Given an array arr[] consisting of N positive integers, the task is to rotate the digits of array elements in an anti-clockwise direction such that elements of the array elements are in alternate even-odd or odd-even form. If multiple solutions exists, then print any one of them. Otherwise, print -1.
Examples:
Input: arr[] = { 143, 251, 534, 232, 854 }
Output: 143 512 345 232 485
Explanation:
Rotating arr[1] by 1 in anticlockwise direction modifies arr[1] to 512.
Rotating arr[2] by 1 in anticlockwise direction modifies arr[2] to 345.
Rotating arr[4] by 2 in anticlockwise direction modifies arr[4] to 485.Input: arr[] = { 44, 23, 21, 33, 14 }
Output: 44 23 12 33 14
Approach: The above problem can be solved by either modifying the first array element to an odd number or an even number. The array element can be modified by converting the number into a string and then rotate the characters of the string left as needed. Follow the steps below to solve the problem:
- Rearrange the first array element as an even number and check if the remaining array elements can be rearranged into odd-even alternately or not. If found to be true, then rearrange the array elements into odd-even alternately and print the array elements.
- Otherwise, rearrange the first array element as an odd number and check if the remaining array elements can be rearranged into even-odd alternately or not. If found to be true, then rearrange the remaining array elements into even-odd alternately and print the array elements.
- Otherwise, print -1.
Below is the implementation of the above approach:
// c++ program of the above approach #include <bits/stdc++.h> using namespace std;
// Utility function to rotate the digits of // array elements such that array elements are // in placed even-odd or odd-even alternately bool is_possible(vector< int >& arr, bool check)
{ // Checks if array can be converted
// into even-odd or odd-even form
bool exists = true ;
// Store array elements
vector< int > cpy = arr;
bool flag;
// Traverse the array
for ( int i = 0; i < arr.size(); i++) {
// Check if arr[i] is already
// at correct position
if (arr[i] % 2 == check) {
check = !(check);
continue ;
}
// Checks if it is possible
// to modify the number arr[i]
// by rotating the digits of
// the number anticlockwise
flag = false ;
// Stores the number arr[i] as
// string
string strEle = to_string(arr[i]);
// Traverse over the digits of
// the current element
for ( int j = 0; j < strEle.size(); j++) {
// Checks if parity of check and
// current digit is same or not
if ( int (strEle[j]) % 2 == check) {
// Rotates the string by j + 1 times
// in anticlockwise
arr[i] = stoi(strEle.substr(j + 1)
+ strEle.substr(0, j + 1));
// Marks the flag
// as true and break
flag = true ;
break ;
}
}
// If flag is false
if (flag == false ) {
// Update exists
exists = false ;
break ;
}
// Changes the
// parity of check
check = !(check);
}
// Checks if arr[] cannot be
// modified, then returns false
if (!exists) {
arr = cpy;
return false ;
}
// Otherwise, return true
else
return true ;
} // Function to rotate the digits of array // elements such that array elements are // in the form of even-odd or odd-even form void convert_arr(vector< int >& arr)
{ // If array elements can be arranged
// in even-odd manner alternately
if (is_possible(arr, 0)) {
for ( auto & i : arr)
cout << i << " " ;
}
// If array elements can be arranged
// in odd-even manner alternately
else if (is_possible(arr, 1)) {
for ( auto & i : arr)
cout << i << " " ;
}
// Otherwise, prints -1
else
cout << "-1" << endl;
} // Driver Code int main()
{ vector< int > arr = { 143, 251, 534, 232, 854 };
convert_arr(arr);
} // This code is contributed by grand_master. |
314 251 534 223 854
Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(N), as we are using extra space for cpy.
Please refer complete article on Rearrange array elements into alternate even-odd sequence by anticlockwise rotation of digits for more details!