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Rearrange given binary strings to maximize their Bitwise XOR value

Last Updated : 15 Jun, 2021
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Given three binary strings S1, S2, and S3 each of length N, the task is to find the maximum possible Bitwise XOR that can be obtained by rearranging the characters of the given strings.

Examples:

Input: S1 = "1001", S2 = "0010", S3 = "1110"
Output: 15
Explanation:
Rearrange the digits of S1 as "1010", S2 as "1000" and S3 as "1101".
The XOR of these strings is "1111" which is 15 in decimal form.

Input: S1 = "11111", S2 = "11111", S3 = "11111" 
Output: 31
Explanation:
There is no other way to arrange the digits. Hence, XOR is "11111" which is 31 in decimal form. 

Naive Approach: The simplest approach is to generate all possible ways to rearrange S1, S2, and S3. Suppose there are O1, O2 and O3 set bits in the strings S1, S2, and S3 respectively. The total number of rearrangements to check to get the maximum Bitwise XOR value is as follows:

Total ways to rearrange S1 = NCO1  
Total ways to rearrange S2 = NCO2  
Total ways to rearrange S3 = NCO3  

Hence, total possible rearrangements to check = NCO1*NCO2 * NCO3

Time Complexity: O((N!)3), where N is the length of the given strings.
Auxiliary Space: O(N)

Efficient Approach: The idea is to find a suitable rearrangement of S1, S2, and S3 such that their Bitwise XOR value is maximized using Dynamic Programming. The subproblems can be stored in a dp[][][][] table where dp[i][o1][o2][o3] stores the maximum XOR value up to position N-1 starting from the index i, where o1 is, o2 and o3 are the number of 1s still remaining to be placed in strings S1, S2 and S3 respectively.

There can be four cases possible at any position i from 0 to (N - 1):

  1. Assign 1s to all the three strings
  2. Assign 1s to any two strings
  3. Assign 1s to any one of the strings.
  4. Assign 0s to all the strings.

From the above possible cases for each position, calculate the maximum Bitwise XOR obtainable from the four possibilities:

Follow the steps below to solve the problem:

  • Initialize a table dp[][][][] to store the number of ones in S1, S2 and S3 for the positions i from 0 to N-1.
  • The transition states is as follows:

dp[i][o1][o2][o3] = max(dp(assign 1s to all three strings), dp(assign 1s to any of the two strings), dp(assign 1s to any one string), dp(do not assign 1 to any string)) where,

i = current position
o1 = remaining ones to be placed in the string S1
o2 = remaining ones to be placed in the string S2
o3 = remaining ones to be placed in the string S3

  • Solve the subproblems for all cases using the above transition and print the maximum XOR value amongst them.

Below is the implementation of the above approach:

C++ 
// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Dp table to store the sub-problems
int dp[20][20][20][20];

// Function to find the maximum XOR
// value after rearranging the digits
int maxXorValue(int i, string& s1,
                string& s2, string& s3,
                int ones1, int ones2,
                int ones3, int n)
{
    // Base Case
    if (i >= n)
        return 0;

    // Return if already calculated
    if (dp[i][ones1][ones2][ones3] != -1)
        return dp[i][ones1][ones2][ones3];

    int option1 = 0, option2 = 0,
        option3 = 0, option4 = 0,
        option5 = 0, option6 = 0,
        option7 = 0, option8 = 0;

    // Assigning 1's to all string at
    // position 'i'.
    if (ones1 > 0 && ones2 > 0
        && ones3 > 0)

        // 2^(n-1-i) is the value
        // added to the total
        option1
            = (1 << ((n - 1) - i))
              + maxXorValue(i + 1, s1,
                            s2, s3, ones1 - 1,
                            ones2 - 1,
                            ones3 - 1, n);

    // Assigning 1's to strings 1 & 2
    if (ones1 > 0 && ones2 > 0
        && (n - i > ones3))
        option2
            = 0
              + maxXorValue(i + 1, s1,
                            s2, s3, ones1 - 1,
                            ones2 - 1,
                            ones3, n);

    // Assigning 1's to strings 2 & 3
    if (ones2 > 0 && ones3 > 0
        && (n - i > ones1))
        option3 = 0
                  + maxXorValue(i + 1, s1,
                                s2, s3,
                                ones1,
                                ones2 - 1,
                                ones3 - 1, n);

    // Assigning 1's to strings 3 & 1
    if (ones3 > 0 && ones1 > 0
        && (n - i > ones2))
        option4
            = 0
              + maxXorValue(i + 1, s1,
                            s2, s3,
                            ones1 - 1,
                            ones2,
                            ones3 - 1, n);

    // Assigning 1 to string 1
    if (ones1 > 0 && (n - i > ones2)
        && (n - i > ones3))
        option5 = (1 << ((n - 1) - i))
                  + maxXorValue(i + 1, s1,
                                s2, s3,
                                ones1 - 1,
                                ones2,
                                ones3, n);

    // Assigning 1 to string 2
    if (ones2 > 0 && (n - i > ones3)
        && (n - i > ones1))
        option6 = (1 << ((n - 1) - i))
                  + maxXorValue(i + 1, s1,
                                s2, s3, ones1,
                                ones2 - 1,
                                ones3, n);

    // Assigning 1 to string 3.
    if (ones3 > 0 && (n - i > ones2)
        && (n - i > ones1))
        option7 = (1 << ((n - 1) - i))
                  + maxXorValue(i + 1, s1,
                                s2, s3, ones1,
                                ones2,
                                ones3 - 1, n);

    // Assigning 0 to all the strings
    if ((n - i > ones2) && (n - i > ones3)
        && (n - i > ones1))
        option8 = 0
                  + maxXorValue(i + 1, s1,
                                s2, s3,
                                ones1,
                                ones2,
                                ones3, n);

    // Take the maximum amongst all of
    // the above solutions
    return dp[i][ones1][ones2][ones3]
           = max(option1,
                 max(option2,
                     max(option3,
                         max(option4,
                             max(option5,
                                 max(option6,
                                     max(option7,
                                         option8)))))));
}

// Function to get the count of ones
// in the string s
int onesCount(string& s)
{
    int count = 0;

    // Traverse the string
    for (auto x : s) {
        if (x == '1')
            ++count;
    }

    // Return the count
    return count;
}

// Utility Function to find the maximum
// XOR value after rearranging the digits
void maxXORUtil(string s1, string s2,
                string s3, int n)
{

    // Find the count of ones in
    // each of the strings
    int ones1 = onesCount(s1);
    int ones2 = onesCount(s2);
    int ones3 = onesCount(s3);

    // Initialize dp table with -1
    memset(dp, -1, sizeof dp);

    // Function Call
    cout << maxXorValue(0, s1, s2, s3,
                        ones1, ones2,
                        ones3, n);
}

// Driver code
int main()
{
    string s1 = "11110";
    string s2 = "10101";
    string s3 = "00111";

    int n = s1.size();

    // Function Call
    maxXORUtil(s1, s2, s3, n);

    return 0;
}
Java 
// Java program for the above approach 
import java.util.*;
import java.lang.*;
class GFG{

// Dp table to store the sub-problems 
static int[][][][]  dp = new int[20][20][20][20]; 
  
// Function to find the maximum XOR 
// value after rearranging the digits 
static int maxXorValue(int i, String s1, 
                       String s2, String s3, 
                       int ones1, int ones2, 
                       int ones3, int n) 
{ 
    
    // Base Case 
    if (i >= n) 
        return 0; 
  
    // Return if already calculated 
    if (dp[i][ones1][ones2][ones3] != -1) 
        return dp[i][ones1][ones2][ones3]; 
  
    int option1 = 0, option2 = 0, 
        option3 = 0, option4 = 0, 
        option5 = 0, option6 = 0, 
        option7 = 0, option8 = 0; 
  
    // Assigning 1's to all string at 
    // position 'i'. 
    if (ones1 > 0 && ones2 > 0 && 
        ones3 > 0) 
  
        // 2^(n-1-i) is the value 
        // added to the total 
        option1 = (1 << ((n - 1) - i)) +
              maxXorValue(i + 1, s1, s2, 
                          s3, ones1 - 1, 
                              ones2 - 1, 
                              ones3 - 1, n); 
  
    // Assigning 1's to strings 1 & 2 
    if (ones1 > 0 && ones2 > 0 && 
       (n - i > ones3)) 
        option2 = 0 + maxXorValue(i + 1, s1, s2,
                                  s3, ones1 - 1, 
                                      ones2 - 1, 
                                      ones3, n); 
  
    // Assigning 1's to strings 2 & 3 
    if (ones2 > 0 && ones3 > 0 && 
       (n - i > ones1)) 
        option3 = 0 + maxXorValue(i + 1, s1, s2,
                                  s3, ones1, 
                                  ones2 - 1, 
                                  ones3 - 1, n); 
  
    // Assigning 1's to strings 3 & 1 
    if (ones3 > 0 && ones1 > 0 && 
       (n - i > ones2)) 
        option4 = 0 + maxXorValue(i + 1, s1, s2,
                                  s3, ones1 - 1, 
                                  ones2, 
                                  ones3 - 1, n); 
  
    // Assigning 1 to string 1 
    if (ones1 > 0 && (n - i > ones2) &&
       (n - i > ones3)) 
        option5 = (1 << ((n - 1) - i)) + 
                  maxXorValue(i + 1, s1, s2,
                              s3, ones1 - 1, 
                              ones2, ones3, n); 
  
    // Assigning 1 to string 2 
    if (ones2 > 0 && (n - i > ones3) && 
       (n - i > ones1)) 
        option6 = (1 << ((n - 1) - i)) + 
                  maxXorValue(i + 1, s1, 
                              s2, s3, ones1, 
                              ones2 - 1, 
                              ones3, n); 
  
    // Assigning 1 to string 3. 
    if (ones3 > 0 && (n - i > ones2) && 
       (n - i > ones1)) 
        option7 = (1 << ((n - 1) - i)) + 
                  maxXorValue(i + 1, s1, 
                              s2, s3, ones1, 
                              ones2, 
                              ones3 - 1, n); 
  
    // Assigning 0 to all the strings 
    if ((n - i > ones2) && (n - i > ones3) &&
        (n - i > ones1)) 
        option8 = 0 + maxXorValue(i + 1, s1, 
                                  s2, s3, ones1, 
                                  ones2, ones3, n); 
  
    // Take the maximum amongst all of 
    // the above solutions 
    return dp[i][ones1][ones2][ones3] = 
        Math.max(option1, 
                 Math.max(option2, 
                     Math.max(option3, 
                        Math.max(option4, 
                             Math.max(option5, 
                                 Math.max(option6, 
                                     Math.max(option7, 
                                              option8))))))); 
} 
  
// Function to get the count of ones 
// in the string s 
static int onesCount(String s) 
{ 
    int count = 0; 
  
    // Traverse the string 
    for(char x : s.toCharArray())
    { 
        if (x == '1') 
            ++count; 
    } 
  
    // Return the count 
    return count; 
} 
  
// Utility Function to find the maximum 
// XOR value after rearranging the digits 
static void maxXORUtil(String s1, String s2, 
                       String s3, int n) 
{ 
  
    // Find the count of ones in 
    // each of the strings 
    int ones1 = onesCount(s1); 
    int ones2 = onesCount(s2); 
    int ones3 = onesCount(s3); 
  
    // Initialize dp table with -1 
    for(int[][][] i : dp)
        for(int[][] j : i)
            for(int[] k : j)
                Arrays.fill(k, -1);
        
    // Function Call 
    System.out.println(maxXorValue(0, s1, s2, s3, 
                                   ones1, ones2, 
                                   ones3, n)); 
} 

// Driver code
public static void main (String[] args)
{
    String s1 = "11110"; 
    String s2 = "10101"; 
    String s3 = "00111"; 
    
    int n = s1.length(); 
    
    // Function call 
    maxXORUtil(s1, s2, s3, n); 
}
}

// This code is contributed by offbeat
Python3 
# Python3 program for the
# above approach

# Dp table to store the 
# sub-problems
dp = [[[[-1 for x in range(20)]
            for y in range(20)]
            for z in range(20)]
            for p in range(20)]

# Function to find the maximum
# XOR value after rearranging 
# the digits
def maxXorValue(i, s1, s2, s3, 
                ones1, ones2, 
                ones3, n):

    # Base Case
    if (i >= n):
        return 0

    # Return if already 
     #calculated
    if (dp[i][ones1][ones2][ones3] != -1):
        return dp[i][ones1][ones2][ones3]

    option1 = 0
    option2 = 0
    option3 = 0
    option4 = 0
    option5 = 0
    option6 = 0
    option7 = 0
    option8 = 0

    # Assigning 1's to all 
    # string at position 'i'.
    if (ones1 > 0 and ones2 > 0 and 
        ones3 > 0):

        # 2^(n-1-i) is the value
        # added to the total
        option1 = ((1 << ((n - 1) - i)) + 
                    maxXorValue(i + 1, s1,
                                s2, s3, 
                                ones1 - 1,
                                ones2 - 1,
                                ones3 - 1, n))

    # Assigning 1's to strings 
    # 1 & 2
    if (ones1 > 0 and ones2 > 0 and 
       (n - i > ones3)):
        option2 = (0 + maxXorValue(i + 1, s1,
                                   s2, s3, 
                                   ones1 - 1,
                                   ones2 - 1,
                                   ones3, n))

    # Assigning 1's to strings 
    # 2 & 3
    if (ones2 > 0 and ones3 > 0 and 
       (n - i > ones1)):
        option3 = (0 + maxXorValue(i + 1, s1,
                                   s2, s3,
                                   ones1,
                                   ones2 - 1,
                                   ones3 - 1, n))

    # Assigning 1's to strings 
    # 3 & 1
    if (ones3 > 0 and ones1 > 0 and 
       (n - i > ones2)):
        option4 = (0 + maxXorValue(i + 1, s1,
                                   s2, s3,
                                   ones1 - 1,
                                   ones2,
                                   ones3 - 1, n))

    # Assigning 1 to string 1
    if (ones1 > 0 and (n - i > ones2)
            and (n - i > ones3)):
        option5 = ((1 << ((n - 1) - i))
                   + maxXorValue(i + 1, s1,
                                 s2, s3,
                                 ones1 - 1,
                                 ones2,
                                 ones3, n))

    # Assigning 1 to string 2
    if (ones2 > 0 and (n - i > ones3) and 
       (n - i > ones1)):
        option6 = ((1 << ((n - 1) - i)) + 
                    maxXorValue(i + 1, s1,
                                s2, s3, 
                                ones1,
                                ones2 - 1,
                                ones3, n))

    # Assigning 1 to string 3.
    if (ones3 > 0 and (n - i > ones2) and 
        (n - i > ones1)):
        option7 = ((1 << ((n - 1) - i)) + 
                    maxXorValue(i + 1, s1,
                                s2, s3, 
                                ones1, ones2,
                                ones3 - 1, n))

    # Assigning 0 to all the strings
    if ((n - i > ones2) and (n - i > ones3) and 
        (n - i > ones1)):
        option8 = (0 + maxXorValue(i + 1, s1,
                                   s2, s3,
                                   ones1,
                                   ones2,
                                   ones3, n))

    # Take the maximum amongst all of
    # the above solutions
    dp[i][ones1][ones2][ones3] = max(option1,
                                 max(option2,
                                 max(option3,
                                 max(option4,
                                 max(option5,
                                 max(option6,
                                 max(option7,
                                 option8)))))))
    return dp[i][ones1][ones2][ones3]

# Function to get the count
# of ones in the string s
def onesCount(s):
  
    count = 0

    # Traverse the string
    for x in s:
        if (x == '1'):
            count += 1

    # Return the count
    return count

# Utility Function to find 
# the maximum XOR value after
# rearranging the digits
def maxXORUtil(s1, s2,
               s3, n):

    # Find the count of ones in
    # each of the strings
    ones1 = onesCount(s1)
    ones2 = onesCount(s2)
    ones3 = onesCount(s3)

    global dp

    # Function Call
    print(maxXorValue(0, s1, s2, s3,
                      ones1, ones2,
                      ones3, n))

# Driver code
if __name__ == "__main__":

    s1 = "11110"
    s2 = "10101"
    s3 = "00111"
    n = len(s1)

    # Function Call
    maxXORUtil(s1, s2, s3, n)

# This code is contributed by Chitranayal
C# 
// C# program for the 
// above approach 
using System;
class GFG{

// Dp table to store 
// the sub-problems 
static int[,,,] dp = 
       new int[20, 20, 
               20, 20]; 
  
// Function to find the 
// maximum XOR value after 
// rearranging the digits 
static int maxXorValue(int i, String s1, 
                       String s2, String s3, 
                       int ones1, int ones2, 
                       int ones3, int n) 
{     
  // Base Case 
  if (i >= n) 
    return 0; 

  // Return if already calculated 
  if (dp[i, ones1, 
         ones2, ones3] != -1) 
    return dp[i, ones1,
              ones2, ones3]; 

  int option1 = 0, option2 = 0, 
  option3 = 0, option4 = 0, 
  option5 = 0, option6 = 0, 
  option7 = 0, option8 = 0; 

  // Assigning 1's to all 
  // string at position 'i'. 
  if (ones1 > 0 && ones2 > 0 && 
      ones3 > 0) 

    // 2^(n-1-i) is the value 
    // added to the total 
    option1 = (1 << ((n - 1) - i)) +
               maxXorValue(i + 1, s1, 
                           s2, s3, 
                           ones1 - 1, 
                           ones2 - 1, 
                           ones3 - 1, n); 

  // Assigning 1's to 
  // strings 1 & 2 
  if (ones1 > 0 && ones2 > 0 && 
     (n - i > ones3)) 
    option2 = 0 + maxXorValue(i + 1, s1, s2,
                              s3, ones1 - 1, 
                              ones2 - 1, 
                              ones3, n); 

  // Assigning 1's to strings 2 & 3 
  if (ones2 > 0 && ones3 > 0 && 
     (n - i > ones1)) 
    option3 = 0 + maxXorValue(i + 1, s1, s2,
                              s3, ones1, 
                              ones2 - 1, 
                              ones3 - 1, n); 

  // Assigning 1's to strings 3 & 1 
  if (ones3 > 0 && ones1 > 0 && 
     (n - i > ones2)) 
    option4 = 0 + maxXorValue(i + 1, s1, s2,
                              s3, ones1 - 1, 
                              ones2, 
                              ones3 - 1, n); 

  // Assigning 1 to string 1 
  if (ones1 > 0 && (n - i > ones2) &&
     (n - i > ones3)) 
    option5 = (1 << ((n - 1) - i)) + 
               maxXorValue(i + 1, s1, 
                           s2,s3, 
                           ones1 - 1,
                           ones2, ones3, n); 

  // Assigning 1 to string 2 
  if (ones2 > 0 && (n - i > ones3) && 
     (n - i > ones1)) 
    option6 = (1 << ((n - 1) - i)) + 
               maxXorValue(i + 1, s1, 
                           s2, s3, 
                           ones1, 
                           ones2 - 1, 
                           ones3, n); 

  // Assigning 1 to string 3. 
  if (ones3 > 0 && (n - i > ones2) && 
     (n - i > ones1)) 
    option7 = (1 << ((n - 1) - i)) + 
               maxXorValue(i + 1, s1, 
                           s2, s3, 
                           ones1, 
                           ones2, 
                           ones3 - 1, n); 

  // Assigning 0 to all the strings 
  if ((n - i > ones2) &&
      (n - i > ones3) &&
      (n - i > ones1)) 
    option8 = 0 + maxXorValue(i + 1, 
                              s1, s2, 
                              s3, ones1, 
                              ones2, ones3, n); 

  // Take the maximum amongst all of 
  // the above solutions 
  return dp[i, ones1,
            ones2, ones3] = Math.Max(option1, 
                            Math.Max(option2, 
                            Math.Max(option3, 
                            Math.Max(option4, 
                            Math.Max(option5, 
                            Math.Max(option6, 
                            Math.Max(option7, 
                                     option8))))))); 
} 
  
// Function to get the count 
// of ones in the string s 
static int onesCount(String s) 
{ 
  int count = 0; 

  // Traverse the string 
  foreach(char x in s.ToCharArray())
  { 
    if (x == '1') 
      ++count; 
  } 

  // Return the count 
  return count; 
} 
  
// Utility Function to find the maximum 
// XOR value after rearranging the digits 
static void maxXORUtil(String s1, String s2, 
                       String s3, int n) 
{   
  // Find the count of ones in 
  // each of the strings 
  int ones1 = onesCount(s1); 
  int ones2 = onesCount(s2); 
  int ones3 = onesCount(s3); 

  // Initialize dp table with -1 
  for(int i = 0; i < 20; i++)
  {
    for(int j = 0; j < 20; j++)
    {
      for(int l = 0; l < 20; l++)
        for(int k = 0; k < 20; k++)
          dp[i, j, l, k] =- 1;
    }
  }
  
  // Function Call 
  Console.WriteLine(maxXorValue(0, s1, s2, s3, 
                                ones1, ones2, 
                                ones3, n)); 
} 

// Driver code
public static void Main(String[] args)
{
  String s1 = "11110"; 
  String s2 = "10101"; 
  String s3 = "00111"; 

  int n = s1.Length; 

  // Function call 
  maxXORUtil(s1, s2, s3, n); 
}
}

// This code is contributed by 29AjayKumar
JavaScript 
<script>

// Javascript program for the above approach

// Dp table to store the sub-problems
let dp = new Array(20);
for(let i = 0; i < 20; i++)
{
    dp[i] = new Array(20);
    for(let j = 0; j < 20; j++)
    {
        dp[i][j] = new Array(20);
        for(let k = 0; k < 20; k++)
        {
            dp[i][j][k] = new Array(20);
            for(let l = 0; l < 20; l++)
            {
                dp[i][j][k][l] = -1;
            }
        }
    }
}

// Function to find the maximum XOR
// value after rearranging the digits
function maxXorValue(i, s1, s2, s3, ones1, 
                     ones2, ones3, n)
{
    
    // Base Case
    if (i >= n)
        return 0;
   
    // Return if already calculated
    if (dp[i][ones1][ones2][ones3] != -1)
        return dp[i][ones1][ones2][ones3];
   
    let option1 = 0, option2 = 0,
        option3 = 0, option4 = 0,
        option5 = 0, option6 = 0,
        option7 = 0, option8 = 0;
   
    // Assigning 1's to all string at
    // position 'i'.
    if (ones1 > 0 && ones2 > 0 &&
        ones3 > 0)
   
        // 2^(n-1-i) is the value
        // added to the total
        option1 = (1 << ((n - 1) - i)) +
              maxXorValue(i + 1, s1, s2,
                          s3, ones1 - 1,
                              ones2 - 1,
                              ones3 - 1, n);
   
    // Assigning 1's to strings 1 & 2
    if (ones1 > 0 && ones2 > 0 &&
       (n - i > ones3))
        option2 = 0 + maxXorValue(i + 1, s1, s2,
                                  s3, ones1 - 1,
                                      ones2 - 1,
                                      ones3, n);
   
    // Assigning 1's to strings 2 & 3
    if (ones2 > 0 && ones3 > 0 &&
       (n - i > ones1))
        option3 = 0 + maxXorValue(i + 1, s1, s2,
                                  s3, ones1,
                                  ones2 - 1,
                                  ones3 - 1, n);
   
    // Assigning 1's to strings 3 & 1
    if (ones3 > 0 && ones1 > 0 &&
       (n - i > ones2))
        option4 = 0 + maxXorValue(i + 1, s1, s2,
                                  s3, ones1 - 1,
                                  ones2,
                                  ones3 - 1, n);
   
    // Assigning 1 to string 1
    if (ones1 > 0 && (n - i > ones2) &&
       (n - i > ones3))
        option5 = (1 << ((n - 1) - i)) +
                   maxXorValue(i + 1, s1, s2,
                               s3, ones1 - 1,
                               ones2, ones3, n);
   
    // Assigning 1 to string 2
    if (ones2 > 0 && (n - i > ones3) &&
       (n - i > ones1))
        option6 = (1 << ((n - 1) - i)) +
                   maxXorValue(i + 1, s1,
                               s2, s3, ones1,
                               ones2 - 1,
                               ones3, n);
   
    // Assigning 1 to string 3.
    if (ones3 > 0 && (n - i > ones2) &&
       (n - i > ones1))
        option7 = (1 << ((n - 1) - i)) +
                  maxXorValue(i + 1, s1,
                              s2, s3, ones1,
                              ones2,
                              ones3 - 1, n);
   
    // Assigning 0 to all the strings
    if ((n - i > ones2) && (n - i > ones3) &&
        (n - i > ones1))
        option8 = 0 + maxXorValue(i + 1, s1,
                                  s2, s3, ones1,
                                  ones2, ones3, n);
   
    // Take the maximum amongst all of
    // the above solutions
    return dp[i][ones1][ones2][ones3] =
        Math.max(option1,
                 Math.max(option2,
                     Math.max(option3,
                        Math.max(option4,
                             Math.max(option5,
                                 Math.max(option6,
                                     Math.max(option7,
                                              option8)))))));
}

// Function to get the count of ones
// in the string s
function onesCount(s)
{
    let count = 0;
   
    // Traverse the string
    for(let x = 0; x < s.length; x++)
    {
        if (s[x] == '1')
            ++count;
    }
   
    // Return the count
    return count;
}

// Utility Function to find the maximum
// XOR value after rearranging the digits
function maxXORUtil(s1, s2, s3, n)
{
    
    // Find the count of ones in
    // each of the strings
    let ones1 = onesCount(s1);
    let ones2 = onesCount(s2);
    let ones3 = onesCount(s3);
    
    // Function Call
    document.write(maxXorValue(0, s1, s2, s3,
                               ones1, ones2,
                               ones3, n));
}

// Driver code
let s1 = "11110";
let s2 = "10101";
let s3 = "00111";
let n = s1.length;

// Function call
maxXORUtil(s1, s2, s3, n);

// This code is contributed by avanitrachhadiya2155

</script>

Output
30

Time Complexity: O(N4)
Auxiliary Space: O(N)


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