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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 array``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)``        ``return` `n - 1;` `    ``return` `n;``}` `// Driver Code``int` `main()``{``    ``int` `a[] = { 1, 4, 1 };` `    ``// size of an array``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);` `    ``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 array``int` `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 Code``int` `main()``{``  ``int` `a[] = { 1, 4, 1 };` `  ``// Size of the array``  ``int` `n = ``sizeof``(a) / ``sizeof``(a);` `  ``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 array``def` `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 Code``a ``=` `[``1``, ``4``, ``1` `];` `# size of an array``n ``=` `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 array``using` `System;` `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``static` `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`

## PHP

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## Javascript

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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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