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

• Last Updated : 02 Jun, 2022

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

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 ``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

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## Javascript

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Output:

`2 1 1 1 2`

Time Complexity: O(n), where n is the size of the given array
Auxiliary Space: O(1), as no extra space is used

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