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C++ Program to Check if all array elements can be converted to pronic numbers by rotating digits

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  • Last Updated : 27 Jan, 2022

Given an array arr[] of size N, the task is to check if it is possible to convert all of the array elements to a pronic number by rotating the digits of array elements any number of times.

Examples:

Input: {321, 402, 246, 299} 
Output: True 
Explanation: 
arr[0] → Right rotation once modifies arr[0] to 132 (= 11 × 12). 
arr[1] → Right rotation once modifies arr[0] to 240 (= 15 × 16). 
arr[2] → Right rotation twice modifies arr[2] to 462 (= 21 × 22). 
arr[3] → Right rotation twice modifies arr[3] to 992 (= 31 × 32).

Input: {433, 653, 402, 186}
Output: False

Approach: Follow the steps below to solve the problem:

  • Traverse the array and check for each array element, whether it is possible to convert it to a pronic number.
  • For each array element, apply all the possible rotations and check after each rotation, whether the generated number is pronic or not.
  • If it is not possible to convert any array element to a pronic number, print “False”.
  • Otherwise, print “True”.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// function to check Pronic Number
bool isPronic(int x)
{
    for (int i = 0; i < (int)(sqrt(x)) + 1; i++) 
    {
  
        // Checking Pronic Number
        // by multiplying consecutive
        // numbers
        if (x == i * (i + 1)) 
        {
            return true;
        }
    }
    return false;
}
  
// Function to check if any permutation
// of val is a pronic number or not
bool checkRot(int val)
{
  
    string temp = to_string(val);
    for (int i = 0; i < temp.length(); i++) 
    {
        if (isPronic(stoi(temp)) == true)
        {
            return true;
        }
        temp = temp.substr(1, temp.size() - 1) + temp[0];
    }
    return false;
}
  
// Function to check if all array
// elements can be converted to
// a pronic number or not
bool check(int arr[], int N)
{
  
    // Traverse the array
    for (int i = 0; i < N; i++) 
    {
  
        // If current element
        // cannot be converted
        // to a pronic number
        if (checkRot(arr[i]) == false
        {
            return false;
        }
    }
    return true;
}
  
// Driven Program
int main()
{
    
    // Given array
    int arr[] = { 321, 402, 246, 299 };
    int N = sizeof(arr) / sizeof(arr[0]);
  
    // function call
    cout << (check(arr, N) ? "True" : "False");
  
    return 0;
}
  
// This code is contributed by Kingash.

Output: 

True

 

Time Complexity: O(N3/2)
Auxiliary Space: O(1)

Please refer complete article on Check if all array elements can be converted to pronic numbers by rotating digits for more details!


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