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:
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 operationvoid 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 Codeint 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 approachimport 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 operationdef 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 Codeif __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 approachusing 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 operationfunction 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 Codelet A = [0, 2, 2];let N = A.length;// Function CallpossibleEqualArray(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 operationvoid 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 Codeint 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 approachimport java.util.*;class GFG{// Function to check if it is possible// to make all the array elements equal// using the given operationstatic 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 Codepublic 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 operationdef 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 Codeif __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 approachusing System;class GFG{// Function to check if it is possible// to make all the array elements equal// using the given operationstatic 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 Codepublic 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 operationfunction 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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