Given an array, arr[] of N integers, the task is to find the maximum possible count of adjacent pairs with an even sum, rearranging the array arr[].
Examples:
Input: arr[] = {5, 5, 1}
Output: 2
Explanation:
The given array is already arranged to give the maximum count of adjacent pairs with an even sum.
- {arr[0](= 5), arr[1](= 5}, the sum of the elements is 10, which is even.
- {arr[1](= 5), arr[2](= 1}, the sum of the elements is 6, which is even.
Therefore, there are totals of 2 adjacent pairs with an even sum. And it is also the maximum possible count.
Input: arr[] = {9, 13, 15, 3, 16, 9, 13, 18}
Output: 6
Explanation:
One way to obtain the maximum count is to rearrange the array as {9, 9, 3, 13, 13, 15, 16, 18}.
- {arr[0](= 9), arr[1](= 9}, the sum of the elements is 18, which is even.
- {arr[1](= 9), arr[2](= 3}, the sum of the elements is 12, which is even.
- {arr[2](= 3), arr[3](= 13}, the sum of the elements is 16, which is even.
- {arr[3](= 13), arr[4](= 13}, the sum of the elements is 26, which is even.
- {arr[4](= 13), arr[5](= 15}, the sum of the elements is 28, which is even.
- {arr[5](= 15), arr[6](= 16}, the sum of the elements is 31, which is not even.
- {arr[6](= 16), arr[7](= 18}, the sum of the elements is 34, which is even.
Therefore, there are a total of 6 adjacent pairs with an even sum. And it is also the maximum possible count.
Naive Approach: The simplest approach is to try every possible arrangement of the elements and then count the number of the adjacent pairs with an even sum.
Time Complexity: O(N*N!)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized based on the following observations:
- It is known that:
- Odd + Odd = Even
- Even + Even = Even
- Even + Odd = Odd
- Odd + Even = Odd
- The total count of adjacent pairs is N-1.
- Therefore, the maximum count can be obtained by putting all even numbers together and then all odd numbers or vice versa.
- Rearranging in the above-mentioned way, there will be only one pair of adjacent elements with an odd sum which will be at the junction of even numbers and odd numbers.
Follow the steps below to solve the problem:
- Find the count of odd numbers and even numbers in an array and then store them in variables say odd and even.
- If odd and even both are greater than 0, then print the total count N-2 as the answer.
- Otherwise, print N-1 as the answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find maximum count// pair of adjacent elements with// even sumint maximumCount(int arr[], int N){ // Stores count of odd numbers int odd = 0; // Stores count of even numbers int even = 0; // Traverse the array arr[] for (int i = 0; i < N; i++) { // If arr[i]%2 is 1 if (arr[i] % 2) odd++; // Else else even++; } // If odd and even both // are greater than 0 if (odd and even) return N - 2; // Otherwise else return N - 1;}// Driver Codeint main(){ int arr[] = { 9, 13, 15, 3, 16, 9, 13, 18 }; int N = sizeof(arr) / sizeof(arr[0]); cout << maximumCount(arr, N); return 0;} |
Java
/*package whatever //do not write package name here */import java.io.*;class GFG { // Function to find maximum count // pair of adjacent elements with // even sum static int maximumCount(int arr[], int N) { // Stores count of odd numbers int odd = 0; // Stores count of even numbers int even = 0; // Traverse the array arr[] for (int i = 0; i < N; i++) { // If arr[i]%2 is 1 if (arr[i] % 2 == 1) odd++; // Else else even++; } // If odd and even both // are greater than 0 if (odd > 0 && even > 0) return N - 2; // Otherwise else return N - 1; } // Driver Code public static void main(String[] args) { int arr[] = { 9, 13, 15, 3, 16, 9, 13, 18 }; int N = arr.length; System.out.println(maximumCount(arr, N)); }} // This code is contributed by Potta Lokesh |
Python3
# Python 3 program for the above approach# Function to find maximum count# pair of adjacent elements with# even sumdef maximumCount(arr, N): # Stores count of odd numbers odd = 0 # Stores count of even numbers even = 0 # Traverse the array arr[] for i in range(N): # If arr[i]%2 is 1 if (arr[i] % 2): odd += 1 # Else else: even += 1 # If odd and even both # are greater than 0 if (odd and even): return N - 2 # Otherwise else: return N - 1# Driver Codeif __name__ == '__main__': arr = [9, 13, 15, 3, 16, 9, 13, 18] N = len(arr) print(maximumCount(arr, N)) # This code is contributed by bgangwar59. |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function to find maximum count// pair of adjacent elements with// even sumstatic int maximumCount(int []arr, int N){ // Stores count of odd numbers int odd = 0; // Stores count of even numbers int even = 0; // Traverse the array arr[] for (int i = 0; i < N; i++) { // If arr[i]%2 is 1 if (arr[i] % 2 !=0) odd++; // Else else even++; } // If odd and even both // are greater than 0 if (odd!=0 && even!=0) return N - 2; // Otherwise else return N - 1;}// Driver Codepublic static void Main(){ int []arr = { 9, 13, 15, 3, 16, 9, 13, 18 }; int N = arr.Length; Console.Write(maximumCount(arr, N));}}// This code is contributed by ipg2016107. |
Javascript
<script> // JavaScript program for the above approach // Function to find maximum count // pair of adjacent elements with // even sum function maximumCount(arr, N) { // Stores count of odd numbers let odd = 0; // Stores count of even numbers let even = 0; // Traverse the array arr[] for (let i = 0; i < N; i++) { // If arr[i]%2 is 1 if (arr[i] % 2) odd++; // Else else even++; } // If odd and even both // are greater than 0 if (odd && even) return N - 2; // Otherwise else return N - 1; } // Driver Code let arr = [9, 13, 15, 3, 16, 9, 13, 18]; let N = arr.length; document.write(maximumCount(arr, N)); // This code is contributed by Potta Lokesh </script> |
Output:
6
Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.
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!
