Check if Array elements can be made equal by adding unit digit to the number

Last Updated : 12 Sep, 2022

Given an array arr[] of N integers, the task is to check whether it is possible to make all the array elements identical by applying the following option any number of times:

Choose any number and add its unit digit to it.

Examples:

Input: arr[] = {1, 2, 4, 8, 24}
Output: Possible
Explanation:
For 1: 1 -> 2 -> 4 -> 8 -> 16- > 22 -> 24
For 2: 2 -> 4 -> 8 -> 16 -> 22 -> 24
For 4: 4 -> 8 -> 16 -> 22 -> 24
For 8: 16- > 22 -> 24
For 24: 24 (it's already 24, so, no need to change)

Input: arr[] = {5, 10}
Output: Possible

Approach: The problem can be solved based on the following observation.

Any number can be converted to have either 0 or 2 as its unit digit by repetitively adding its unit digit to itself at most 10 times.

The idea is to

Initially make the unit digit of every number either 2 or 0.
Then for the numbers that have 0 as their unit digit, all of those numbers must be same(because x+0 = 0), and
For every number with 2 as the unit digit, the difference between them must be multiple of 20 so that they can be made equal.

Follow the steps to solve this problem:

Add the unit digit of each number to itself until its unit digit becomes 2 or 0.
For every number that has 0 as a unit digit, check whether all the numbers are equal.
For numbers with 2 as the unit digit, the difference between them must be a multiple of 20.

Below is the implementation of the above approach:

C++

// C++ code to implement the approach  #include <bits/stdc++.h> using namespace std;  // Function to check whether the elements // of array can be equal or not string solve(int* arr, int n) {     // Initialize flag as 1     bool flag = 1;      // Make unit digit of every element of array     // with either 0 or 2     for (int i = 0; i < n; i++) {         while (arr[i] % 10 != 0 && arr[i] % 10 != 2)             arr[i] += arr[i] % 10;     }      // Check if it is possible to make     // the array elements identical or not     for (int i = 0; i < n; i++) {          // Elements for 0 as unit digit         if ((arr[0] % 10 == 0) && (arr[i] != arr[0])) {             flag = 0;             break;         }          // Elements for 2 as unit digit         if ((arr[i] - arr[0]) % 20) {             flag = 0;             break;         }     }      if (flag)         return "Possible";     else         return "Impossible"; }  // Driver Code int main() {     int arr[] = { 1, 2, 4, 8, 24 };     int N = sizeof(arr) / sizeof(arr[0]);      // Function call     cout << solve(arr, N);     return 0; }

Java

// Java code for the above approach import java.io.*;  class GFG {  // Function to check whether the elements // of array can be equal or not static String solve(int[] arr, int n) {     // Initialize flag as 1     boolean flag = true;      // Make unit digit of every element of array     // with either 0 or 2     for (int i = 0; i < n; i++) {         while (arr[i] % 10 != 0 && arr[i] % 10 != 2)             arr[i] += arr[i] % 10;     }      // Check if it is possible to make     // the array elements identical or not     for (int i = 0; i < n; i++) {          // Elements for 0 as unit digit         if ((arr[0] % 10 == 0) && (arr[i] != arr[0])) {             flag = false;             break;         }          // Elements for 2 as unit digit         if (((arr[i] - arr[0]) % 20) != 0) {             flag = false;             break;         }     }      if (flag )         return "Possible";     else         return "Impossible"; }      // Driver code     public static void main(String[] args)     {         int arr[] = { 1, 2, 4, 8, 24 };     int N = arr.length;      // Function call     System.out.println( solve(arr, N));     } }   // This code is contributed by sanjoy_62.

Python3

# Python3 code for the above approach  # Function to check whether the elements of array can # be equal or not def solve(arr, n):          # Initialize flag as true     flag = True      # Make unit digit of every element of array with     # either 0 or 2     # for (int i = 0; i < n; i++) {     for i in range(0, n):         while ((arr[i] % 10 != 0) and (arr[i] % 10 != 2)):             arr[i] += arr[i] % 10              # Check if it is possible to make the array             # elements identical or not             # for (int i = 0; i < n; i++) {     for i in range(0, n):               # Elements for 0 as unit digit         if ((arr[0] % 10) == 0 and (arr[i] != arr[0])):             flag = False             break              # Elements for 2 as unit digit         if ((arr[i] - arr[0]) % 20 != 0):              flag = False             break      if (flag):         return "Possible"     else:         return "Impossible"  # Driver Code arr = [1, 2, 4, 8, 24] N = len(arr)  # Function call ans = solve(arr, N) print(ans)  # This code is contributed by akashish.

// C# code for the above approach  using System;  public class GFG {      // Function to check whether the elements of array can     // be equal or not     static string solve(int[] arr, int n)     {         // Initialize flag as true         bool flag = true;          // Make unit digit of every element of array with         // either 0 or 2         for (int i = 0; i < n; i++) {             while (arr[i] % 10 != 0 && arr[i] % 10 != 2) {                 arr[i] += arr[i] % 10;             }         }          // Check if it is possible to make the array         // elements identical or not         for (int i = 0; i < n; i++) {             // Elements for 0 as unit digit             if ((arr[0] % 10) == 0 && (arr[i] != arr[0])) {                 flag = false;                 break;             }              // Elements for 2 as unit digit             if ((arr[i] - arr[0]) % 20 != 0) {                 flag = false;                 break;             }         }          if (flag)             return "Possible";         else             return "Impossible";     }      static public void Main()     {          // Code         int[] arr = { 1, 2, 4, 8, 24 };         int N = arr.Length;          // Function call         Console.Write(solve(arr, N));     } }  // This code is contributed by lokeshmvs21.

JavaScript

<script>     // JavaScript code for the approach  // Function to check whether the elements // of array can be equal or not function solve(arr, n) {     // Initialize flag as 1     let flag = true;       // Make unit digit of every element of array     // with either 0 or 2     for (let i = 0; i < n; i++) {         while (arr[i] % 10 != 0 && arr[i] % 10 != 2)             arr[i] += arr[i] % 10;     }       // Check if it is possible to make     // the array elements identical or not     for (let i = 0; i < n; i++) {           // Elements for 0 as unit digit         if ((arr[0] % 10 == 0) && (arr[i] != arr[0])) {             flag = false;             break;         }           // Elements for 2 as unit digit         if (((arr[i] - arr[0]) % 20) != 0) {             flag = false;             break;         }     }       if (flag )         return "Possible";     else         return "Impossible"; }      // Driver code      // Given input     let arr = [ 1, 2, 4, 8, 24 ];     let N = arr.length;       // Function call     document.write( solve(arr, N));  // This code is contributed by code_hunt. </script>

Output

Possible

Time Complexity: O(N)
Auxiliary Space: O(1)

Check if elements of array can be made equal by multiplying given prime numbers

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Check if Array elements can be made equal by adding unit digit to the number

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