Given a String ph[], the task is to find the absolute difference of consecutive elements and insert the result in between the consecutive elements. By doing this size of phone numbers will increase from 10 to 19. Now we have to compare digits from first and last and select maximum in this way we will get a new phone number.
Examples:
Input: ph = “9647253846”
Output: 9364857553
Explanation:
Input: ph = “7635892563”
Output: 7363535791
Approach: The idea is to use two pointers approach to solve this problem along with using a string variable to add the phone numbers with a difference. Follow the steps below to solve the problem:
- Initialize a string variable S[] as an empty string.
- Iterate over the range [0, N-1) using the variable i and perform the following tasks:
- Add ph[i] to the variable S.
- Initialize the variables s as int(ph[i]) – int(ph[i+1]) and x as abs(s).
- Add str(x) to the variable S.
- Add ph[9] to the variable S.
- Initialize the variables s as 0 and e as len(S)-1.
- Initialize a string variable ph2[] as an empty string.
- Traverse in a while loop till s is not equals to e and perform the following tasks:
- Add the max of S[s] or S[e] to the variable ph2[].
- Increase the value of s by 1 and decrease the value of e by 1.
- Add S[e] to the variable ph2[].
- After performing the above steps, print the value of ph2[] as the answer.
Below is the implementation of the above approach.
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to create the phone numberstring phNumber(string ph, int n){ // Empty string to store extended number string S = ""; // Loop for extend phone number for (int i = 0; i < n - 1; i++) { // Add character form string S += ph[i]; // Finding absolute difference int x = abs(ph[i] - ph[i + 1]); // Adding the difference to S S += to_string(x); } // Adding last digit of ph S += ph[9]; // Using 2 pointer algorithm to form comparison // s is start index int s = 0; // e is end index int e = S.length() - 1; // ph2 an empty string for // storing the phone number string ph2 = ""; // Loop till e become equal to s while (s != e) { // Comparing element at s and e // index and adding maximum if (S[s] > S[e]) { ph2 += S[s]; } else { ph2 += S[e]; } // Increment start index s += 1; // Decrement end index e -= 1; } // When s = e loop will terminate // so adding element at index s = e ph2 += S[e]; // Return phone number return ph2;}// Driver Codeint main(){ // Original phone number string ph = "9647253846"; // Calling function to create new number string num = phNumber(ph, ph.length()); // Print the new number cout << num;}// This code is contributed by Samim Hossain Mondal. |
Java
// Java program for the above approachclass GFG { // Function to create the phone number static String phNumber(String ph, int n) { // Empty string to store extended number String S = ""; // Loop for extend phone number for (int i = 0; i < n - 1; i++) { // Add character form string S += ph.charAt(i); // Finding absolute difference int x = Math.abs(ph.charAt(i) - ph.charAt(i + 1)); // Adding the difference to S S += Integer.toString(x); } // Adding last digit of ph S += ph.charAt(9); // Using 2 pointer algorithm to form comparison // s is start index int s = 0; // e is end index int e = S.length() - 1; // ph2 an empty string for // storing the phone number String ph2 = ""; // Loop till e become equal to s while (s != e) { // Comparing element at s and e // index and adding maximum if (S.charAt(s) > S.charAt(e)) { ph2 += S.charAt(s); } else { ph2 += S.charAt(e); } // Increment start index s += 1; // Decrement end index e -= 1; } // When s = e loop will terminate // so adding element at index s = e ph2 += S.charAt(e); // Return phone number return ph2; } // Driver Code public static void main(String args[]) { // Original phone number String ph = "9647253846"; // Calling function to create new number String num = phNumber(ph, ph.length()); // Print the new number System.out.println(num); }}// This code is contributed by gfgking |
Python3
# Function to create the phone numberdef phNumber(ph, n): # Empty string to store extended number S ="" # Loop for extend phone number for i in range(n-1): # Add character form string S+= ph[i] # Finding absolute difference s = int(ph[i])-int(ph[i + 1]) x = abs(s) # Adding the difference to S S+= str(x) # Adding last digit of ph S+= ph[9] # Using 2 pointer algorithm to form comparison # s is start index s = 0 # e is end index e = len(S)-1 # ph2 an empty string for # storing the phone number ph2 ="" # Loop till e become equal to s while s != e: # Comparing element at s and e # index and adding maximum ph2+= max(S[s], S[e]) # Increment start index s+= 1 # Decrement end index e-= 1 # When s = e loop will terminate # so adding element at index s = e ph2+= S[e] # Return phone number return ph2 # Original phone numberph = "9647253846"# Calling function to create new numbernum = phNumber(ph, len(ph))# Print the new numberprint(num) |
C#
// C# program for the above approachusing System;class GFG { // Function to create the phone number static string phNumber(string ph, int n) { // Empty string to store extended number string S = ""; // Loop for extend phone number for (int i = 0; i < n - 1; i++) { // Add character form string S += ph[i]; // Finding absolute difference int x = Math.Abs(ph[i] - ph[i + 1]); // Adding the difference to S S += x.ToString(); } // Adding last digit of ph S += ph[9]; // Using 2 pointer algorithm to form comparison // s is start index int s = 0; // e is end index int e = S.Length - 1; // ph2 an empty string for // storing the phone number string ph2 = ""; // Loop till e become equal to s while (s != e) { // Comparing element at s and e // index and adding maximum if (S[s] > S[e]) { ph2 += S[s]; } else { ph2 += S[e]; } // Increment start index s += 1; // Decrement end index e -= 1; } // When s = e loop will terminate // so adding element at index s = e ph2 += S[e]; // Return phone number return ph2; } // Driver Code public static void Main() { // Original phone number string ph = "9647253846"; // Calling function to create new number string num = phNumber(ph, ph.Length); // Print the new number Console.Write(num); }}// This code is contributed by ukasp. |
Javascript
<script> // JavaScript code for the above approach // Function to create the phone number function phNumber(ph, n) { // Empty string to store extended number let S = "" // Loop for extend phone number for (let i = 0; i < n - 1; i++) { // Add character form string S += ph[i] // Finding absolute difference let s = ph[i].charCodeAt(0) - ph[i + 1].charCodeAt(0) let x = Math.abs(s) // Adding the difference to S S += x.toString(); } // Adding last digit of ph S += ph[9] // Using 2 pointer algorithm to form comparison // s is start index s = 0 // e is end index e = S.length - 1 // ph2 an empty string for // storing the phone number ph2 = "" // Loop till e become equal to s while (s != e) { // Comparing element at s and e // index and adding maximum if (S[s].charCodeAt(0) > S[e].charCodeAt(0)) { ph2 += S[s]; } else { ph2 += S[e]; } // Increment start index s += 1 // Decrement end index e -= 1 } // When s = e loop will terminate // so adding element at index s = e ph2 += S[e] // Return phone number return ph2 } // Original phone number let ph = "9647253846" // Calling function to create new number let num = phNumber(ph, ph.length) // Print the new number document.write(num) // This code is contributed by Potta Lokesh </script> |
Output
9364857553
Time Complexity: O(N)
Auxiliary Space: O(N)
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