C++ Program to Rearrange array elements into alternate even-odd sequence by anticlockwise rotation of digits

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++

// 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.

Output:

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.

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