Given two arrays arr1[] and arr2[] of N integers. We can choose any two adjacent elements from array arr1[] and swap them if they are of opposite parity, the task is to check if it is possible to convert array arr1[] to array arr2[] by performing the given operation on arr1[]. Print “Yes” if it is possible to convert array arr1[] to arr2[] Else print “No”.
Examples:
Input: arr1[] = {5, 7, 8, 2, 10, 13}, arr2[] = {8, 5, 2, 7, 13, 10}
Output: Yes
Explanation:
At first, swap 10 and 13 so arr1[] = [5, 7, 8, 2, 13, 10].
Now, swap 7 and 8 so arr1[] = [5, 8, 7, 2, 13, 10].
Now, swap 5 and 8 so arr1[] = [8, 5, 7, 2, 13, 10].
Now, swap 7 and 2 so arr1[] = [8, 5, 2, 7, 13, 10] = arr2[].
In each operation, we swap adjacent elements with different parity.
Input: arr1[] = {0, 1, 13, 3, 4, 14, 6}, arr2[] = {0, 1, 14, 3, 4, 13, 6}
Output: No
Explanation:
It is not possible to swap 13, 14 because they are not adjacent.
Approach: The problem can be solved using Greedy Approach. Since we cannot swap any two even or odd numbers. So the relative position of both even and odd numbers in the arrays arr1[] and arr2[] must be exactly the same to make both the arrays equal with the given operation. Below are the steps:
- Create two arrays(say even[] and odd[]) insert all the even and odd numbers from arr1[] in even[] and odd[] respectively.
- Now check whether the even and odd numbers in arr2[] are in the same order as in even[] and odd[].
- If the above steps doesn’t gives any number from arr2[] which are not in the order of numbers in even[] and odd[] arrays respectively then, print “Yes” else print “No”.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function which checks if it is// possible to convert arr1[] to// arr2[] by given operationsvoid convert(int a[], int b[], int n){ // even[] will store the even // elements of a[] // odd[] will store the odd // elements of a[] vector<int> even, odd; // Traverse a[] and insert the even // and odd element respectively for (int x = 0; x < n; x++) { if (a[x] % 2 == 0) even.push_back(a[x]); else odd.push_back(a[x]); } // ei points to the next // available even element // oi points to the next // available odd element int ei = 0, oi = 0; // poss will store whether the // given transformation // of a[] to b[] is possible bool poss = true; // Traverse b[] for (int x = 0; x < n; x++) { if (b[x] % 2 == 0) { // Check if both even // elements are equal if (ei < even.size() && b[x] == even[ei]) { ei++; } else { poss = false; break; } } else { // Check if both odd // elements are equal if (oi < odd.size() && b[x] == odd[oi]) { oi++; } else { poss = false; break; } } } // If poss is true, then we can // transform a[] to b[] if (poss) cout << "Yes" << endl; else cout << "No" << endl;}// Driver Codeint main(){ // Given arrays int arr1[] = { 0, 1, 13, 3, 4, 14, 6 }; int arr2[] = { 0, 1, 14, 3, 4, 13, 6 }; int N = sizeof(arr1) / sizeof(arr1[0]); // Function Call convert(arr1, arr2, N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function which checks if it is// possible to convert arr1[] to// arr2[] by given operationsstatic void convert(int a[], int b[], int n){ // even[] will store the even // elements of a[] // odd[] will store the odd // elements of a[] Vector<Integer> even = new Vector<Integer>(), odd = new Vector<Integer>(); // Traverse a[] and insert the even // and odd element respectively for(int x = 0; x < n; x++) { if (a[x] % 2 == 0) even.add(a[x]); else odd.add(a[x]); } // ei points to the next // available even element // oi points to the next // available odd element int ei = 0, oi = 0; // poss will store whether the // given transformation // of a[] to b[] is possible boolean poss = true; // Traverse b[] for(int x = 0; x < n; x++) { if (b[x] % 2 == 0) { // Check if both even // elements are equal if (ei < even.size() && b[x] == even.get(ei)) { ei++; } else { poss = false; break; } } else { // Check if both odd // elements are equal if (oi < odd.size() && b[x] == odd.get(oi)) { oi++; } else { poss = false; break; } } } // If poss is true, then we can // transform a[] to b[] if (poss) System.out.print("Yes" + "\n"); else System.out.print("No" + "\n");}// Driver Codepublic static void main(String[] args){ // Given arrays int arr1[] = { 0, 1, 13, 3, 4, 14, 6 }; int arr2[] = { 0, 1, 14, 3, 4, 13, 6 }; int N = arr1.length; // Function Call convert(arr1, arr2, N);}}// This code is contributed by gauravrajput1 |
Python3
# Python3 program for the above approach# Function which checks if it is# possible to convert arr1[] to# arr2[] by given operationsdef convert(a, b, n): # even[] will store the even # elements of a[] # odd[] will store the odd # elements of a[] even = [] odd = [] # Traverse a[] and insert the even # and odd element respectively for x in range(n): if (a[x] % 2 == 0): even.append(a[x]) else: odd.append(a[x]) # ei points to the next # available even element # oi points to the next # available odd element ei, oi = 0, 0 # poss will store whether the # given transformation # of a[] to b[] is possible poss = True # Traverse b[] for x in range(n): if (b[x] % 2 == 0): # Check if both even # elements are equal if (ei < len(even) and b[x] == even[ei]): ei += 1 else: poss = False break else: # Check if both odd # elements are equal if (oi < len(odd) and b[x] == odd[oi]): oi += 1 else: poss = False break # If poss is true, then we can # transform a[] to b[] if (poss): print("Yes") else: print("No")# Driver Codeif __name__ == "__main__": # Given arrays arr1 = [ 0, 1, 13, 3, 4, 14, 6 ] arr2 = [ 0, 1, 14, 3, 4, 13, 6 ] N = len(arr1) # Function call convert(arr1, arr2, N)# This code is contributed by chitranayal |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function which checks if it is// possible to convert arr1[] to// arr2[] by given operationsstatic void convert(int []a, int []b, int n){ // even[] will store the even // elements of []a // odd[] will store the odd // elements of []a List<int> even = new List<int>(), odd = new List<int>(); // Traverse []a and insert the even // and odd element respectively for(int x = 0; x < n; x++) { if (a[x] % 2 == 0) even.Add(a[x]); else odd.Add(a[x]); } // ei points to the next // available even element // oi points to the next // available odd element int ei = 0, oi = 0; // poss will store whether the // given transformation // of []a to []b is possible bool poss = true; // Traverse []b for(int x = 0; x < n; x++) { if (b[x] % 2 == 0) { // Check if both even // elements are equal if (ei < even.Count && b[x] == even[ei]) { ei++; } else { poss = false; break; } } else { // Check if both odd // elements are equal if (oi < odd.Count && b[x] == odd[oi]) { oi++; } else { poss = false; break; } } } // If poss is true, then we can // transform []a to []b if (poss) Console.Write("Yes" + "\n"); else Console.Write("No" + "\n");}// Driver Codepublic static void Main(String[] args){ // Given arrays int []arr1 = { 0, 1, 13, 3, 4, 14, 6 }; int []arr2 = { 0, 1, 14, 3, 4, 13, 6 }; int N = arr1.Length; // Function call convert(arr1, arr2, N);}}// This code is contributed by gauravrajput1 |
Javascript
<script>// JavaScript program for the above approach// Function which checks if it is// possible to convert arr1[] to// arr2[] by given operationsfunction convert(a, b, n){ // even[] will store the even // elements of a[] // odd[] will store the odd // elements of a[] var even = [], odd = []; // Traverse a[] and insert the even // and odd element respectively for (var x = 0; x < n; x++) { if (a[x] % 2 == 0) even.push(a[x]); else odd.push(a[x]); } // ei points to the next // available even element // oi points to the next // available odd element var ei = 0, oi = 0; // poss will store whether the // given transformation // of a[] to b[] is possible var poss = true; // Traverse b[] for (var x = 0; x < n; x++) { if (b[x] % 2 == 0) { // Check if both even // elements are equal if (ei < even.length && b[x] == even[ei]) { ei++; } else { poss = false; break; } } else { // Check if both odd // elements are equal if (oi < odd.length && b[x] == odd[oi]) { oi++; } else { poss = false; break; } } } // If poss is true, then we can // transform a[] to b[] if (poss) document.write( "Yes" ); else document.write( "No" );}// Driver Code// Given arraysvar arr1 = [0, 1, 13, 3, 4, 14, 6 ];var arr2 = [0, 1, 14, 3, 4, 13, 6 ];var N = arr1.length;// Function Callconvert(arr1, arr2, N);</script> |
Output:
No
Time Complexity: O(N), where N is the number of elements in the array.
Auxiliary Space: O(N), where N is the number of elements in the array.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
