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XOR of all substrings of a given Binary String

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  • Last Updated : 31 May, 2022

Given a binary string str of size N, the task is to calculate the bitwise XOR of all substrings of str.

Examples:

Input: str = “11”
Output: 11
Explanation: The substrings of “11” are: 1, 1,  and 11.
Their XOR = 1 ⊕ 1 ⊕ 11 = 11

Input: str = “110”
Output: 111
Explanation: The substrings of 110 are: 1, 1, 0, 11, 10, 110. 
Their XOR = 1 ⊕ 1 ⊕ 0 ⊕ 11 ⊕ 10 ⊕ 110 = 111

Input: str = “10101”
Output: 11001
Explanation: The substrings of 10101 are: 1, 10, 101, 1010, 10101, 0, 01, 010, 0101, 1, 10, 101, 0, 01, and 1.
Their XOR = 1 ⊕ 10 ⊕ 101 ⊕ 1010 ⊕ 10101 ⊕ 0 ⊕ 01 ⊕ 010 ⊕ 0101 ⊕ 1 ⊕ 10 ⊕ 101 ⊕ 0 ⊕ 01 ⊕ 1 = 11001

 

Approach: This problem can be solved based on the following observation:

XOR of odd number of 1s is always 1. Otherwise, the XOR is 0. 
Each jth bit can be the ith bit in a substring when 0 ≤ j ≤ N-i.
So each character has contribution for the last bit (LSB) of the result. 
All characters from i = 0 to N-2 has contribution for 2nd last bit and so on.

Follow the steps mentioned below to utilize the above observation to solve this problem:

  • Create an array (say occurrence[]) to store the total count of 1s having contribution for ith index from LSB end in resultant XOR.
  • Now iterate from the LSB end (iterator i):
    • If the total number of 1s at ith index is odd then the resultant XOR will have ith bit from the LSB end as 1.
    • Otherwise, the value in ith bit will be 0.
  • Return the resultant XOR.

Below is the implementation of the above approach:

C++




// C++ code to implement above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the XOR
string totalXOR(string s, int n)
{
    // Vector for storing the occurrences
    vector<int> occurrence;
    for (int i = 0; i < n; i++) {
        if (s[i] == '1')
            occurrence.push_back(i + 1);
        else
            occurrence.push_back(0);
    }
 
    // Calculating the total occurrences
    // of nth bit
    int sums
        = accumulate(occurrence.begin(),
                     occurrence.end(), 0);
    string ans = "";
    for (int i = 0; i < n; i++) {
 
        // Checking if the total occurrences
        // are odd
        if (sums % 2 == 1)
            ans = "1" + ans;
        else
            ans = "0" + ans;
        sums -= occurrence.back();
        occurrence.pop_back();
    }
    return ans;
}
 
// Driver Code
int main()
{
    int N = 5;
    string str = "10101";
    cout << totalXOR(str, N);
    return 0;
}


Java




// Java program to implement the approach
import java.io.*;
import java.util.ArrayList;
 
public class GFG {
 
  // function to find XOR
  static String totalXOR(String s, int n)
  {
 
    // initializing occurrence list
    ArrayList<Integer> occurrence
      = new ArrayList<Integer>();
    for (int i = 0; i < n; i++) {
      if ((s.charAt(i)) == '1')
        occurrence.add(i + 1);
      else
        occurrence.add(0);
    }
 
    // calculating sum of occurrence
    int sums = 0;
    for (int i : occurrence)
      sums += i;
 
    String ans = "";
    for (int i = 0; i < n; i++) {
 
      // Checking if the total occurrences
      // are odd
      if (sums % 2 == 1)
        ans = "1" + ans;
      else
        ans = "0" + ans;
 
      // subtracting last element of occurrence from
      // sums
      sums -= occurrence.get(occurrence.size() - 1);
 
      // removing last element of occurrence
      occurrence.remove(occurrence.size() - 1);
    }
 
    return ans;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int N = 5;
    String str = "10101";
    System.out.print(totalXOR(str, N));
  }
}
 
// This code is contributed by phasing17


Python3




# Python Program to implement
# above approach
def totalXOR(string, n):
   
    # Storing the occurrences of 1s
    occurrences = [i + 1 if string[i] ==
           '1' else 0 for i in range(n)]
     
    # Calculating the occurrences for nth bit
    sums = sum(occurrences)
    ans = ''
    for i in range(n):
        ans \
            = ['0', '1'][sums % 2 == 1] + ans
        # Subtracting the occurrences of
        # (i + 1)th bit from ith bit
        sums -= occurrences[-1]
        del occurrences[-1]
    return ans
 
# Driver Code
if __name__ == '__main__':
    N = 5
    str = "10101"
    print(totalXOR(str, N))


C#




// C# program to implement the approach
using System;
using System.Collections;
 
public class GFG{
 
  // function to find XOR
  static string totalXOR(string s, int n)
  {
 
    // initializing occurrence list
    ArrayList occurrence
      = new ArrayList();
    for (int i = 0; i < n; i++) {
      if (s[i] == '1')
        occurrence.Add(i + 1);
      else
        occurrence.Add(0);
    }
 
    // calculating sum of occurrence
    int sums = 0;
    foreach (var i in occurrence)
      sums = sums + (int)i;
 
    string ans = "";
    for (int i = 0; i < n; i++) {
 
      // Checking if the total occurrences
      // are odd
      if (sums % 2 == 1)
        ans = "1" + ans;
      else
        ans = "0" + ans;
 
      // subtracting last element of occurrence from
      // sums
      sums = sums - (int)occurrence[occurrence.Count - 1];
 
      // removing last element of occurrence
      occurrence.RemoveAt(occurrence.Count - 1);
    }
 
    return ans;
  }
 
  // Driver Code
  static public void Main (){
 
    int N = 5;
    string str = "10101";
    Console.Write(totalXOR(str, N));
  }
}
 
// This code is contributed by hrithikgarg03188.


Javascript




<script>
    // JavaScript code to implement above approach
 
    // Function to find the XOR
    const totalXOR = (s, n) => {
     
        // Vector for storing the occurrences
        let occurrence = [];
        for (let i = 0; i < n; i++) {
            if (s[i] == '1')
                occurrence.push(i + 1);
            else
                occurrence.push(0);
        }
 
        // Calculating the total occurrences
        // of nth bit
        let sums = 0;
        for (let itm in occurrence) sums += occurrence[itm];
 
        let ans = "";
        for (let i = 0; i < n; i++) {
 
            // Checking if the total occurrences
            // are odd
            if (sums % 2 == 1)
                ans = "1" + ans;
            else
                ans = "0" + ans;
            sums -= occurrence[occurrence.length - 1];
            occurrence.pop();
        }
        return ans;
    }
 
    // Driver Code
    let N = 5;
    let str = "10101";
    document.write(totalXOR(str, N));
 
// This code is contributed by rakeshsahni
 
</script>


Output

11001

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


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