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# Maximum count of equal numbers in an array after performing given operations

Given an array of integers. The task is to find the maximum count of equal numbers in an array after applying the given operation any number of times.
In an operation:

1. Choose two elements of the array a[i], a[j] (such that i is not equals to j) and,
2. Increase number a[i] by 1 and decrease number a[j] by 1 i.e., a[i] = a[i] + 1 and a[j] = a[j] – 1

Examples

Input: a = { 1, 4, 1 }
Output: 3
after first step { 1, 3, 2}
after second step { 2, 2, 2}

Input: a = { 1, 2 }
Output: 1

Approach :

1. Calculate the sum of the array elements.
2. If the sum is divisible by n, where n is the number of elements in the array then the answer will also be n.
3. Otherwise, the answer will be n-1.

Below is the implementation of the above approach:

## C++

 // CPP program to find the maximum// number of equal numbers in an array #include using namespace std; // Function to find the maximum number of// equal numbers in an arrayint EqualNumbers(int a[], int n){    // to store sum of elements    int sum = 0;    for (int i = 0; i < n; i++)        sum += a[i];     // if sum of numbers is not divisible    // by n    if (sum % n)        return n - 1;     return n;} // Driver Codeint main(){    int a[] = { 1, 4, 1 };     // size of an array    int n = sizeof(a) / sizeof(a[0]);     cout << EqualNumbers(a, n);     return 0;}

## C

 //C program to find the maximum// number of equal numbers in an array#include  // Function to find the maximum number of// equal numbers in an arrayint EqualNumbers(int a[], int n){  // storing the sum of elements  int sum = 0;  for (int i = 0; i < n; i++)    sum += a[i];   // checking if sum of numbers is not divisible  // by n  if (sum % n)    return n - 1;   return n;} // Driver Codeint main(){  int a[] = { 1, 4, 1 };   // Size of the array  int n = sizeof(a) / sizeof(a[0]);   printf("%d\n", EqualNumbers(a, n));   return 0;} // This code is contributed by phalashi

## Java

 // Java program to find the maximum// number of equal numbers in an array public class GFG{         // Function to find the maximum number of    // equal numbers in an array    static int EqualNumbers(int a[], int n)    {        // to store sum of elements        int sum = 0;        for (int i = 0; i < n; i++)            sum += a[i];         // if sum of numbers is not divisible        // by n        if (sum % n != 0)            return n - 1;         return n;    }          // Driver code    public static void main(String args[])    {            int a[] = { 1, 4, 1 };             // size of an array            int n = a.length;             System.out.println(EqualNumbers(a, n));    }    // This code is contributed by ANKITRAI1}

## Python3

 # Python3 program to find the maximum# number of equal numbers in an array # Function to find the maximum# number of equal numbers in an arraydef EqualNumbers(a, n):     # to store sum of elements    sum = 0;    for i in range(n):        sum += a[i];     # if sum of numbers is not    # divisible by n    if (sum % n):        return n - 1;     return n; # Driver Codea = [1, 4, 1 ]; # size of an arrayn = len(a); print(EqualNumbers(a, n)); # This code is contributed by mits

## C#

 // C# program to find the maximum// number of equal numbers in an arrayusing System; class GFG{// Function to find the maximum// number of equal numbers in an arraystatic int EqualNumbers(int []a, int n){    // to store sum of elements    int sum = 0;    for (int i = 0; i < n; i++)        sum += a[i];     // if sum of numbers is not    // divisible by n    if (sum % n != 0)        return n - 1;     return n;} // Driver codestatic public void Main (){    int []a = { 1, 4, 1 };     // size of an array    int n = a.Length;     Console.WriteLine(EqualNumbers(a, n));}} // This code is contributed by jit_t



## Javascript



Output

3

Complexity Analysis:

• Time Complexity: O(n), where n represents the size of the given array.
• Auxiliary complexity: O(1), no extra space is required, so it is a constant.

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