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:
- Choose two elements of the array a[i], a[j] (such that i is not equals to j) and,
- 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 :
- Calculate the sum of the array elements.
- If the sum is divisible by n, where n is the number of elements in the array then the answer will also be n.
- 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 <bits/stdc++.h>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 <stdio.h>// 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 arraypublic 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 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
<?php// PHP program to find the maximum// number of equal numbers in an array// Function to find the maximum // number of equal numbers in an arrayfunction EqualNumbers($a, $n){ // to store sum of elements $sum = 0; for ($i = 0; $i < $n; $i++) $sum += $a[$i]; // if sum of numbers is not // divisible by n if ($sum % $n) return $n - 1; return $n;}// Driver Code$a = array(1, 4, 1 );// size of an array$n = sizeof($a);echo EqualNumbers($a, $n);// This code is contributed// by Akanksha Rai |
Javascript
<script>// JavaScript program to find the maximum// number of equal numbers in an array// Function to find the maximum number of// equal numbers in an arrayfunction EqualNumbers(a, n){ // to store sum of elements let sum = 0; for (let 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 let a = [ 1, 4, 1 ]; // size of an array let n = a.length; document.write(EqualNumbers(a, n));// This code is contributed by Surbhi Tyagi.</script> |
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.
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!
