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 <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 ?> |
2 1 1 1 2
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