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 alternatelybool 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 formvoid 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 Codeint main(){ vector<int> arr = { 143, 251, 534, 232, 854 }; convert_arr(arr);}// This code is contributed by grand_master. |
Java
// java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;class GFG { // Utility function to rotate the digits of // array elements such that array elements are // in placed even-odd or odd-even alternately static boolean is_possible(int arr[], int check) { // Checks if array can be converted // into even-odd or odd-even form boolean exists = true; // Store array elements int cpy[] = arr.clone(); boolean flag; // Traverse the array for (int i = 0; i < arr.length; i++) { // Check if arr[i] is already // at correct position if (arr[i] % 2 == check) { // Changes the // parity of check check = (check == 0 ? 1 : 0); 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 = Integer.toString(arr[i]); // Traverse over the digits of // the current element for (int j = 0; j < strEle.length(); j++) { // Checks if parity of check and // current digit is same or not if ((strEle.charAt(j) - '0') % 2 == check) { // Rotates the string by j + 1 times // in anticlockwise arr[i] = Integer.parseInt( strEle.substring(j + 1) + strEle.substring(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 == 0 ? 1 : 0); } // 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 static void convert_arr(int arr[]) { // If array elements can be arranged // in even-odd manner alternately if (is_possible(arr, 0)) { for (int v : arr) { System.out.print(v + " "); } } // If array elements can be arranged // in odd-even manner alternately else if (is_possible(arr, 1)) { for (int v : arr) { System.out.print(v + " "); } } // Otherwise, prints -1 else System.out.println(-1); } // Driver code public static void main(String[] args) { // Given array int arr[] = { 143, 251, 534, 232, 854 }; // FUnction call convert_arr(arr); }}// This code is contributed by Kingash. |
Python3
# Python program of the above approach# Utility function to rotate the digits of # array elements such that array elements are# in placed even-odd or odd-even alternatelydef is_possible(arr, check): # Checks if array can be converted # into even-odd or odd-even form exists = True # Store array elements cpy = arr # Traverse the array for i in range(len(arr)): # Check if arr[i] is already # at correct position if (arr[i] % 2 == check): check = not(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 strEle = str(arr[i]) # Traverse over the digits of # the current element for j in range(len(strEle)): # 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] = int(strEle[j + 1:] + strEle[: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 = not(check) # Checks if arr[] cannot be # modified, then returns false if not 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 formdef convert_arr(arr): # If array elements can be arranged # in even-odd manner alternately if(is_possible(arr, 0)): print(*arr) # If array elements can be arranged # in odd-even manner alternately elif(is_possible(arr, 1)): print(*arr) # Otherwise, prints -1 else: print(-1)# Driver Codeif __name__ == '__main__': arr = [143, 251, 534, 232, 854] convert_arr(arr) # This code is contributed by ipg2016107. |
C#
// C# program of the above approachusing System;using System.Collections.Generic;class GFG{ // Utility function to rotate the digits of// array elements such that array elements are// in placed even-odd or odd-even alternatelystatic bool ispossible(List<int> arr, bool check){ // Checks if array can be converted // into even-odd or odd-even form bool exists = true; // Store array elements List<int> cpy = arr; bool flag; // Traverse the array for (int i=0;i<arr.Count;i++) { // Check if arr[i] is already // at correct position int temp = check ? 1 : 0; if (arr[i] % 2 == temp) { 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 int p = arr[i]; string strEle = p.ToString(); // Traverse over the digits of // the current element for (int j = 0; j < strEle.Length; j++) { // Checks if parity of check and // current digit is same or not temp = check ? 1 : 0; if ((int)(strEle[j] - '0')% 2 == temp) { // Rotates the string by j + 1 times // in anticlockwise string s = strEle.Substring(j + 1) + strEle.Substring(0, j + 1); arr[i] = Int32.Parse(s); // 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==false) { 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 formstatic void convert_arr(List<int> arr){ // If array elements can be arranged // in even-odd manner alternately if (ispossible(arr, false)) { foreach (int i in arr) Console.Write(i +" "); } // If array elements can be arranged // in odd-even manner alternately else if (ispossible(arr, true)) { foreach (int i in arr) Console.Write(i + " "); } // Otherwise, prints -1 else Console.Write("-1");}// Driver Codepublic static void Main(){ List<int> arr = new List<int>(){143, 251, 534, 232, 854}; convert_arr(arr);}} |
Javascript
<script> // JavaScript program of the above approach // Utility function to rotate the digits of // array elements such that array elements are // in placed even-odd or odd-even alternately function is_possible(arr,check) { // Checks if array can be converted // into even-odd or odd-even form let exists = true; // Store array elements let cpy = arr; let flag; // Traverse the array for (let i = 0; i < arr.length; 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 let strEle = arr[i].toString(); // Traverse over the digits of // the current element for (let j = 0; j < strEle.length; j++) { // Checks if parity of check and // current digit is same or not if (strEle[j].charCodeAt() % 2 == check) { // Rotates the string by j + 1 times // in anticlockwise arr[i] = parseInt(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 function convert_arr(arr) { // If array elements can be arranged // in even-odd manner alternately if (is_possible(arr, 0)) { for (let i = 0; i < arr.length; i++) document.write(arr[i] + " "); } // If array elements can be arranged // in odd-even manner alternately else if (is_possible(arr, 1)) { for (let i = 0; i < arr.length; i++) document.write(arr[i] + " "); } // Otherwise, prints -1 else document.write("-1"); } // Driver Code let arr = [ 143, 251, 534, 232, 854 ]; convert_arr(arr); </script> |
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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