Find the Mobile Number formed using first digits of arrays of absolute differences of consecutive numbers

Given a String ph[], the task is to find a new number for the user, based on following conditions: 

1. The new number will also start from the same digit as of original number.
2. The digits of the new number will be the first digits of a series of arrays of absolute differences of the consecutive elements.

Examples:

Input: ph = “9827218706”
Output: 9154301011
Explanation:

Input: ph =”9647253846″
Output: 9310100011

Approach: Consider the following steps to solve this problem: 

1. Convert every character from the string into integer and store it into the array ph1[] using list comprehension.
2. Declare an empty string ph2.
3. Convert first element of array ph1[ ] into a string and add it to ph2.
4. Using List comprehension create an array by storing the absolute difference of consecutive elements.
5. Assign this array to ph1.
6. Repeat step 3-5, ten times as the phone number have ten digits.

Below is the implementation of the above approach.

Java

// Java program for above approach
public class Mobile
{

  // Function to find lucky phone number
  static String phone(String ph, int n)
  {

    // Converting char to int and storing into array.
    int[] ph1 = new int[n];
    for (int i = 0; i < n; i++)
      ph1[i] = ph.charAt(i) - '0';

    // Empty string to store lucky number.
    String ph2 = "";

    // Loop for performing action
    // and adding digit to ph2.
    for (int i = 0; i < n; i++) {
      // Convert first element into
      // string and adding to ph2.
      ph2 += ph1[0];

      // Creating new ph1 by subtracting
      // consecutive element.
      int ph3[] = new int[ph1.length - 1];
      for (int j = 0; j < ph1.length - 1; j++)
        ph3[j] = Math.abs(ph1[j] - ph1[j + 1]);
      ph1 = ph3;
    }

    // Return lucky number ph2
    return ph2;
  }
  // Driver code
  public static void main(String[] args)
  {

    // Original number
    String ph = "9827218706";

    // Calling phone function.
    String num = phone(ph, ph.length());

    // Print the lucky number
    System.out.println(num);
  }
}

// This code is contributed by Lovely Jain

Python3

# Function to find lucky phone number
def phone(ph, n):

    # Converting char to int and storing into array.
    ph1 = [int(i) for i in ph]

    # Empty string to store lucky number.
    ph2 = ""

    # Loop for performing action
    # and adding digit to ph2.
    for _ in range(n):

        # Convert first element into
        # string and adding to ph2.
        ph2 += str(ph1[0])

        # Creating new ph1 by subtracting
        # consecutive element.
        ph1 = [abs(ph1[j]-ph1[j + 1]) \
              for j in range(len(ph1)-1)]

    # Return lucky number ph2
    return ph2


# Original number
ph = "9827218706"

# Calling phone function.
num = phone(ph, len(ph))

# Print the lucky number
print(num)

C#

// C# program for above approach
using System;
public class Mobile
{

  // Function to find lucky phone number
  static String phone(String ph, int n)
  {

    // Converting char to int and storing into array.
    int[] ph1 = new int[n];
    for (int i = 0; i < n; i++)
      ph1[i] = ph[i] - '0';

    // Empty string to store lucky number.
    String ph2 = "";

    // Loop for performing action
    // and adding digit to ph2.
    for (int i = 0; i < n; i++) {
      // Convert first element into
      // string and adding to ph2.
      ph2 += ph1[0];

      // Creating new ph1 by subtracting
      // consecutive element.
      int[] ph3 = new int[ph1.Length - 1];
      for (int j = 0; j < ph1.Length - 1; j++)
        ph3[j] = Math.Abs(ph1[j] - ph1[j + 1]);
      ph1 = ph3;
    }

    // Return lucky number ph2
    return ph2;
  }
  // Driver code
  public static void Main()
  {

    // Original number
    String ph = "9827218706";

    // Calling phone function.
    String num = phone(ph, ph.Length);

    // Print the lucky number
    Console.Write(num);
  }
}

// This code is contributed by Saurabh Jaiswal

Javascript

<script>
// Function to find lucky phone number
function phone(ph, n) {

    // Converting char to int and storing into array.
    let ph1 = [];
    for (i of ph)
        ph1.push(i)

    // Empty string to store lucky number.
    let ph2 = ""

    // Loop for performing action
    // and adding digit to ph2.
    for (let _ = 0; _ < n; _++) {

        // Convert first element into
        // string and adding to ph2.
        ph2 += new String(ph1[0])

        // Creating new ph1 by subtracting
        // consecutive element.
        let temp = []
        for (let j = 0; j < ph1.length - 1; j++) {
            temp.push(Math.abs(ph1[j] - ph1[j + 1]))
        }
        ph1 = temp
    }

    // Return lucky number ph2
    return ph2
}

// Original number
let ph = "9827218706"

// Calling phone function.
let num = phone(ph, ph.length)

// Print the lucky number
document.write(num)

// This code is contributed by gfgking.
</script>

C++

#include <iostream>
#include <cmath>
#include <string>
#include <vector>

using namespace std;

// Function to find lucky phone number
string phone(string ph, int n) {
    // Vector to store the digits of the phone number
    vector<int> ph1;
    for (int i = 0; i < n; i++) {
        // Convert each character to its numerical value
        ph1.push_back(ph[i] - '0');
    }

    // Empty string to store the lucky phone number
    string ph2 = "";

    // Loop to perform the operations and add the digits to ph2
    for (int i = 0; i < n; i++) {
        // Convert the first element to a string and add it to ph2
        ph2 += to_string(ph1[0]);

        // Create a new vector ph3 by subtracting consecutive elements
        vector<int> ph3;
        for (int j = 0; j < ph1.size() - 1; j++) {
            ph3.push_back(abs(ph1[j] - ph1[j + 1]));
        }
        ph1 = ph3;
    }

    // Return the lucky phone number
    return ph2;
}

int main() {
    // Original phone number
    string ph = "9827218706";

    // Call the phone function
    string num = phone(ph, ph.length());

    // Print the lucky phone number
    cout << num << endl;

    return 0;
}

#This code is contributed by Edula Vinay Kumar Reddy

Output

9154301011

Time Complexity: O(N*N), The time complexity is O(N*N) because there are two nested loops in the “phone()” function. The first loop iterates over each digit of the input phone number, which takes O(N) time. The second loop also iterates over each digit of the phone number, but it creates a new vector “ph3” by subtracting consecutive elements. This operation takes O(N) time because it performs N-1 subtractions. Therefore, the overall time complexity of the algorithm is O(N*N).
Auxiliary Space: O(N), The space complexity is O(N) because the algorithm uses two vectors to store the digits of the input phone number and the difference between consecutive digits. Both vectors have a maximum length of N, which is the length of the input phone number. Therefore, the space complexity is O(N).

Efficient Approach: In this approach, no extra space is required for storing elements in the array. First, declare an empty string ph2 in which lucky number will be stored, now create a for loop in which the first character of the string will be added to ph2 and again another for loop to find the absolute difference of consecutive element. Now the string of absolute difference will be assigned to ph1 which is the original number and the same steps will be followed. Follow the steps below to solve the problem:

• Initialize a string variable ph2[] as an empty string.
• Iterate over the range [0, N) using the variable i and perform the following tasks:
  • Add ph[0] to the variable ph2[].
  • Initialize a string variable S[] as an empty string.
  • Iterate over the range [0, N-1) using the variable j and perform the following tasks:
    • Add the value of str(abs(int(ph[j])-int(ph[j+1]))) to the variable S[].
  • Set the value of ph as S[].
• After performing the above steps, print the value of ph2[] as the answer.

Below is the implementation of the above approach.

C++

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find lucky number.
string phone(string ph, int n)
{

    // ph2 is empty string to store lucky number.
    string ph2 = "";

    // For loop for finding lucky number
    for (int i = 0; i < ph.length(); i++) {

        // Add first element of ph to ph2
        ph2 += ph[0];

        // S for storing the difference
        string S = "";

        // Loop to calculate the absolute difference
        for (int j = 0; j < ph.length(); j++) {
            int x = abs(int(ph[j]) - int(ph[j + 1]));
            S += x + '0';
        }

        // Assigning S to ph.
        ph = S;
    }

    // Return the lucky number
    return ph2;
}

// Driver Code
int main()
{

    // Original number
    string ph = "9827218706";

    // Call phone function
    string num = phone(ph, ph.length());

    // Printing lucky number
    cout << (num);
}

// This code is contributed by Potta Lokesh

Java

// Java program for the above approach
import java.util.*;
 class GFG {

  // Function to find lucky number.
  static String phone(String ph, int n)
  {

    // ph2 is empty string to store lucky number.
    String ph2 = "";

    // For loop for finding lucky number
    for (int i = 0; i < ph.length(); i++)
    {

      // Add first element of ph to ph2
      ph2 += ph.charAt(0);

      // S for storing the difference
      String S = "";

      // Loop to calculate the absolute difference
      for (int j = 0; j < ph.length()-1; j++)
      {
        int x = Math.abs(ph.charAt(j) - ph.charAt(j+1));
        S += (char)(x + '0');
      }

      // Assigning S to ph.
      ph = S;
    }

    // Return the lucky number
    return ph2;

  }

  // Driver Code
  public static void main(String args[])
  {
    // Original number
    String ph = "9827218706";

    // Call phone function
    String num = phone(ph, ph.length());

    // Printing lucky number
    System.out.println(num);
  }
}

// This code is contributed by avijitmondal1998

Python3

# Function to find lucky number.
def phone(ph, n):

    # ph2 is empty string to store lucky number.
    ph2 = ""

    # For loop for finding lucky number
    for i in range(len(ph)):

        # Add first element of ph to ph2
        ph2 += ph[0]

        # S for storing the difference
        S = ""

        # Loop to calculate the absolute difference
        for j in range(len(ph)-1):
            x = abs(int(ph[j])-int(ph[j + 1]))
            S += str(x)

        # Assigning S to ph.
        ph = S

    # Return the lucky number
    return ph2


# Original number
ph = "9827218706"

# Call phone function
num = phone(ph, len(ph))

# Printing lucky number
print(num)

C#

// C# code for the above approach
using System;
class GFG
{

  // Function to find lucky number.
  static string phone(string ph, int n)
  {

    // ph2 is empty string to store lucky number.
    string ph2 = "";

    // For loop for finding lucky number
    for (int i = 0; i < ph.Length; i++)
    {

      // Add first element of ph to ph2
      ph2 += ph[0];

      // S for storing the difference
      string S = "";

      // Loop to calculate the absolute difference
      for (int j = 0; j < ph.Length; j++)
      {
        int x = Math.Abs(ph[j] - ph[j + 1]);
        S += x + '0';
      }

      // Assigning S to ph.
      ph = S;
    }

    // Return the lucky number
    return ph2;
  }

  // Driver Code
  public static void Main()
  {

    // Original number
    string ph = "9827218706";

    // Call phone function
    string num = phone(ph, ph.Length);

    // Printing lucky number
    Console.WriteLine (num);
  }
}

// This code is contributed by ukasp.

Javascript

<script>
// javascript program for the above approach

    // Function to find lucky number.
     function phone( ph , n) {

        // ph2 is empty string to store lucky number.
         // Converting char to int and storing into array.
    let ph1 = [];
    for (i of ph)
        ph1.push(i)

    // Empty string to store lucky number.
    let ph2 = ""

    // Loop for performing action
    // and adding digit to ph2.
    for (let _ = 0; _ < n; _++) {

        // Convert first element into
        // string and adding to ph2.
        ph2 += new String(ph1[0])

        // Creating new ph1 by subtracting
        // consecutive element.
        let S = []
        for (let j = 0; j < ph1.length - 1; j++) {
            S.push(Math.abs(ph1[j] - ph1[j + 1]))
        }
        ph1 = S
    }

        // Return the lucky number
        return ph2;

    }

    // Driver Code

        // Original number
        var ph = "9827218706";

        // Call phone function
        var num = phone(ph, ph.length);

        // Printing lucky number
        document.write(num);

// This code is contributed by umadevi9616
</script>

Output

9154301011

Time Complexity: O(N*N), The code has a nested for loop. The outer loop runs N times, where N is the length of the input string. The inner loop also runs N times, where for each iteration, the difference between two adjacent characters of the input string is calculated. Therefore, the total time complexity of the code is O(N*N).
Auxiliary Space: O(N),The code uses a few variables to store the input and output strings. The input string and the lucky number string are stored as strings, and therefore, they use O(N) space, where N is the length of the input string. The difference string S also uses O(N) space. Therefore, the total space complexity of the code is O(N).