# Find the maximum number of elements divisible by 3

Given an array of size N. The task to find the maximum possible number of elements divisible by 3 that are in the array after performing the operation an arbitrary (possibly, zero) number of times. In each operation, one can add any two elements of the array.

Examples:

Input : a[] = {1, 2, 3}
Output : 2
After applying the operation once (on elements 1, 2), the array becomes {3, 3}.
It contains 2 numbers which are divisible by 3 which are maximum possible.

Input : a[] = {1, 1, 1, 1, 1, 2, 2}
Output : 3

Approach :
Let cnti be the number of elements of a with the remainder i modulo 3. Then the initial answer can be represented as cnt0 and we have to compose numbers with remainders 1 and 2 somehow optimally. It can be shown that the best way to do it is the following:

• Firstly, while there is at least one remainder of 1 and at least one remainder of 2 then compose them into one 0. After this, at least one of the numbers cnt1, cnt2 will be zero, then we have to compose remaining numbers into numbers divisible by 3.
• If cnt1=0 then the maximum remaining number of elements we can obtain is [cnt2/3] (because 2+2+2=6), and in the other case (cnt2=0) the maximum number of elements is [cnt1/3] (because 1+1+1=3).

Below is the implementation of the above approach :

## C++

 `// C++ program to find the maximum ` `// number of elements divisible by 3 ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the maximum ` `// number of elements divisible by 3 ` `int` `MaxNumbers(``int` `a[], ``int` `n) ` `{ ` `    ``// To store frequency of each number ` `    ``int` `fre = { 0 }; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``// Store modulo value ` `        ``a[i] %= 3; ` ` `  `        ``// Strore frequency ` `        ``fre[a[i]]++; ` `    ``} ` ` `  `    ``// Add numbers with zero modulo to answer ` `    ``int` `ans = fre; ` ` `  `    ``// Find minimum of elements with modulo ` `    ``// frequency one and zero ` `    ``int` `k = min(fre, fre); ` ` `  `    ``// Add k to the answer ` `    ``ans += k; ` ` `  `    ``// Remove them from frequency ` `    ``fre -= k; ` `    ``fre -= k; ` ` `  `    ``// Add numbers possible with ` `    ``// remaining frequency ` `    ``ans += fre / 3 + fre / 3; ` ` `  `    ``// Return the required answer ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `a[] = { 1, 4, 10, 7, 11, 2, 8, 5, 9 }; ` ` `  `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` ` `  `    ``// Function call ` `    ``cout << MaxNumbers(a, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find the maximum  ` `// number of elements divisible by 3  ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `     `  `    ``// Function to find the maximum  ` `    ``// number of elements divisible by 3  ` `    ``static` `int` `MaxNumbers(``int` `a[], ``int` `n)  ` `    ``{  ` `        ``// To store frequency of each number  ` `        ``int` `[]fre = { ``0``,``0``,``0` `};  ` `     `  `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{  ` `            ``// Store modulo value  ` `            ``a[i] %= ``3``;  ` `     `  `            ``// Strore frequency  ` `            ``fre[a[i]]++;  ` `        ``}  ` `     `  `        ``// Add numbers with zero modulo to answer  ` `        ``int` `ans = fre[``0``];  ` `     `  `        ``// Find minimum of elements with modulo  ` `        ``// frequency one and zero  ` `        ``int` `k = Math.min(fre[``1``], fre[``2``]);  ` `     `  `        ``// Add k to the answer  ` `        ``ans += k;  ` `     `  `        ``// Remove them from frequency  ` `        ``fre[``1``] -= k;  ` `        ``fre[``2``] -= k;  ` `     `  `        ``// Add numbers possible with  ` `        ``// remaining frequency  ` `        ``ans += fre[``1``] / ``3` `+ fre[``2``] / ``3``;  ` `     `  `        ``// Return the required answer  ` `        ``return` `ans;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `a[] = { ``1``, ``4``, ``10``, ``7``, ``11``, ``2``, ``8``, ``5``, ``9` `};  ` `     `  `        ``int` `n = a.length;  ` `     `  `        ``// Function call  ` `        ``System.out.println(MaxNumbers(a, n));  ` `    ``} ` `} ` ` `  `// This code is contributed by @@ajit.. `

## Python3

 `# Python3 program to find the maximum ` `# number of elements divisible by 3 ` ` `  `# Function to find the maximum ` `# number of elements divisible by 3 ` `def` `MaxNumbers(a, n): ` `     `  `    ``# To store frequency of each number ` `    ``fre ``=` `[``0` `for` `i ``in` `range``(``3``)] ` ` `  `    ``for` `i ``in` `range``(n): ` `         `  `        ``# Store modulo value ` `        ``a[i] ``%``=` `3` ` `  `        ``# Strore frequency ` `        ``fre[a[i]] ``+``=` `1` ` `  `    ``# Add numbers with zero modulo to answer ` `    ``ans ``=` `fre[``0``] ` ` `  `    ``# Find minimum of elements with modulo ` `    ``# frequency one and zero ` `    ``k ``=` `min``(fre[``1``], fre[``2``]) ` ` `  `    ``# Add k to the answer ` `    ``ans ``+``=` `k ` ` `  `    ``# Remove them from frequency ` `    ``fre[``1``] ``-``=` `k ` `    ``fre[``2``] ``-``=` `k ` ` `  `    ``# Add numbers possible with ` `    ``# remaining frequency ` `    ``ans ``+``=` `fre[``1``] ``/``/` `3` `+` `fre[``2``] ``/``/` `3` ` `  `    ``# Return the required answer ` `    ``return` `ans ` ` `  `# Driver code ` `a ``=` `[``1``, ``4``, ``10``, ``7``, ``11``, ``2``, ``8``, ``5``, ``9``] ` ` `  `n ``=` `len``(a) ` ` `  `# Function call ` `print``(MaxNumbers(a, n)) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# program to find the maximum  ` `// number of elements divisible by 3  ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Function to find the maximum  ` `    ``// number of elements divisible by 3  ` `    ``static` `int` `MaxNumbers(``int` `[]a, ``int` `n)  ` `    ``{  ` `        ``// To store frequency of each number  ` `        ``int` `[]fre = { 0,0,0 };  ` `     `  `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{  ` `            ``// Store modulo value  ` `            ``a[i] %= 3;  ` `     `  `            ``// Strore frequency  ` `            ``fre[a[i]]++;  ` `        ``}  ` `     `  `        ``// Add numbers with zero modulo to answer  ` `        ``int` `ans = fre;  ` `     `  `        ``// Find minimum of elements with modulo  ` `        ``// frequency one and zero  ` `        ``int` `k = Math.Min(fre, fre);  ` `     `  `        ``// Add k to the answer  ` `        ``ans += k;  ` `     `  `        ``// Remove them from frequency  ` `        ``fre -= k;  ` `        ``fre -= k;  ` `     `  `        ``// Add numbers possible with  ` `        ``// remaining frequency  ` `        ``ans += fre / 3 + fre / 3;  ` `     `  `        ``// Return the required answer  ` `        ``return` `ans;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``static` `public` `void` `Main () ` `    ``{ ` `         `  `        ``int` `[]a = { 1, 4, 10, 7, 11, 2, 8, 5, 9 };  ` `     `  `        ``int` `n = a.Length;  ` `     `  `        ``// Function call  ` `        ``Console.WriteLine(MaxNumbers(a, n));  ` `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```5
```

Time Complexity : O(N)

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