Given an array arr[] of N elements, the task is to check if the array has an element which is equal to the XOR of all the remaining elements.
Examples:
Input: arr[] = { 8, 2, 4, 15, 1 }
Output: Yes
8 is the required element as 2 ^ 4 ^ 15 ^ 1 = 8.Input: arr[] = {4, 2, 3}
Output: No
Approach: First, take the XOR of all the elements of the array and store it in a variable xorArr. Now traverse the complete array again and for every element, calculate the xor of the array elements excluding the current element arr[i] i.e. x = xorArr ^ arr[i].
If x = arr[i] then arr[i] is the required element and hence print Yes. If no such element is found then print No.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function that returns true if the array // contains an element which is equal to // the XOR of the remaining elements bool containsElement(int arr[], int n) { // To store the XOR of all // the array elements int xorArr = 0; for (int i = 0; i < n; ++i) xorArr ^= arr[i]; // For every element of the array for (int i = 0; i < n; ++i) { // Take the XOR after excluding // the current element int x = xorArr ^ arr[i]; // If the XOR of the remaining elements // is equal to the current element if (arr[i] == x) return true; } // If no such element is found return false; } // Driver code int main() { int arr[] = { 8, 2, 4, 15, 1 }; int n = sizeof(arr) / sizeof(arr[0]); if (containsElement(arr, n)) cout << "Yes"; else cout << "No"; return 0; } |
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Java
// Java implementation of the approach class GFG { // Function that returns true if the array // contains an element which is equal to // the XOR of the remaining elements static boolean containsElement(int [] arr, int n) { // To store the XOR of all // the array elements int xorArr = 0; for (int i = 0; i < n; ++i) xorArr ^= arr[i]; // For every element of the array for (int i = 0; i < n; ++i) { // Take the XOR after excluding // the current element int x = xorArr ^ arr[i]; // If the XOR of the remaining elements // is equal to the current element if (arr[i] == x) return true; } // If no such element is found return false; } // Driver code public static void main (String[] args) { int [] arr = { 8, 2, 4, 15, 1 }; int n = arr.length; if (containsElement(arr, n)) System.out.println("Yes"); else System.out.println("No"); } } // This code is contributed by ihritik |
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Python3
# Python3 implementation of the approach # Function that returns true if the array # contains an element which is equal to # the XOR of the remaining elements def containsElement(arr, n): # To store the XOR of all # the array elements xorArr = 0 for i in range(n): xorArr ^= arr[i] # For every element of the array for i in range(n): # Take the XOR after excluding # the current element x = xorArr ^ arr[i] # If the XOR of the remaining elements # is equal to the current element if (arr[i] == x): return True # If no such element is found return False # Driver Code arr = [8, 2, 4, 15, 1] n = len(arr) if (containsElement(arr, n)): print("Yes") else: print("No") # This code is contributed by Mohit Kumar |
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C#
// C# implementation of the approach using System; class GFG { // Function that returns true if the array // contains an element which is equal to // the XOR of the remaining elements static bool containsElement(int [] arr, int n) { // To store the XOR of all // the array elements int xorArr = 0; for (int i = 0; i < n; ++i) xorArr ^= arr[i]; // For every element of the array for (int i = 0; i < n; ++i) { // Take the XOR after excluding // the current element int x = xorArr ^ arr[i]; // If the XOR of the remaining elements // is equal to the current element if (arr[i] == x) return true; } // If no such element is found return false; } // Driver code public static void Main () { int [] arr = { 8, 2, 4, 15, 1 }; int n = arr.Length; if (containsElement(arr, n)) Console.WriteLine("Yes"); else Console.WriteLine("No"); } } // This code is contributed by ihritik |
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Output:
Yes
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