Given an array arr[] containing natural numbers from 1 to N, the task is to find the maximum number of elements that can be made equal after the below operations:
- Remove any pair of elements from the array and insert their sum to an array.
- Repeat the above operation any numbers of times to maximize the count of equal elements.
Examples:
Input: arr[] = {1, 2, 3, 4}
Output: 2
Explanation:
We can perform following operations:
{1, 2, 3, 4} -> {3, 3, 4} -> 2 elements are equalInput: arr[] = {1 2 3 4 5 6}
Output: 3
Explanation:
{1, 2, 3, 4, 5, 6} -> {7, 2, 3, 4, 5} -> {7, 7, 3, 4} -> {7, 7, 37} -> 3 elements are equal
Approach: The key observation in the problem is that:
- If N is even, we can make a maximum count of equal elements by
- If N is odd, we can make the maximum count of equal elements by
Therefore, the answer will always be
Below is the implementation of the above approach:
C++
// C++ implementation of// the above approach#include <bits/stdc++.h>using namespace std;// Function to count maximum number// of array elements equalint countEqual(int n){ return (n + 1) / 2;}// Driver Codeint main(){ int arr[] = { 1, 2, 3, 4, 5, 6 }; int n = sizeof(arr) / sizeof(arr[0]); // Function Call cout << countEqual(n); return 0;} |
Java
// Java implementation of// the above approachimport java.io.*;class GFG{// Function to count maximum number// of array elements equalstatic int countEqual(int n){ return (n + 1) / 2;}// Driver Codepublic static void main (String[] args){ int arr[] = { 1, 2, 3, 4, 5, 6 }; int n = arr.length; // Function call System.out.println(countEqual(n));}}// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of # the above approach # Function to count maximum number # of array elements equal def countEqual(n): return (n + 1) // 2# Driver Code lst = [ 1, 2, 3, 4, 5, 6 ]n = len(lst) # Function callprint(countEqual(n))# This code is contributed by vishu2908 |
C#
// C# implementation of// the above approachusing System;class GFG{// Function to count maximum number// of array elements equalstatic int countEqual(int n){ return (n + 1) / 2;}// Driver Codepublic static void Main(String[] args){ int []arr = {1, 2, 3, 4, 5, 6}; int n = arr.Length; // Function call Console.WriteLine(countEqual(n));}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of// the above approach// Function to count maximum number// of array elements equalfunction countEqual(n){ return parseInt((n + 1) / 2);}// Driver Codevar arr = [ 1, 2, 3, 4, 5, 6 ];var n = arr.length;// Function Calldocument.write( countEqual(n));// This code is contributed by rrrtnx.</script> |
Output:
3
Performance Analysis:
Time Complexity: O(1), as we are not using any loops or recursion.
Auxiliary Space: O(1), as we are not using any extra space.
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