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 = NCO3Hence, 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):
- Assign 1s to all the three strings
- Assign 1s to any two strings
- Assign 1s to any one of the strings.
- 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-problemsint dp[20][20][20][20];// Function to find the maximum XOR// value after rearranging the digitsint 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 sint 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 digitsvoid 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 codeint 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 codepublic 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-problemsdp = [[[[-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 digitsdef 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 sdef 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 digitsdef 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 codeif __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 codepublic 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-problemslet 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 digitsfunction 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 sfunction 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 digitsfunction 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 codelet s1 = "11110";let s2 = "10101";let s3 = "00111";let n = s1.length;// Function callmaxXORUtil(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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