Find same contacts in a list of contacts in Python

Python Server Side Programming Programming

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], [1] 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 = [0] * 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

Arnab Chakraborty

Published on 19-Aug-2020 11:15:57

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