C++ Program to Check if all array elements can be converted to pronic numbers by rotating digits
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}Â
Output: TrueÂ
Explanation:Â
arr[0] ? Right rotation once modifies arr[0] to 132 (= 11 × 12).Â
arr[1] ? Right rotation once modifies arr[0] to 240 (= 15 × 16).Â
arr[2] ? Right rotation twice modifies arr[2] to 462 (= 21 × 22).Â
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++
#include <bits/stdc++.h>
using namespace std;
bool isPronic( int x)
{
for ( int i = 0; i < ( int )( sqrt (x)) + 1; i++)
{
if (x == i * (i + 1))
{
return true ;
}
}
return false ;
}
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 ;
}
bool check( int arr[], int N)
{
for ( int i = 0; i < N; i++)
{
if (checkRot(arr[i]) == false )
{
return false ;
}
}
return true ;
}
int main()
{
int arr[] = { 321, 402, 246, 299 };
int N = sizeof (arr) / sizeof (arr[0]);
cout << (check(arr, N) ? "True" : "False" );
return 0;
}
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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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