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# Rearrange given binary strings to maximize their Bitwise XOR value

• Difficulty Level : Expert
• Last Updated : 15 Jun, 2021

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 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



Output

30

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

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