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Make Array elements equal by replacing adjacent elements with their XOR

  • Difficulty Level : Hard
  • Last Updated : 11 Sep, 2021

Given an array A[] consisting of N integers, the task is to check if it is possible to reduce array of at least length 2 such that all the elements in the array are equal. In an operation, choose any index i, and replace A[i] and A[i+1] with their XOR value.

Example: 

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Input: A[] = {0, 2, 2}
Output: YES
Explanation: Apply the given operation for i=0 (zero based indexing). Therefore, replace A[0] and A[1] to A[0]^A[1] i.e., 0^2->2. The resulting array is {2, 2} having all the elements equal.



Input: A[] = {2, 3, 1, 10}
Output: NO
Explanation: There is no possible way for the array to have all equal elements

Naive Approach: The above problem can be solved using the observation that the given array can be reduced to an array with either two equal elements or three equal elements. Below are the steps by step approach to solve the above problem:

  • Create a prefix array in which the ith index stores the XOR of elements from index 0 to i of the array A[].
  • Case 1 where the array can be reduced to two equal elements
    • Suppose the resulting array is {X, X}. Since the XOR of all elements of the array will remain constant, therefore X^X=0=XOR( A[0…N-1]). This case can be easily handled by checking if the XOR of all the elements of the array is 0.
  • Case 2 where the array can be reduced to three equal elements
    • This case can be handled by iterating over all the possible values of (i, j) such that 0<= i < j <=N-1 and check if there exist a value of (i, j) such that XOR(A[0…i]) = XOR(A[i+1…j]) = XOR(A[j+1…N]).

Below is the implementation of the above approach:

C++




// C++ Program of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if it is possible
// to make all the array elements equal
// using the given operation
void possibleEqualArray(int A[], int N)
{
    // Stores the prefix XOR array
    vector<int> pref(N);
    pref[0] = A[0];
 
    for (int i = 1; i < N; i++) {
 
        // Calculate prefix[i]
        pref[i] = pref[i - 1] ^ A[i];
    }
 
    // Case 1, check if the XOR of
    // the input array is 0
    if (pref[N - 1] == 0) {
        cout << "YES";
        return;
    }
 
    // Case 2
    // Iterate over all the ways to
    // divide the array into three
    // non empty subarrays
    int cur_xor = 0;
 
    for (int i = N - 1; i >= 0; i--) {
 
        cur_xor ^= A[i];
 
        for (int j = 0; j < i; j++) {
 
            if (j) {
 
                // XOR of Middle Block
                int middle_xor
 = pref[j - 1] ^ pref[i - 1];
 
                // XOR of Left Block
                int left_xor = pref[j - 1];
 
                // XOR of Right Block
                int right_xor = cur_xor;
 
                if (left_xor == middle_xor
                    && middle_xor == right_xor) {
                    cout << "YES";
                    return;
                }
            }
        }
    }
 
    // Not Possible
    cout << "NO";
}
 
// Driver Code
int main()
{
    int A[] = { 0, 2, 2 };
    int N = sizeof(A) / sizeof(int);
 
    // Function Call
    possibleEqualArray(A, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
 
class GFG
{
 
  // Function to check if it is possible
  // to make all the array elements equal
  // using the given operation
  static void possibleEqualArray(int A[], int N)
  {
 
    // Stores the prefix XOR array
    int[] pref= new int[N];
    pref[0] = A[0];
 
    for (int i = 1; i < N; i++) {
 
      // Calculate prefix[i]
      pref[i] = pref[i - 1] ^ A[i];
    }
 
    // Case 1, check if the XOR of
    // the input array is 0
    if (pref[N - 1] == 0) {
      System.out.println("YES");
      return;
    }
 
    // Case 2
    // Iterate over all the ways to
    // divide the array into three
    // non empty subarrays
    int cur_xor = 0;
 
    for (int i = N - 1; i >= 0; i--) {
 
      cur_xor ^= A[i];
 
      for (int j = 0; j < i; j++) {
 
        if (j!=0) {
 
          // XOR of Middle Block
          int middle_xor
            = pref[j - 1] ^ pref[i - 1];
 
          // XOR of Left Block
          int left_xor = pref[j - 1];
 
          // XOR of Right Block
          int right_xor = cur_xor;
 
          if (left_xor == middle_xor
              && middle_xor == right_xor) {
            System.out.println( "YES");
            return;
          }
        }
      }
    }
 
    // Not Possible
    System.out.println( "NO");
  }
 
  // Driver code
  public static void main (String[] args)
  {
    int A[] = { 0, 2, 2 };
    int N = A.length;
 
    // Function Call
    possibleEqualArray(A, N);
  }
}
 
// This code is contributed by Potta Lokesh

Python3




# Python 3 Program of the above approach
 
# Function to check if it is possible
# to make all the array elements equal
# using the given operation
def possibleEqualArray(A, N):
   
    # Stores the prefix XOR array
    pref = [0 for i in range(N)]
    pref[0] = A[0]
 
    for i in range(1, N, 1):
       
        # Calculate prefix[i]
        pref[i] = pref[i - 1] ^ A[i]
 
    # Case 1, check if the XOR of
    # the input array is 0
    if (pref[N - 1] == 0):
        print("YES")
        return
 
    # Case 2
    # Iterate over all the ways to
    # divide the array into three
    # non empty subarrays
    cur_xor = 0
    i = N - 1
    while(i >= 0):
        cur_xor ^= A[i]
 
        for j in range(i):
            if (j):
                # XOR of Middle Block
                middle_xor = pref[j - 1] ^ pref[i - 1]
 
                # XOR of Left Block
                left_xor = pref[j - 1]
 
                # XOR of Right Block
                right_xor = cur_xor
 
                if (left_xor == middle_xor and middle_xor == right_xor):
                    print("YES")
                    return
 
        i -= 1
 
    # Not Possible
    print("NO")
 
# Driver Code
if __name__ == '__main__':
    A = [0, 2, 2]
    N = len(A)
     
    # Function Call
    possibleEqualArray(A, N)
     
    # This code is contributed by ipg2016107.

C#




// C# program for the above approach
using System;
 
public class GFG
{
 
  // Function to check if it is possible
  // to make all the array elements equal
  // using the given operation
  static void possibleEqualArray(int []A, int N)
  {
 
    // Stores the prefix XOR array
    int[] pref= new int[N];
    pref[0] = A[0];
 
    for (int i = 1; i < N; i++) {
 
      // Calculate prefix[i]
      pref[i] = pref[i - 1] ^ A[i];
    }
 
    // Case 1, check if the XOR of
    // the input array is 0
    if (pref[N - 1] == 0) {
      Console.WriteLine("YES");
      return;
    }
 
    // Case 2
    // Iterate over all the ways to
    // divide the array into three
    // non empty subarrays
    int cur_xor = 0;
 
    for (int i = N - 1; i >= 0; i--) {
 
      cur_xor ^= A[i];
 
      for (int j = 0; j < i; j++) {
 
        if (j!=0) {
 
          // XOR of Middle Block
          int middle_xor
            = pref[j - 1] ^ pref[i - 1];
 
          // XOR of Left Block
          int left_xor = pref[j - 1];
 
          // XOR of Right Block
          int right_xor = cur_xor;
 
          if (left_xor == middle_xor
              && middle_xor == right_xor) {
            Console.WriteLine( "YES");
            return;
          }
        }
      }
    }
 
    // Not Possible
    Console.WriteLine( "NO");
  }
 
  // Driver code
  public static void Main(String[] args)
  {
    int []A = { 0, 2, 2 };
    int N = A.Length;
 
    // Function Call
    possibleEqualArray(A, N);
  }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
// Javascript Program of the above approach
 
// Function to check if it is possible
// to make all the array elements equal
// using the given operation
function possibleEqualArray(A, N) {
  // Stores the prefix XOR array
  let pref = new Array(N);
  pref[0] = A[0];
 
  for (let i = 1; i < N; i++) {
    // Calculate prefix[i]
    pref[i] = pref[i - 1] ^ A[i];
  }
 
  // Case 1, check if the XOR of
  // the input array is 0
  if (pref[N - 1] == 0) {
    document.write("YES");
    return;
  }
 
  // Case 2
  // Iterate over all the ways to
  // divide the array into three
  // non empty subarrays
  let cur_xor = 0;
 
  for (let i = N - 1; i >= 0; i--) {
    cur_xor ^= A[i];
 
    for (let j = 0; j < i; j++) {
      if (j) {
        // XOR of Middle Block
        let middle_xor = pref[j - 1] ^ pref[i - 1];
 
        // XOR of Left Block
        let left_xor = pref[j - 1];
 
        // XOR of Right Block
        let right_xor = cur_xor;
 
        if (left_xor == middle_xor && middle_xor == right_xor) {
          document.write("YES");
          return;
        }
      }
    }
  }
 
  // Not Possible
  document.write("NO");
}
 
// Driver Code
 
let A = [0, 2, 2];
let N = A.length;
 
// Function Call
possibleEqualArray(A, N);
 
// This code is contributed by _saurabh_jaiswal.
</script>
Output: 
YES

 

Time Complexity: O(N2)
Space Complexity: O(N)

Efficient Approach: The above approach can be optimized using the observation that in the case where the array can be reduced to the three equal elements, the resultant array can be represented as {X, X, X}. Since, (X^X^X) = XOR(A[0…N-1]) it implies that X = XOR(A[0…N-1]). Case 1 can be handled the same as that of the naive approach. Case 2 can be solved as follows:

  • Initialize cnt and cur_XOR by 0, and store the XOR of all elements of A[] in tot_XOR.
  • Iterate over array A[] and keep track of the XOR till the current element in cur_XOR.
  • If cur_XOR = tot_XOR, increment the cnt by 1 and initialize cur_XOR = 0.
  • After traversing the whole array, if the value of cnt > 2, it is possible to make all the elements of the array equal using the given operation. 

Below is the implementation of the above approach:

C++




// C++ Program of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if it is possible
// to make all the array elements equal
// using the given operation
void possibleEqualArray(int A[], int N)
{
    // Stores the XOR of all
    // elements of array A[]
    int tot_XOR = 0;
    for (int i = 0; i < N; i++) {
        tot_XOR ^= A[i];
    }
 
    // Case 1, check if the XOR of
    // the array A[] is 0
    if (tot_XOR == 0) {
        cout << "YES";
        return;
    }
 
    // Case 2
 
    // Maintains the XOR till
    // the current element
    int cur_XOR = 0;
    int cnt = 0;
 
    // Iterate over the array
    for (int i = 0; i < N; i++) {
        cur_XOR ^= A[i];
 
        // If the current XOR is equal
        // to the total XOR increment
        // the count and initialize
        // current XOR as 0
        if (cur_XOR == tot_XOR) {
            cnt++;
            cur_XOR = 0;
        }
    }
 
    // Print Answer
    if (cnt > 2) {
        cout << "YES";
    }
    else {
        cout << "NO";
    }
}
 
// Driver Code
int main()
{
    int A[] = { 0, 2, 2 };
    int N = sizeof(A) / sizeof(int);
 
    // Function Call
    possibleEqualArray(A, N);
 
    return 0;
}

Java




// Java Program of the above approach
import java.util.*;
 
class GFG{
 
// Function to check if it is possible
// to make all the array elements equal
// using the given operation
static void possibleEqualArray(int A[], int N)
{
   
    // Stores the XOR of all
    // elements of array A[]
    int tot_XOR = 0;
    for (int i = 0; i < N; i++) {
        tot_XOR ^= A[i];
    }
 
    // Case 1, check if the XOR of
    // the array A[] is 0
    if (tot_XOR == 0) {
        System.out.print("YES");
        return;
    }
 
    // Case 2
 
    // Maintains the XOR till
    // the current element
    int cur_XOR = 0;
    int cnt = 0;
 
    // Iterate over the array
    for (int i = 0; i < N; i++) {
        cur_XOR ^= A[i];
 
        // If the current XOR is equal
        // to the total XOR increment
        // the count and initialize
        // current XOR as 0
        if (cur_XOR == tot_XOR) {
            cnt++;
            cur_XOR = 0;
        }
    }
 
    // Print Answer
    if (cnt > 2) {
        System.out.print("YES");
    }
    else {
        System.out.print("NO");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int A[] = { 0, 2, 2 };
    int N =( A.length);
 
    // Function Call
    possibleEqualArray(A, N);
 
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python 3 Program of the above approach
 
# Function to check if it is possible
# to make all the array elements equal
# using the given operation
def possibleEqualArray(A, N):
   
    # Stores the XOR of all
    # elements of array A[]
    tot_XOR = 0
    for i in range(N):
        tot_XOR ^= A[i]
 
    # Case 1, check if the XOR of
    # the array A[] is 0
    if (tot_XOR == 0):
        print("YES")
        return
    # Case 2
 
    # Maintains the XOR till
    # the current element
    cur_XOR = 0
    cnt = 0
 
    # Iterate over the array
    for i in range(N):
        cur_XOR ^= A[i]
 
        # If the current XOR is equal
        # to the total XOR increment
        # the count and initialize
        # current XOR as 0
        if (cur_XOR == tot_XOR):
            cnt += 1
            cur_XOR = 0
 
    # Print Answer
    if (cnt > 2):
        print("YES")
    else:
        print("NO")
 
# Driver Code
if __name__ == '__main__':
    A = [0, 2, 2]
    N = len(A)
 
    # Function Call
    possibleEqualArray(A, N)
     
    # This code is contributed by SURENDRA_GANGWAR.

C#




// C# Program of the above approach
using System;
 
class GFG{
 
// Function to check if it is possible
// to make all the array elements equal
// using the given operation
static void possibleEqualArray(int []A, int N)
{
   
    // Stores the XOR of all
    // elements of array A[]
    int tot_XOR = 0;
    for (int i = 0; i < N; i++) {
        tot_XOR ^= A[i];
    }
 
    // Case 1, check if the XOR of
    // the array A[] is 0
    if (tot_XOR == 0) {
        Console.Write("YES");
        return;
    }
 
    // Case 2
 
    // Maintains the XOR till
    // the current element
    int cur_XOR = 0;
    int cnt = 0;
 
    // Iterate over the array
    for (int i = 0; i < N; i++) {
        cur_XOR ^= A[i];
 
        // If the current XOR is equal
        // to the total XOR increment
        // the count and initialize
        // current XOR as 0
        if (cur_XOR == tot_XOR) {
            cnt++;
            cur_XOR = 0;
        }
    }
 
    // Print Answer
    if (cnt > 2) {
        Console.Write("YES");
    }
    else {
        Console.Write("NO");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int []A = { 0, 2, 2 };
    int N =( A.Length);
 
    // Function Call
    possibleEqualArray(A, N);
 
}
}
 
// This code is contributed by shivanisinghss2110.

Javascript




<script>
// JavaScript Program of the above approach
// Function to check if it is possible
// to make all the array elements equal
// using the given operation
function possibleEqualArray( A,  N)
{
   
    // Stores the XOR of all
    // elements of array A[]
    var tot_XOR = 0;
    for (var i = 0; i < N; i++) {
        tot_XOR ^= A[i];
    }
 
    // Case 1, check if the XOR of
    // the array A[] is 0
    if (tot_XOR == 0) {
       document.write("YES");
        return;
    }
 
    // Case 2
 
    // Maintains the XOR till
    // the current element
    var cur_XOR = 0;
    var cnt = 0;
 
    // Iterate over the array
    for (var i = 0; i < N; i++) {
        cur_XOR ^= A[i];
 
        // If the current XOR is equal
        // to the total XOR increment
        // the count and initialize
        // current XOR as 0
        if (cur_XOR == tot_XOR) {
            cnt++;
            cur_XOR = 0;
        }
    }
 
    // Print Answer
    if (cnt > 2) {
       document.write("YES");
    }
    else {
       document.write("NO");
    }
}
 
// Driver Code
    var A = [ 0, 2, 2 ];
    var N =( A.length);
 
    // Function Call
    possibleEqualArray(A, N);
 
// This code is contributed by shivanisinghss2110
</script>
Output: 
YES

 

Time Complexity: O(N)
Auxiliary Space: O(1)




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