Rearrange the array to maximize the number of primes in prefix sum of the array

Given an array arr[] of 1’s and 2’s, the task is to re-arrange the array in such a way that the prefix sum of the rearranged array has the maximum number of primes. Note that there can be multiple answers to it.

Examples:

Input: arr[] = {1, 2, 1, 2, 1}
Output: 2 1 1 1 2
The prefix sum array is {2, 3, 4, 5, 7} which has {2, 3, 5, 7} as primes
which is the maximum possible.

Input: arr[] = {1, 1, 2, 1, 1, 1, 2, 1, 1}
Output: 2 1 1 1 1 1 1 1 2

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The problem can be solved with two observations, one is the first prime is 2, and all other primes after that are odd numbers (All odd numbers are not prime). Hence simply fill the first position with 2 if there are any, and then fill an odd number of ones, and then fill the remaining 2’s. At the end insert the only 1 left (if the initial number of ones were even).

In doing so, we start from 2 and end at an odd number by adding an odd number of 1’s and then by adding 2’s to it, we jump from an odd number to another odd number which maximizes the probability of primes.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to print the re-arranged array 
void solve(int a[], int n) 
{ 
    int ones = 0, twos = 0; 
  
    // Count the number of 
    // ones and twos in a[] 
    for (int i = 0; i < n; i++) { 
  
        // If the array element is 1 
        if (a[i] == 1) 
            ones++; 
  
        // Array element is 2 
        else
            twos++; 
    } 
    int ind = 0; 
  
    // If it has at least one 2 
    // Fill up first 2 
    if (twos) 
        a[ind++] = 2; 
  
    // Decrease the cnt of 
    // ones if even 
    bool evenOnes = (ones % 2 == 0) ? true : false; 
    if (evenOnes) 
        ones -= 1; 
  
    // Fill up with odd count of ones 
    for (int i = 0; i < ones; i++) 
        a[ind++] = 1; 
  
    // Fill up with remaining twos 
    for (int i = 0; i < twos - 1; i++) 
        a[ind++] = 2; 
  
    // If even ones, then fill last position 
    if (evenOnes) 
        a[ind++] = 1; 
  
    // Print the rearranged array 
    for (int i = 0; i < n; i++) 
        cout << a[i] << " "; 
} 
  
// Driver code 
int main() 
{ 
    int a[] = { 1, 2, 1, 2, 1 }; 
    int n = sizeof(a) / sizeof(a[0]); 
    solve(a, n); 
  
    return 0; 
} 

Java

// Java implementation of the approach 
import java.io.*; 
  
class GFG { 
  
    // Function to print the re-arranged array 
    static void solve(int a[], int n) 
    { 
        int ones = 0, twos = 0; 
  
        // Count the number of 
        // ones and twos in a[] 
        for (int i = 0; i < n; i++) { 
  
            // If the array element is 1 
            if (a[i] == 1) 
                ones++; 
  
            // Array element is 2 
            else
                twos++; 
        } 
        int ind = 0; 
  
        // If it has at least one 2 
        // Fill up first 2 
        if (twos > 0) 
            a[ind++] = 2; 
  
        // Decrease the cnt of 
        // ones if even 
        boolean evenOnes = (ones % 2 == 0) ? true : false; 
        if (evenOnes) 
            ones -= 1; 
  
        // Fill up with odd count of ones 
        for (int i = 0; i < ones; i++) 
            a[ind++] = 1; 
  
        // Fill up with remaining twos 
        for (int i = 0; i < twos - 1; i++) 
            a[ind++] = 2; 
  
        // If even ones, then fill last position 
        if (evenOnes) 
            a[ind++] = 1; 
  
        // Print the rearranged array 
        for (int i = 0; i < n; i++) 
            System.out.print(a[i] + " "); 
    } 
  
    // Driver code 
    public static void main(String[] args) 
    { 
  
        int a[] = { 1, 2, 1, 2, 1 }; 
        int n = a.length; 
        solve(a, n); 
    } 
} 
  
// This code is contributed by ajit. 

Python

# Python3 implementation of the approach 
  
# Function to print the re-arranged array 
def solve(a, n): 
  
    ones, twos = 0, 0
  
    # Count the number of 
    # ones and twos in a[] 
    for i in range(n): 
  
        # If the array element is 1 
        if (a[i] == 1): 
            ones+=1
  
        # Array element is 2 
        else: 
            twos+=1
  
    ind = 0
  
    # If it has at least one 2 
    # Fill up first 2 
    if (twos): 
        a[ind] = 2
        ind+=1
  
    # Decrease the cnt of 
    # ones if even 
    if ones % 2 == 0: 
        evenOnes = True
    else: 
        evenOnes = False
  
  
    if (evenOnes): 
        ones -= 1
  
    # Fill up with odd count of ones 
    for i in range(ones): 
        a[ind] = 1
        ind += 1
  
  
    # Fill up with remaining twos 
    for i in range(twos-1): 
        a[ind] = 2
        ind += 1
  
    # If even ones, then fill last position 
    if (evenOnes): 
        a[ind] = 1
        ind += 1
  
    # Print the rearranged array 
    for i in range(n): 
        print(a[i],end = " ") 
  
# Driver code 
  
a = [1, 2, 1, 2, 1] 
n = len(a) 
solve(a, n) 
  
# This code is contributed by mohit kumar 29 

C#

// C# implementation of the approach 
using System; 
  
class GFG 
{ 
      
    // Function to print the re-arranged array 
    static void solve(int []a, int n) 
    { 
        int ones = 0, twos = 0; 
  
        // Count the number of 
        // ones and twos in a[] 
        for (int i = 0; i < n; i++) 
        { 
  
            // If the array element is 1 
            if (a[i] == 1) 
                ones++; 
  
            // Array element is 2 
            else
                twos++; 
        } 
        int ind = 0; 
  
        // If it has at least one 2 
        // Fill up first 2 
        if (twos > 0) 
            a[ind++] = 2; 
  
        // Decrease the cnt of 
        // ones if even 
        bool evenOnes = (ones % 2 == 0) ? true : false; 
        if (evenOnes) 
            ones -= 1; 
  
        // Fill up with odd count of ones 
        for (int i = 0; i < ones; i++) 
            a[ind++] = 1; 
  
        // Fill up with remaining twos 
        for (int i = 0; i < twos - 1; i++) 
            a[ind++] = 2; 
  
        // If even ones, then fill last position 
        if (evenOnes) 
            a[ind++] = 1; 
  
        // Print the rearranged array 
        for (int i = 0; i < n; i++) 
            Console.Write(a[i] + " "); 
    } 
  
    // Driver code 
    static public void Main () 
    { 
        int []a = { 1, 2, 1, 2, 1 }; 
        int n = a.Length; 
        solve(a, n); 
    } 
} 
  
// This code is contributed by Tushil. 

PHP

<?php 
// PHP implementation of the approach  
  
// Function to print the re-arranged array  
function solve($a, $n)  
{  
    $ones = 0; $twos = 0;  
  
    // Count the number of  
    // ones and twos in a[]  
    for ($i = 0; $i < $n; $i++)  
    {  
  
        // If the array element is 1  
        if ($a[$i] == 1)  
            $ones++;  
  
        // Array element is 2  
        else
            $twos++;  
    }  
    $ind = 0;  
  
    // If it has at least one 2  
    // Fill up first 2  
    if ($twos)  
        $a[$ind++] = 2;  
  
    // Decrease the cnt of  
    // ones if even  
    $evenOnes = ($ones % 2 == 0) ? true : false;  
    if ($evenOnes)  
        $ones -= 1;  
  
    // Fill up with odd count of ones  
    for ($i = 0; $i < $ones; $i++)  
        $a[$ind++] = 1;  
  
    // Fill up with remaining twos  
    for ($i = 0; $i < $twos - 1; $i++)  
        $a[$ind++] = 2;  
  
    // If even ones, then fill last position  
    if ($evenOnes)  
        $a[$ind++] = 1;  
  
    // Print the rearranged array  
    for ($i = 0; $i < $n; $i++)  
        echo $a[$i], " ";  
}  
  
// Driver code  
$a = array( 1, 2, 1, 2, 1 );  
$n = count($a);  
solve($a, $n);  
  
// This code is contributed by AnkitRai01 
  
?> 

Output:

2 1 1 1 2

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python

C#

PHP

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