts together. We have to keep in mind that −

• A contact can store username, email and phone fields according to any order.

• Two contacts are same if they have either same username or same email or same phone number.

So, if the input is like Contacts = [{"Amal", "amal@gmail.com", "+915264"},{ "Bimal", "bimal321@yahoo.com", "+1234567"},{ "Amal123", "+1234567", "amal_new@gmail.com"},{ "AmalAnother", "+962547", "amal_new@gmail.com"}], then the output will be [0,2,3],  as contacts at index [0,2,3] are same, and another contact at index 1.

To solve this, we will follow these steps −

• Define a function generate_graph() . This will take cnt, n, matrix

• for i in range 0 to n, do

• for j in range 0 to n, do

• matrix[i, j] := 0

• for i in range 0 to n, do

• for j in range i + 1 to n, do

• if cnt[i].slot1 is same as cnt[j].slot1 or cnt[i].slot1 is same as cnt[j].slot2 or cnt[i].slot1 is same as cnt[j].slot3 or cnt[i].slot2 is same as cnt[j].slot1 or cnt[i].slot2 is same as cnt[j].slot2 or cnt[i].slot2 is same as cnt[j].slot3 or cnt[i].slot3 is same as cnt[j].slot1 or cnt[i].slot3 is same as cnt[j].slot2 or cnt[i].slot3 is same as cnt[j].slot3, then

• matrix[i, j] := 1

• matrix[j, i] := 1

• come out from the loop

• Define a function visit_using_dfs() . This will take i, matrix, visited, sol, n

• visited[i] := True

• insert i at the end of sol

• for j in range 0 to n, do

• if matrix[i][j] is non-zero and not visited[j] is non-zero, then

• visit_using_dfs(j, matrix, visited, sol, n)

• From the main method, do the following −

• n := size of cnt

• sol := a new list

• matrix := make a square matrix of size n x n

• visited := make an array of size n, and fill with 0

• generate_graph(cnt, n, matrix)

• for i in range 0 to n, do

• if not visited[i] is non-zero, then

• visit_using_dfs(i, matrix, visited, sol, n)

• insert -1 at the end of sol

• for i in range 0 to size of sol, do

• if sol[i] is same as -1, then

• go to the next line

• otherwise,

• display sol[i]

## Example

Let us see the following implementation to get better understanding −

Live Demo

class contact:
def __init__(self, slot1, slot2, slot3):
self.slot1 = slot1
self.slot2 = slot2
self.slot3 = slot3
def generate_graph(cnt, n, matrix):
for i in range(n):
for j in range(n):
matrix[i][j] = 0
for i in range(n):
for j in range(i + 1, n):
if (cnt[i].slot1 == cnt[j].slot1 or cnt[i].slot1 == cnt[j].slot2 or cnt[i].slot1 == cnt[j].slot3 or cnt[i].slot2 == cnt[j].slot1 or cnt[i].slot2 == cnt[j].slot2 or cnt[i].slot2 == cnt[j].slot3 or cnt[i].slot3 == cnt[j].slot1 or cnt[i].slot3 == cnt[j].slot2 or cnt[i].slot3 == cnt[j].slot3):
matrix[i][j] = 1
matrix[j][i] = 1
break
def visit_using_dfs(i, matrix, visited, sol, n):
visited[i] = True
sol.append(i)
for j in range(n):
if (matrix[i][j] and not visited[j]):
visit_using_dfs(j, matrix, visited, sol, n)
def get_similar_contacts(cnt):
n = len(cnt)
sol = []
matrix = [[None] * n for i in range(n)]
visited =  * n
generate_graph(cnt, n, matrix)
for i in range(n):
if (not visited[i]):
visit_using_dfs(i, matrix, visited, sol, n)
sol.append(-1)
for i in range(len(sol)):
if (sol[i] == -1):
print()
else:
print(sol[i], end = " ")

cnt = [contact("Amal", "amal@gmail.com", "+915264"),
contact("Bimal", "bimal321@yahoo.com", "+1234567"),
contact("Amal123", "+915264", "amal_new@gmail.com"),
contact("AmalAnother", "+962547", "amal_new@gmail.com")]
get_similar_contacts(cnt)

## Input

cnt = [contact("Amal", "amal@gmail.com", "+915264"),
contact("Bimal", "bimal321@yahoo.com", "+1234567"),
contact("Amal123", "+915264", "amal_new@gmail.com"),
contact("AmalAnother", "+962547", "amal_new@gmail.com")]

## Output

0 2 3
1